A nonempty subset of a vector space \(V\) over a field \(F\) is a subspace if it is closed under addition and scalar multiplication. Every finite spanning set of \(V\) contains a basis, and for a linear map \(T\colon V\to W\) with \(V\) finite-dimensional, \(\dim\ker T+\dim\im T=\dim V\). The eigenvalues of an operator \(T\) on \(V\) are the roots in \(F\) of its characteristic polynomial \(\det(xI-T)\). The operator \(T\) is diagonalizable if and only if \(V\) has a basis of eigenvectors of \(T\), if and only if the minimal polynomial of \(T\) is a product of distinct monic linear factors in \(F[x]\). The rational canonical form of \(T\) is the block diagonal matrix of the companion matrices of its invariant factors and exists over every field; the Jordan form is determined by the elementary divisors and exists when the characteristic polynomial splits over \(F\). A nilpotent operator on an \(n\)-dimensional space has characteristic polynomial \(x^n\), and its Jordan form is determined by the ranks of its powers.
A spanning set with no proper spanning subset is linearly independent (Tag P-4STQR)
Define what it means for vectors \(v_1, \ldots, v_n\) in a vector space \(V\) to be linearly independent.
Suppose that vectors \(v_1, \ldots, v_n\) in a vector space \(V\) span \(V\) and that no proper subset of \(\{v_1, \ldots, v_n\}\) spans \(V\). Prove that \(v_1, \ldots, v_n\) are linearly independent.
Solution
(a).
<1>1. The vectors \(v_1, \ldots, v_n\) are linearly independent if the only scalars \(c_1, \ldots, c_n\) with \(c_1 v_1 + \cdots + c_n v_n = 0\) are \(c_1 = \cdots = c_n = 0\).
definition of linear independence.
(b).
<1>1. Suppose for contradiction that \(v_1, \ldots, v_n\) are linearly dependent.
assume the conclusion fails.
<1>2. Then there is a nontrivial relation \(c_1 v_1 + \cdots + c_n v_n = 0\) with some \(c_j \neq 0\).
<1>1 (a).
<1>3. Hence \(v_j = -\sum_{i \neq j} \frac{c_i}{c_j} v_i\), so \(v_j\) is a linear combination of the other vectors.
<1>2, solving for \(v_j\).
<1>4. Therefore \(\{v_1, \ldots, v_n\} \setminus \{v_j\}\) still spans \(V\) (any linear combination using \(v_j\) can be rewritten using the other vectors).
<1>3.
<1>5. This contradicts the hypothesis that no proper subset spans \(V\).
<1>4.
<1>6. Hence \(v_1, \ldots, v_n\) are linearly independent.
<1>5.
<1>7. Q.E.D.
<1>1 (a) and <1>6 (b).
The plane $3x+4y+5z=0$ as kernel or image of an endomorphism of $\mathbb{R}^3$ (Tag P-CWELY)
Let \[ V=\{(x,y,z)\in\mathbb R^3:3x+4y+5z=0\}. \]
- Show that \(V\) is a linear subspace of \(\mathbb R^3\).
- Prove or disprove that some linear map \(S:\mathbb R^3\to\mathbb R^3\) has kernel \(V\).
- Prove or disprove that some linear map \(T:\mathbb R^3\to\mathbb R^3\) has image \(V\).
- Prove or disprove that some linear map \(U:\mathbb R^3\to\mathbb R^3\) has both kernel and image equal to \(V\).
Solution
Let \(\varphi\colon\mathbb R^3\to\mathbb R\) be \(\varphi(x,y,z)=3x+4y+5z\), so that \(V=\ker\varphi\).
<1>1. \(V\) is a linear subspace of \(\mathbb R^3\) of dimension \(2\), with basis \(v_1=(4,-3,0)\), \(v_2=(5,0,-3)\).
\(V\) is the kernel of the linear map \(\varphi\), hence a subspace. Since \(\varphi(1,0,0)=3\neq0\), the image of \(\varphi\) is \(\mathbb R\), and rank–nullity gives \(\dim V=\dim\mathbb R^3-\dim\operatorname{im}\varphi=3-1=2\). Both \(v_1\) and \(v_2\) satisfy \(3x+4y+5z=0\), and they are linearly independent: if \(av_1+bv_2=0\), the third coordinate gives \(-3b=0\) and then the second gives \(-3a=0\). Two independent vectors in a \(2\)-dimensional space form a basis.
<1>2. Some linear map \(S\colon\mathbb R^3\to\mathbb R^3\) has kernel \(V\).
Let \[ S=\begin{pmatrix}3&4&5\\0&0&0\\0&0&0\end{pmatrix}, \qquad S(x,y,z)^T=(3x+4y+5z,0,0)^T. \] Then \(S\mathbf x=0\) exactly when \(3x+4y+5z=0\), so \(\ker S=V\).
<1>3. Some linear map \(T\colon\mathbb R^3\to\mathbb R^3\) has image \(V\).
Let \(T\) be the matrix with columns \(v_1\), \(v_2\), \(0\): \[ T=\begin{pmatrix}4&5&0\\-3&0&0\\0&-3&0\end{pmatrix}. \] The image of \(T\) is its column space, \(\operatorname{span}\{v_1,v_2\}=V\) by <1>1.
<1>4. No linear map \(U\colon\mathbb R^3\to\mathbb R^3\) has \(\ker U=\operatorname{im}U=V\).
Rank–nullity gives \(\dim\ker U+\dim\operatorname{im}U=\dim\mathbb R^3=3\). If both equalled \(V\), the left side would be \(2\dim V=4\) by <1>1, a contradiction.
The nullspace of a given $3\times 3$ matrix is $\mathrm{span}\{(1,3,5)\}$ (Tag P-KBQZ5)
Let \[ A = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 5 & -3 \\ 0 & 0 & 0 \end{pmatrix}. \] Show that \(\ker A = \operatorname{span}\{(1,3,5)\}\).
Solution
Row-reducing \(A\) (already nearly echelon) gives the system \(x-2y+z=0\) and \(5y-3z=0\). Thus \(y=\frac{3}{5}z\) and \(x=2y-z=\frac{1}{5}z\). Taking \(z=5\) yields the basis vector \((1,3,5)\). The kernel is one-dimensional, so \(\ker A = \operatorname{span}\{(1,3,5)\}\).
$\{v_1+v_2,\, v_2-v_3,\, v_2+2v_3\}$ is a basis whenever $\{v_1,v_2,v_3\}$ is (Tag P-QJE7B)
Let \(v_1\), \(v_2\), \(v_3\) form a basis for a vector space \(V\). Prove that \(v_1+v_2\), \(v_2-v_3\), \(v_2+2v_3\) form a basis for \(V\).
Solution
Write \(w_1 = v_1+v_2\), \(w_2 = v_2-v_3\), \(w_3 = v_2+2v_3\). In the ordered basis \((v_1,v_2,v_3)\) these have coordinate matrix \[ A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & -1 & 2 \end{pmatrix}. \] A short expansion gives \(\det A = 3 \neq 0\), so \(A\) is invertible. Thus \((w_1,w_2,w_3)\) is obtained from a basis by an invertible linear change of coordinates, hence is itself a basis of \(V\).
A finite spanning set in $\RR^n$ contains a basis of its span (Tag P-TGH5F)
Let \(v_1, \dots, v_k\) be vectors in \(\mathbb{R}^n\). Define the span \(V = \langle v_1, \dots, v_k \rangle\) of the vectors. Prove that there is a subset of the vectors \(v_1, \dots, v_k\) that forms a basis for \(V\).
Solution
The span of \(v_1,\ldots,v_k\) is \[ V=\langle v_1,\ldots,v_k\rangle =\left\{\sum_{i=1}^k a_i v_i:a_i\in\mathbb R\right\}. \]
Among all linearly independent subsets of the finite set \(\{v_1,\ldots,v_k\}\), choose one of maximal cardinality, say \[ \{v_{i_1},\ldots,v_{i_r}\}. \] It is linearly independent by construction. We claim that it spans \(V\). If some \(v_j\) were not in its span, then \[ \{v_{i_1},\ldots,v_{i_r},v_j\} \] would still be linearly independent, contradicting maximality. Thus every \(v_j\) lies in \[ \operatorname{span}(v_{i_1},\ldots,v_{i_r}), \] so this span equals \(V\). Hence the chosen subset is a basis of \(V\).
Basis for $\operatorname{span}\{(1,1,1,1),(3,4,6,7),(5,6,8,9)\}$ (Tag P-KOTV2)
Find a basis for the subspace of \(\mathbb R^4\) spanned by \[ (1,1,1,1),\qquad (3,4,6,7),\qquad (5,6,8,9). \]
Solution
- It will exactly be the row space of \[ A = \left(\begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 3 & 4 & 6 & 7 \\ 5 & 6 & 8 & 9 \end{array}\right), \] where we note that \(R_3 = 2R_1 + R_2\) but \(R_1 \neq \lambda R_2\) and so the first two rows span the correct subspace. We can also compute the RREF, which has the same rowspace, \[\tilde A = \left(\begin{array}{rrrr} 1 & 0 & -2 & -3 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 0 & 0 \end{array}\right)\] from which we find that \(\vector v_1 = \thevector{1,0,-2,-3}\) and \(\vector v_2 = \thevector{0,1,3,4}\) also do the job. \(\qed\)
Solution
Goal: Find a basis for the subspace \(W = \operatorname{span}\{(1,1,1,1), \, (3,4,6,7), \, (5,6,8,9)\} \subseteq \mathbb{R}^4\).
<1>1. Let \(v_1 = (1,1,1,1)\), \(v_2 = (3,4,6,7)\), and \(v_3 = (5,6,8,9)\). The subspace \(W\) equals the row space \(\operatorname{Row}(A)\) of the matrix: \[A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 3 & 4 & 6 & 7 \\ 5 & 6 & 8 & 9 \end{pmatrix}.\]
By definition of the row space of a matrix.
<1>2. The reduced row echelon form (RREF) of \(A\) is: \[R = \begin{pmatrix} 1 & 0 & -2 & -3 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 0 & 0 \end{pmatrix}.\] Proof: <2>1. Apply elementary row operations: \(R_2 \gets R_2 - 3R_1\) and \(R_3 \gets R_3 - 5R_1\): \[\begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 4 \\ 0 & 1 & 3 & 4 \end{pmatrix}.\] <2>2. Apply \(R_3 \gets R_3 - R_2\) and \(R_1 \gets R_1 - R_2\): \[\begin{pmatrix} 1 & 0 & -2 & -3 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 0 & 0 \end{pmatrix}.\] This matrix is in RREF.
<1>3. Elementary row operations preserve the row space, so \(\operatorname{Row}(A) = \operatorname{Row}(R)\).
Standard linear algebra theorem: each elementary row operation is invertible and preserves linear combinations of the rows.
<1>4. The nonzero rows of \(R\), namely \(w_1 = (1, 0, -2, -3)\) and \(w_2 = (0, 1, 3, 4)\), form a basis for \(W\). Proof: <2>1. \(w_1\) and \(w_2\) span \(\operatorname{Row}(R) = W\) since the third row is zero. <2>2. \(w_1\) and \(w_2\) are linearly independent because their leading 1s occur in distinct columns (columns 1 and 2): \(c_1 w_1 + c_2 w_2 = (c_1, c_2, -2c_1+3c_2, -3c_1+4c_2) = (0,0,0,0) \implies c_1 = 0\) and \(c_2 = 0\). <2>3. Alternatively, the first two rows \(\{v_1, v_2\}\) of the original matrix also form a basis for \(W\) since \(\rank(A) = 2\) and \(v_1, v_2\) are linearly independent with \(v_3 = 2v_1 + v_2\). Hence \(\dim(W) = 2\) and \(\{(1,0,-2,-3), (0,1,3,4)\}\) is a basis for \(W\). Q.E.D.
The projection onto the line $x+2y=0$ along $\operatorname{span}\{(1,2)\}$ (Tag P-JLPVR)
Let \(L \subset \mathbb{R}^2\) be the line spanned by the vector \(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\). Let \(P \subset \mathbb{R}^2\) be the line defined by the equation \(x + 2y = 0\). Find the standard matrix for the linear transformation \(T : \mathbb{R}^2 \to \mathbb{R}^2\) such that \(T(v) = 0\) for \(v \in L\) and \(T(v) = v\) for \(v \in P\).
Solution
<1>1. The vectors \[ \ell=\begin{pmatrix}1\\2\end{pmatrix}, \qquad p=\begin{pmatrix}2\\-1\end{pmatrix} \] form a basis of \(\mathbb R^2\), with \(\ell\in L\) and \(p\in P\).
The vector \(p\) satisfies \(2+2(-1)=0\), so \(p\in P\). Also \[ \det\begin{pmatrix}1&2\\2&-1\end{pmatrix}=-5\ne0, \] so \(\ell,p\) are linearly independent.
<1>2. Every vector \((x,y)^t\) has a unique decomposition \[ \begin{pmatrix}x\\y\end{pmatrix} =a\ell+bp, \qquad b=\frac{2x-y}{5}. \]
The equations are \[ x=a+2b,\qquad y=2a-b. \] Substituting \(a=x-2b\) into the second equation gives \[ y=2x-5b, \] so \(b=(2x-y)/5\).
<1>3. Since \(T\) vanishes on \(L\) and is the identity on \(P\), \[ T\begin{pmatrix}x\\y\end{pmatrix} =b p =\frac{2x-y}{5}\begin{pmatrix}2\\-1\end{pmatrix}. \]
<1>4. Therefore the standard matrix of \(T\) is \[ \boxed{ [T]=\frac15 \begin{pmatrix} 4&-2\\ -2&1 \end{pmatrix}.} \]
Expanding <1>3 gives \[ T(x,y)=\left(\frac{4x-2y}{5},\frac{-2x+y}{5}\right), \] whose coefficient matrix is the displayed matrix.
Diagonalizability of $\begin{pmatrix}0&1\\6&-1\end{pmatrix}$ and a closed form for $A^k\begin{pmatrix}1\\7\end{pmatrix}$ (Tag P-5HVC4)
Let \(A = \begin{bmatrix} 0 & 1 \\ 6 & -1 \end{bmatrix}\). Prove that \(A\) is diagonalizable and find a closed-form expression for \(A^k \begin{bmatrix} 1 \\ 7 \end{bmatrix}\), \(k \in \mathbb{N}\).
Solution
The characteristic polynomial is \[ \det(\lambda I-A)=\lambda^2+\lambda-6=(\lambda-2)(\lambda+3). \] Thus \(A\) has two distinct eigenvalues and is diagonalizable. Corresponding eigenvectors are \[ v_2=\binom12,\qquad v_{-3}=\binom1{-3}. \] Since \[ \binom17=2v_2-v_{-3}, \] we obtain for every \(k\in\mathbb N\) \[ A^k\binom17 =2\,2^k v_2-(-3)^k v_{-3}. \] Equivalently, \[ \boxed{A^k\binom17= \binom{2^{k+1}-(-3)^k}{2^{k+2}+3(-3)^k}}. \]
Rational and Jordan canonical forms of a $3\times 3$ real matrix; abelian groups of order $360$ (Tag P-6ET4N)
Find a Rational Canonical Form and a Jordan Canonical Form for the following \(3 \times 3\) matrix with real entries: \[A = \begin{pmatrix} 0 & -1 & 2 \\ 3 & -4 & 6 \\ 2 & -2 & 3 \end{pmatrix}.\]
Describe all distinct isomorphism classes of Abelian groups of order 360.
Solution
<1>1. The characteristic polynomial of \(A\) is \(p_A(\lambda)=(\lambda-1)(\lambda+1)^2\).
Expanding \(\det(\lambda I-A)\) along the first row of \[ \lambda I-A=\begin{pmatrix}\lambda&1&-2\\-3&\lambda+4&-6\\-2&2&\lambda-3\end{pmatrix} \] gives \[ \lambda(\lambda^2+\lambda)-(-3\lambda-3)-2(2\lambda+2)=\lambda^3+\lambda^2-\lambda-1=(\lambda-1)(\lambda+1)^2. \]
<1>2. The matrix \(A\) is diagonalizable, with minimal polynomial \(m_A(\lambda)=\lambda^2-1\).
The matrix \[ -I-A=\begin{pmatrix}-1&1&-2\\-3&3&-6\\-2&2&-4\end{pmatrix} \] has rank \(1\), since every row is a multiple of \((1,-1,2)\). So the eigenspace for \(-1\) has dimension \(2\), equal to the algebraic multiplicity from step <1>1, and the eigenvalue \(1\) has multiplicity \(1\). Hence \(A\) is diagonalizable, and its minimal polynomial is the product of the distinct factors \((\lambda-1)(\lambda+1)\).
<1>3. In part (a), the Jordan canonical form is \(J=\operatorname{diag}(1,-1,-1)\), and the rational canonical form is \[ R=\begin{pmatrix}-1&0&0\\0&0&1\\0&1&0\end{pmatrix}, \] with invariant factors \(\lambda+1\mid\lambda^2-1\).
By step <1>2, \(A\) is similar to the diagonal matrix of its eigenvalues. The invariant factors \(d_1\mid d_2\) satisfy \(d_2=m_A=\lambda^2-1\) and \(d_1d_2=p_A\), so \(d_1=\lambda+1\). Their companion matrices are \((-1)\) and \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\), and \(R\) is their block sum [@DF04].
<1>4. In part (b), there are six abelian groups of order \(360\): \[ \ZZ_{360},\quad \ZZ_3\times\ZZ_{120},\quad \ZZ_2\times\ZZ_{180},\quad \ZZ_6\times\ZZ_{60},\quad \ZZ_2\times\ZZ_2\times\ZZ_{90},\quad \ZZ_2\times\ZZ_6\times\ZZ_{30}. \]
Since \(360=2^3\cdot3^2\cdot5\), an abelian group of order \(360\) is \(P_2\times P_3\times P_5\) with \(\abs{P_2}=8\), \(\abs{P_3}=9\), \(\abs{P_5}=5\), and its isomorphism type is determined by a partition of each exponent [@DF04]. The partitions \(3\), \(2+1\), \(1+1+1\) give \(P_2=\ZZ_8\), \(\ZZ_4\times\ZZ_2\), \(\ZZ_2^3\); the partitions \(2\), \(1+1\) give \(P_3=\ZZ_9\), \(\ZZ_3^2\); and \(P_5=\ZZ_5\). The \(3\cdot2\cdot1=6\) combinations, rewritten in invariant-factor form by the Chinese remainder theorem, are the listed groups.
<1>5. Q.E.D.
Step <1>3 answers part (a) and step <1>4 answers part (b).
Similarity classes of order-$8$ matrices in $M_n(\mathbb{Q})$ for $n=3,4,5$ (Tag P-LU2L3)
For \(n = 3, 4\), and 5, determine the number of similarity classes of matrices \(A \in M_n(\mathbb{Q})\) such that \(A^8 = I\) but \(A^4 \neq I\).
Solution
The numbers of similarity classes for \(n=3,4,5\) are respectively \(0,1,2\).
<1>1. The condition \(A^8=I\) gives a unique multiplicity description \[ \mathbb Q^n\cong E_1^{\oplus a}\oplus E_2^{\oplus b} \oplus E_4^{\oplus c}\oplus E_8^{\oplus d}, \qquad n=a+b+2c+4d, \] as a \(\mathbb Q[x]\)-module with \(x\) acting by \(A\), where \[ E_1=\mathbb Q[x]/(x-1),\quad E_2=\mathbb Q[x]/(x+1), \quad E_4=\mathbb Q[x]/(x^2+1),\quad E_8=\mathbb Q[x]/(x^4+1). \]
Factor \[ x^8-1=(x-1)(x+1)(x^2+1)(x^4+1). \] The two linear factors are irreducible. The quadratic has no rational root. Substituting \(x=t+1\) in the quartic gives \(t^4+4t^3+6t^2+4t+2\), which is Eisenstein at \(2\), so the quartic is irreducible as well [@DF04]. These are distinct monic irreducibles, so the displayed factorization is squarefree.
The module is finitely generated, since a finite rational basis also generates over \(\mathbb Q[x]\), and is annihilated by \(x^8-1\). The structure theorem over the PID \(\mathbb Q[x]\) expresses it uniquely as a direct sum of modules \(\mathbb Q[x]/(p^e)\) for monic irreducibles \(p\) [@DF04]. On such a summand, annihilation by \(x^8-1\) implies \(p^e\mid x^8-1\) by applying that polynomial to the class of \(1\). Hence \(e=1\) and \(p\) is one of the four displayed factors. Their degrees give the dimension equation. Conversely, every direct sum of this form is annihilated by \(x^8-1\).
An invertible rational linear map intertwines the two matrix actions exactly when it is a \(\mathbb Q[x]\)-module isomorphism. Thus uniqueness of the multiplicities is exactly uniqueness of the similarity class.
<1>2. The condition \(A^4\ne I\) is equivalent to \(d\geq1\).
On \(E_1,E_2,E_4\), the polynomial \(x^4-1\) acts by zero, since their defining polynomials divide it. On \(E_8\) one has \(x^4=-1\), so \(x^4-1\) acts as multiplication by \(-2\). This is nonzero on every nonzero copy of \(E_8\) over \(\mathbb Q\). Therefore \(A^4-I\) is nonzero exactly when at least one such summand occurs.
<1>3. The dimensions \(3,4,5\) yield exactly the stated counts.
For \(n=3\), the inequalities \(d\geq1\) and \(4d\leq n\) are incompatible, so there are no such matrices. For \(n=4\), the only possibility is \((a,b,c,d)=(0,0,0,1)\). For \(n=5\), one must have \(d=1\) and \(a+b+2c=1\). Thus \(c=0\) and \((a,b)=(1,0)\) or \((0,1)\). These cases exhaust the nonnegative integer solutions.
For explicit representatives, put \[ C=\begin{pmatrix} 0&0&0&-1\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0 \end{pmatrix}. \] This is multiplication by \(x\) on \(E_8\) in the basis \(1,x,x^2,x^3\), so \(C^4=-I_4\) and \(C^8=I_4\). It represents the single class in dimension four. In dimension five, take \(\operatorname{diag}(C,1)\) and \(\operatorname{diag}(C,-1)\). Both have eighth power the identity and fourth power different from the identity. They are not similar, since their characteristic polynomials are respectively \((x^4+1)(x-1)\) and \((x^4+1)(x+1)\). Step <1>1 proves that there are no additional classes.
$\mathbb{Q}[x]$-modules from order-$6$ matrices in $M_4(\mathbb{Q})$, and groups of order $24$ (Tag P-OUQ2Q)
Suppose \(A \in M_4(\mathbb{Q})\) is a 4-by-4 matrix of multiplicative order 6. We use \(A\) to give \(\mathbb{Q}^4\) the structure of a \(\mathbb{Q}[x]\)-module in the usual way, by setting \(x \cdot v = Av\) for \(v \in \mathbb{Q}^4\).
What are all of the isomorphism classes of \(\mathbb{Q}[x]\)-modules that can arise this way?
For each of your answers above, write down the rational canonical form of the corresponding matrix \(A\).
What are all the of the abelian groups of size 24 (up to isomorphism)?
Solution
Over \(\QQ\), \[ x^6-1=\Phi_1\Phi_2\Phi_3\Phi_6=(x-1)(x+1)(x^2+x+1)(x^2-x+1), \] a product of distinct monic irreducibles. For a monic polynomial \(p\), write \(C(p)\) for its companion matrix, with ones on the subdiagonal and last column the negated lower coefficients of \(p\).
<1>1. The \(\QQ[x]\)-module \(V=\QQ^4\) is isomorphic to \[ V_{a}=\bigoplus_{d\in\{1,2,3,6\}}\bigl(\QQ[x]/(\Phi_d)\bigr)^{a_d}, \qquad a_1+a_2+2a_3+2a_6=4, \] for unique multiplicities \(a_d\ge0\), and \(A\) has order exactly \(6\) if and only if \(a_6\ge1\), or \(a_2\ge1\) and \(a_3\ge1\).
Since \(A^6=I\), the module is annihilated by the squarefree polynomial \(x^6-1\), so by the structure theorem over the PID \(\QQ[x]\) it is a direct sum of modules \(\QQ[x]/(\Phi_d)\) with \(d\mid6\), with unique multiplicities [@DF04]; dimensions give \(a_1+a_2+2a_3+2a_6=4\). On \(\QQ[x]/(\Phi_d)\), \(x\) acts with order exactly \(d\), so the order of \(A\) is the least common multiple of the \(d\) with \(a_d\ge1\). This least common multiple is \(6\) exactly when \(6\) occurs, or both \(2\) and \(3\) occur.
<1>2. For part (a), there are exactly seven classes, listed by \((a_1,a_2,a_3,a_6)\): \[ (0,0,0,2),\ (2,0,0,1),\ (0,2,0,1),\ (1,1,0,1),\ (0,0,1,1),\ (1,1,1,0),\ (0,2,1,0). \]
If \(a_6=2\), the others vanish. If \(a_6=1\), then \(a_1+a_2+2a_3=2\), with the four solutions \((2,0,0)\), \((1,1,0)\), \((0,2,0)\), \((0,0,1)\). If \(a_6=0\), then \(a_2,a_3\ge1\) and \(a_1+a_2+2a_3=4\) force \(a_3=1\) and \((a_1,a_2)\in\{(1,1),(0,2)\}\). By step <1>1 these seven modules are pairwise nonisomorphic and are exactly the modules that arise.
<1>3. For part (b), the rational canonical forms are given by the invariant factors in the table.
| \((a_1,a_2,a_3,a_6)\) | Invariant factors | Rational canonical form |
|---|---|---|
| \((0,0,0,2)\) | \(x^2-x+1,\ x^2-x+1\) | \(\operatorname{diag}\bigl(C(x^2-x+1),C(x^2-x+1)\bigr)\) |
| \((2,0,0,1)\) | \(x-1,\ x^3-2x^2+2x-1\) | \(\operatorname{diag}\bigl(1,C(x^3-2x^2+2x-1)\bigr)\) |
| \((0,2,0,1)\) | \(x+1,\ x^3+1\) | \(\operatorname{diag}\bigl(-1,C(x^3+1)\bigr)\) |
| \((1,1,0,1)\) | \(x^4-x^3+x-1\) | \(C(x^4-x^3+x-1)\) |
| \((0,0,1,1)\) | \(x^4+x^2+1\) | \(C(x^4+x^2+1)\) |
| \((1,1,1,0)\) | \(x^4+x^3-x-1\) | \(C(x^4+x^3-x-1)\) |
| \((0,2,1,0)\) | \(x+1,\ x^3+2x^2+2x+1\) | \(\operatorname{diag}\bigl(-1,C(x^3+2x^2+2x+1)\bigr)\) |
Explicitly, \[ C(x^2-x+1)=\begin{pmatrix}0&-1\\1&1\end{pmatrix},\quad C(x^3-2x^2+2x-1)=\begin{pmatrix}0&0&1\\1&0&-2\\0&1&2\end{pmatrix},\quad C(x^3+1)=\begin{pmatrix}0&0&-1\\1&0&0\\0&1&0\end{pmatrix}, \] \[ C(x^3+2x^2+2x+1)=\begin{pmatrix}0&0&-1\\1&0&-2\\0&1&-2\end{pmatrix},\quad C(x^4-x^3+x-1)=\begin{pmatrix}0&0&0&1\\1&0&0&-1\\0&1&0&0\\0&0&1&1\end{pmatrix}, \] \[ C(x^4+x^2+1)=\begin{pmatrix}0&0&0&-1\\1&0&0&0\\0&1&0&-1\\0&0&1&0\end{pmatrix},\quad C(x^4+x^3-x-1)=\begin{pmatrix}0&0&0&1\\1&0&0&1\\0&1&0&0\\0&0&1&-1\end{pmatrix}. \]
The invariant factors are obtained by multiplying, from the largest down, one factor \(\Phi_d\) for each \(d\) with \(a_d\) remaining; for instance \((0,2,1,0)\) gives \(\Phi_2\) and \(\Phi_2\Phi_3=(x+1)(x^2+x+1)=x^3+2x^2+2x+1\). The rational canonical form is the block diagonal matrix of the companion matrices of the invariant factors [@DF04].
<1>4. For part (c), the abelian groups of order \(24\) are \(\ZZ/24\), \(\ZZ/12\times\ZZ/2\), and \(\ZZ/6\times(\ZZ/2)^2\).
Since \(24=2^3\cdot3\), an abelian group of order \(24\) is the product of its Sylow \(2\)-subgroup of order \(8\) and its Sylow \(3\)-subgroup \(\ZZ/3\) [@DF04]. The partitions \(3\), \(2+1\), \(1+1+1\) give the Sylow \(2\)-subgroups \(\ZZ/8\), \(\ZZ/4\times\ZZ/2\), \((\ZZ/2)^3\). By the Chinese remainder theorem the products with \(\ZZ/3\) are the three listed groups.
<1>5. Q.E.D.
Steps <1>2, <1>3, and <1>4 answer parts (a), (b), and (c).
A $3\times 3$ matrix with eigenvalues $-1,0,1$ satisfies $A^3=A$ (Tag P-R7TCU)
Suppose \(A\) is a \(3\times3\) matrix with real entries and eigenvalues \(-1\), \(0\), and \(1\). Prove that \(A^3=A\).
Solution
<1>1. The characteristic polynomial of \(A\) is \(p_A(x)=x^3-x\).
\(p_A(x)=\det(xI-A)\) is monic of degree \(3\), and its roots in \(\mathbb C\) are the eigenvalues of \(A\). The three eigenvalues \(-1,0,1\) are distinct, so they are all the roots, each simple, and \(p_A(x)=(x+1)x(x-1)=x^3-x\).
<1>2. \(A^3=A\).
By the Cayley–Hamilton theorem \(p_A(A)=0\), that is, \(A^3-A=0\) by <1>1.
When does a matrix with eigenvalues $\pm 1$ satisfy $A^2=I$? (Tag P-SK4KT)
For a positive integer \(n\), let \(I_n\) denote the \(n\times n\) identity matrix.
Let \(A\) be a \(2\times2\) real matrix with eigenvalues \(1\) and \(-1\). Show that \(A^2=I_2\).
Find a \(3\times3\) real matrix whose only complex eigenvalues are \(1\) and \(-1\), but for which \(A^2\neq I_3\).
Let \(A\) be an \(n\times n\) real symmetric matrix whose only complex eigenvalues are \(1\) and \(-1\). Show that \(A^2=I_n\).
Solution
Since \(A\) is \(2\times 2\) and has 2 eigenvalues, noting that \(\deg \chi_A(x) = 2\), we have \(\chi_A(x) = (x-1)(x+1) = x^2 -1\). The minimal polynomial of \(A\) divides \(\chi_A(x)\), so we have \(\chi_A(A) = 0\) and thus \(A^2 - I_2 = 0 \implies A^2 = I_2\).
This will happen when \(x^2-1 = (x+1)(x-1)\) is not the minimal polynomial, and we can force the minimal polynomial to be degree 3 by inserting a nontrivial Jordan block to a diagonal matrix containing just the eigenvalues \(\pm 1\). An example that works: \[ \left(\begin{array}{rr|r} 1 & 1 & 0 \\ 0 & 1 & 0 \\ \hline 0 & 0 & -1 \end{array}\right),~ A^2 = \left(\begin{array}{rrr} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right), \quad \spec(A) = [-1, 1, 1] \]
Every symmetric matrix with \(A\) real spectrum admits a real eigendecomposition \(\Lambda D \Lambda^T\), where \(D\) is diagonal with entries the eigenvalues of \(A\) and \(\Lambda\) are orthogonal (which are also invertible). Here, we only need the fact that \(A\) is diagonalizable by invertible matrices. In our case we have \([D^2]_{ii} = (\pm 1)^2 = 1\) so \(D^2 = I_n\). Thus we have \[A^2 = (\Lambda D \inverseof{\Lambda})^2 = \Lambda D^2 \inverseof{\Lambda} = \Lambda I_n \inverseof{\Lambda} = I_n. \qed\]
Jordan canonical form of a nilpotent matrix (Tag P-TB7BG)
A square matrix \(N\) over the complex numbers is nilpotent just in case \(N^a=0\) for some positive integer \(a\). Show that every nilpotent \(N\) is similar to a matrix of the form \[\begin{bmatrix}N_1&&&\\&N_2&&\\&&\ddots&\\&&&N_s\end{bmatrix}\] where each \(N_i\) is a square matrix of the form \[\begin{bmatrix}0&&&&\\1&0&&&\\&1&\ddots&&\\&&\ddots&0&\\&&&1&0\end{bmatrix}.\]
Solution
Let \(T\) be the operator represented by \(N\) on \(V=\mathbb C^n\). We prove the assertion by induction on \(n\). For \(n=0\) the empty direct sum is the required form. Suppose \(n>0\) and the assertion holds in smaller dimensions. Let \(r\geq1\) be the least integer with \(T^r=0\).
<1>1. There is an invariant subspace \(U\) with basis \(v,Tv,\ldots,T^{r-1}v\).
Minimality of \(r\) gives \(v\in V\) with \(T^{r-1}v\ne0\). If \(\sum_{j=0}^{r-1}a_jT^jv=0\) is a nonzero linear relation, let \(k\) be the least index with \(a_k\ne0\). Applying \(T^{r-1-k}\) gives \(a_kT^{r-1}v=0\): every term with larger index is killed by \(T^r=0\). This is a contradiction. The chain is therefore independent, and its span \(U\) is invariant because \(T(T^{r-1}v)=0\).
<1>2. The subspace \(U\) has a \(T\)-invariant complement.
Extend the chain in step <1>1 to a basis of \(V\). Define a linear functional \(\ell:V\to\mathbb C\) to be \(1\) on \(T^{r-1}v\) and \(0\) on every other vector of this basis. Define \[ \pi(w)=\sum_{j=0}^{r-1}\ell\bigl(T^{r-1-j}w\bigr)T^jv. \] For \(0\leq k<r\), the coefficient of \(T^jv\) in \(\pi(T^kv)\) is \(\ell(T^{r-1-j+k}v)\). It is \(1\) when \(j=k\), \(0\) when \(k<j\) by the definition of \(\ell\), and \(0\) when \(k>j\) because then the exponent is at least \(r\). Thus \(\pi|_U\) is the identity.
Moreover, \(T^r=0\) gives \[ \pi(Tw)=\sum_{j=1}^{r-1}\ell(T^{r-j}w)T^jv =T\pi(w). \] These sums are empty and both sides are zero when \(r=1\). Consequently \(W=\ker\pi\) is \(T\)-invariant. Since \(\pi\) maps into \(U\) and restricts to its identity, every \(w\in V\) has the unique decomposition \[ w=\pi(w)+(w-\pi(w))\in U\oplus W. \] Indeed, the second term lies in \(\ker\pi\) and \(U\cap\ker\pi=0\).
<1>3. Combining bases of \(U\) and \(W\) produces the asserted blocks.
On the ordered basis \(v,Tv,\ldots,T^{r-1}v\) of \(U\), the matrix of \(T|_U\) has ones immediately below the diagonal and all other entries zero. The restriction \(T|_W\) is nilpotent and \(\dim W=n-r<n\). By induction it has a basis with the required block diagonal matrix. Concatenating these two bases gives a basis of \(V\) in which \(T\) has the stated form. If \(S\) has these basis vectors as columns in the original coordinates, then \(S\) is invertible and \(S^{-1}NS\) is exactly that block diagonal matrix.
Spectral theorem diagonalization with eigenvalues $6$ and $1$, and the minimum of $\langle Ax,x\rangle$ on the unit circle (Tag P-VJBIB)
Take \[ A := \begin{pmatrix} 2 & -2 \\ -2 & 5 \end{pmatrix}. \] Find an orthogonal matrix \(P\) for which \(P^{-1}AP\) is diagonal. Then find the minimum value of the dot products \(Ax \cdot x\) as \(x\) ranges through the unit vectors in \(\mathbb{R}^2\).
Solution
If \(A\) has two distinct eigenvalues, we will have \(A = PD\inverseof{P}\) where \(P\) is the matrix of eigenvectors and \(D\) has eigenvalues on the diagonal. We can compute the characteristic polynomial \[ p_\chi(x) = x^2 - (\Tr A)x + \det A = x^2 - 7x + 6 = (x-6)(x-1), \]
and so \(\spec(A) = \theset{6,1}\). Computing the kernel of \(A-\lambda I\) for each of these yields \[ \vector v_1 = \left[ \begin{array} { c } { 1 } \\ { - 2 } \end{array} \right], \vector v_2 = \left[ \begin{array} { c } { 2 } \\ { 1 } \end{array} \right], \]
And so we can write and check \[ P = \left[ \begin{array} { c c } { 1 } & { 2 } \\ { - 2 } & { 1 } \end{array} \right] \\ D = \left(\begin{array}{rr} 6 & 0 \\ 0 & 1 \end{array}\right) \\ \]
We can compute \(PP^T = \mathrm{diag}(5,5)\), so \(P\) can be made orthogonal by replacing \(P\) with \((1/\sqrt 5) P\). With this replacement, a quick computation shows that \(PDP^T = A\).
We will use the fact that \(A = PD\inverseof{P}\) where since \(A\) is symmetric and \(P\) is orthogonal. We can write \[ \inner{A\vector x}{\vector x} = \vector x^T A^T \vector x = \vector x^T (PDP^T)^T \vector x = (P^T \vector x)^T D (P^T\vector x) = \vector y^T D \vector y = \inner{\vector y}{ D\vector y}, \] where \(\vector x\in S^2 \implies P^T \vector x \definedas \vector y \in S^2\) since \(P^T\) is both orthogonal and full-rank (and thus a bijection \(S^2 \selfmap\)).
We can now expand \[ \inner{\vector y}{D \vector y} = \sum_{i=1}^2 y_i \lambda_i y_i = \sum_{i=1}^2 \lambda_i y_i^2 \]
We now note that we can take \(\vector y = \thevector{0, 1}\), in which case \(D\vector y = \thevector{0, \lambda_2} = \thevector{0, 1}\) and thus \(\inner{\vector y}{D \vector y} = 1\) is a candidate minimum.
We can write this as the constrained optimization problem \[ \text{Minimize } f(y_1, y_2) = 6y_1^2 + 1 y_2^2\\ \text{subject to } g(y_1, y_2) = y_1^2 + y_2^2 = 1 \]
where we note that this constraint is equivalent to the original \(\norm{\vector y} = \sqrt{y_1^2 + y_2^2} = 1\).
This can be approached with Lagrange multipliers, i.e. looking at where \(\nabla f = \lambda \nabla g\). This yields \[ \thevector{12y_1, 2y_2} = \lambda \thevector{2y_1, 2y_2} \implies \\ 6y_1 = \lambda y_1, ~ y_2 = \lambda y_2. \]
The second condition forces \(y_2 \in \theset{0,1}\), and solving for \(\lambda\) in this expression yields \(\lambda = 1\) and so the first condition forces \(y_1 = 0\). This leaves only one possibility, \(\vector y = \thevector{0, 1}\), which is indeed the candidate from above. Thus the minimum value is 1. \(\qed\)
Spectral theorem diagonalization with eigenvalues $1$ and $-2$ (Tag P-VPCLD)
Find an invertible matrix \(A\) and a diagonal matrix \(B\) such that \[ \begin{pmatrix}4&-6\\3&-5\end{pmatrix}=ABA^{-1}. \]
Solution
The eigenvalues are \(1\) and \(-2\). Corresponding eigenvectors are \((2,1)^t\) and \((1,1)^t\). Thus one can take \[ A=\begin{pmatrix}2&1\\1&1\end{pmatrix}, \qquad B=\begin{pmatrix}1&0\\0&-2\end{pmatrix}. \] The columns of \(A\) are linearly independent, and the eigenvector equations give \(MA=AB\) for the given matrix \(M\). Hence \(M=ABA^{-1}\).
Groups of order $120$, $\mathbb{Q}[x]$-modules of dimension $6$ annihilated by $(x-3)^2(x+5)$, and rational canonical forms with that minimal polynomial (Tag P-CSEAZ)
List all isomorphism classes of groups of size 120.
Find all isomorphism classes of \(\mathbf{Q}[x]\) modules \(V\) such that \(\dim_{\mathbf{Q}} V=6\) and \(V\) is annihilated by \((x-3)^2(x+5)\).
Find all possible rational canonical forms for a 6 by 6 matrix over \(\mathbf{Q}\) that has minimal polynomial \((x-3)^2(x+5)\).
Solution
The \(47\) groups of order \(120\) in part (a) are the \(39\) semidirect products specified in the table in step <1>5, the five groups in step <1>6, and the three groups in step <1>7.
<1>1. An extension with finite abelian kernel \(N\) and quotient \(Q\) splits when \(\gcd(|N|,|Q|)=1\).
Write \(N\) additively and choose representatives \(s(q)\) with \(s(1)=1\). Conjugation defines an action of \(Q\) on \(N\), independent of the representatives because \(N\) is abelian. Put \(f(p,q)=s(p)s(q)s(pq)^{-1}\in N\). Associativity gives \[ f(p,q)+f(pq,r)=p\cdot f(q,r)+f(p,qr). \] Let \(S(p)=\sum_{r\in Q}f(p,r)\) and \(m=|Q|\). Summing the identity over \(r\) gives \[ m f(p,q)=S(p)+p\cdot S(q)-S(pq). \] Multiplication by \(m\) is an automorphism of \(N\). Thus, with \(h(p)=m^{-1}S(p)\), replacing \(s(p)\) by \((-h(p))s(p)\) makes its factor set zero. The new representatives form a subgroup isomorphic to \(Q\), so the extension is \(N\rtimes Q\).
<1>2. A central extension of \(A_4\) by \(C_2\) is either \(C_2\times A_4\) or \(\mathrm{SL}_2(\mathbf F_3)\). A central extension of \(A_5\) by \(C_2\) is either \(C_2\times A_5\) or \(\mathrm{SL}_2(\mathbf F_5)\).
<2>1. We use the presentations \[ T_m=\langle a,b\mid a^2=b^3=(ab)^m=1\rangle, \qquad T_3\cong A_4,\quad T_5\cong A_5. \] Coset enumeration gives the orders of the two presented groups.
F := FreeGroup("a", "b");;
a := F.1;; b := F.2;;
orders := List([3,5], m -> Size(F / [a^2,b^3,(a*b)^m]));;
if orders <> [12,60] then Error("unexpected triangle-group order"); fi;
Print(orders, "\n");
# [ 12, 60 ]For \(m=3\), the permutations \((12)(34)\) and \((123)\) generate \(A_4\): the conjugates of the first generate its Klein four subgroup, and the second supplies the quotient of order \(3\). For \(m=5\), use \((12)(34)\) and \((135)\). Their product has order \(5\), so their subgroup has order divisible by \(2\), \(3\), and \(5\). Its order is therefore \(30\) or \(60\); order \(30\) would give a normal subgroup of index \(2\) in the simple group \(A_5\). Consequently the subgroup is \(A_5\). These surjections and the computed orders establish the two presentations. We use the simplicity of \(A_n\) for \(n\geq5\) here and below [@DF04].
<2>2. Let \(z\) generate the central kernel and lift \(a,b\) to \(\alpha,\beta\). Since \(3\) is odd, replacing \(\beta\) by \(z\beta\) when necessary makes \(\beta^3=1\). Since \(m\) is odd, replacing \(\alpha\) by \(z\alpha\) when necessary makes \((\alpha\beta)^m=1\). This replacement does not change \(\alpha^2\), which is either \(1\) or \(z\).
If \(\alpha^2=1\), the subgroup generated by \(\alpha,\beta\) is a quotient of \(T_m\) that surjects onto \(T_m\), hence maps isomorphically to the quotient. It is a complement to the central kernel, giving the direct product.
<2>3. If \(\alpha^2=z\), the extension is a quotient of \[ \widetilde T_m= \langle a,b,z\mid z^2=1,\ [z,a]=[z,b]=1, a^2=z,\ b^3=1,\ (ab)^m=1\rangle. \] Its central subgroup \(\langle z\rangle\) has order at most \(2\), and its quotient is \(T_m\). Thus \(|\widetilde T_m|\leq 2|T_m|\). For \(p=m\in\{3,5\}\), the matrices \[ A=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad B=\begin{pmatrix}0&1\\-1&-1\end{pmatrix} \quad\text{in }\mathrm{SL}_2(\mathbf F_p) \] satisfy \(A^2=-I\), \(B^3=I\), and \(AB=\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)\) of order \(p\). They generate \(\mathrm{SL}_2(\mathbf F_p)\): powers of \(AB\) give all upper elementary matrices, conjugation by \(A\) gives the lower ones, and these perform Gaussian elimination. Diagonal matrices are also generated, as witnessed, for \(t\ne0\), by \[ \begin{pmatrix}1&t\\0&1\end{pmatrix} \begin{pmatrix}1&0\\-t^{-1}&1\end{pmatrix} \begin{pmatrix}1&t\\0&1\end{pmatrix}A =\begin{pmatrix}t&0\\0&t^{-1}\end{pmatrix}. \] There are \(p^2-1\) choices for a nonzero first column and \(p\) choices for a second column having determinant \(1\), so this matrix group has order \(p(p^2-1)=2|T_m|\). It is therefore isomorphic to \(\widetilde T_m\) and to the extension in this case.
<1>3. Every group \(H\) of order \(24\) is solvable. If its Sylow \(3\)-subgroup is not normal, then \[ H\cong S_4,\qquad C_2\times A_4,\qquad \mathrm{SL}_2(\mathbf F_3). \]
If the Sylow \(3\)-subgroup is normal, step <1>1 gives \(H\cong C_3\rtimes P\) for a group \(P\) of order \(8\), so \(H\) is solvable. Otherwise there are four Sylow \(3\)-subgroups. Conjugation on them defines \(H\to S_4\) with kernel \(K\) contained in a normalizer of order \(6\). If \(3\) divided \(|K|\), normality would put all four Sylow \(3\)-subgroups inside \(K\), requiring at least \(1+4(3-1)=9\) elements. Thus \(|K|=1\) or \(2\). In the first case \(H\cong S_4\). In the second, the image is \(A_4\), and \(K\cong C_2\) is central. Step <1>2 gives the other two possibilities. The series \(1\lhd V_4\lhd A_4\lhd S_4\) has abelian factors, and a central extension of a solvable group is solvable, proving the assertion in all cases.
<1>4. A group \(G\) of order \(120\) is solvable if and only if its Sylow \(5\)-subgroup is normal. In that case \(G\cong C_5\rtimes H\) with \(|H|=24\).
We first prove by induction that every solvable group of order dividing \(120\) and divisible by \(5\) has a normal Sylow \(5\)-subgroup. A minimal nontrivial normal subgroup \(N\) of a finite solvable group is elementary abelian: its commutator subgroup must be trivial by minimality and solvability, and its primary subgroups and subgroup of \(p\)-th powers are characteristic. Here \(N\) is therefore \(C_5\), \(C_3\), or \(C_2^r\) with \(1\leq r\leq3\).
If \(N=C_5\), the conclusion follows. Otherwise the induction hypothesis in \(G/N\) gives a normal subgroup \(L/N\) of order \(5\). A Sylow \(5\)-subgroup \(P\) of \(L\) acts trivially on \(N\), because \[ |\operatorname{Aut}(C_3)|=2,\qquad |\operatorname{GL}_r(\mathbf F_2)|\in\{1,6,168\} \] are not divisible by \(5\). Hence \(L=N\times P\), and \(P\) is its unique Sylow \(5\)-subgroup. It is characteristic in \(L\lhd G\), so normal in \(G\). The displayed linear-group orders follow by choosing successive independent columns; for example \((8-1)(8-2)(8-4)=168\).
For \(|G|=120\), a normal Sylow \(5\)-subgroup is cyclic and step <1>1 splits the extension. Its quotient of order \(24\) is solvable by step <1>3, so \(G\) is solvable as well.
<1>5. Exactly \(39\) isomorphism classes have a normal cyclic subgroup of order \(15\). They are the following explicitly specified semidirect products.
<2>1. Put \(N=C_3\times C_5=\langle c\rangle\times\langle d\rangle\). Identify its automorphism group with the additive group \(\mathcal A=C_2\oplus C_4\) by letting \((i,j)\) act as \[ c\longmapsto c^{(-1)^i},\qquad d\longmapsto d^{2^j}. \] Write \(u=(1,0)\), \(v=(0,1)\), and \(t=2v=(0,2)\). Thus \(\mathcal A[2]=\{0,u,t,u+t\}\). For each row below form \(N\rtimes_\phi P\), with \(\phi\) given on the displayed ordered generators. Each choice in a set or each explicitly listed tuple is one group.
| \(P\) and ordered generators | Values of \(\phi\) | Number |
|---|---|---|
| \(C_8=\langle r\rangle\) | \(\phi(r)\in\{0,u,t,u+t,v,u+v\}\) | 6 |
| \(C_4\times C_2=\langle r\rangle\times\langle s\rangle\) | \((\phi(r),\phi(s))=(R,0)\), \(R\in\{0,u,t,u+t\}\) | 4 |
| Same \(P\) | \((R,u)\), \(R\in\{0,t\}\) | 2 |
| Same \(P\) | \((R,t)\), \(R\in\{0,u\}\) | 2 |
| Same \(P\) | \((R,u+t)\), \(R\in\{0,u\}\) | 2 |
| Same \(P\) | \((v,0),(u+v,0),(v,u)\) | 3 |
| \(C_2^3=\langle r,s,w\rangle\) | \((0,0,0)\); \((R,0,0)\) for \(R\in\{u,t,u+t\}\); \((u,t,0)\) | 5 |
| \(D_8=\langle r,s\mid r^4=s^2=1,\ srs^{-1}=r^{-1}\rangle\) | \((0,S)\), \(S\in\{0,u,t,u+t\}\) | 4 |
| Same \(P\) | \((u,S)\), \(S\in\{0,t\}\) | 2 |
| Same \(P\) | \((t,S)\), \(S\in\{0,u\}\) | 2 |
| Same \(P\) | \((u+t,S)\), \(S\in\{0,u\}\) | 2 |
| \(Q_8=\langle r,s\mid r^4=1,\ s^2=r^2,\ srs^{-1}=r^{-1}\rangle\) | \((0,0)\); \((R,0)\) for \(R\in\{u,t,u+t\}\); \((u,t)\) | 5 |
The entries sum to \(6+13+5+10+5=39\).
<2>2. These are all possibilities for \(P\). An abelian group of order \(8\) is \(C_8\), \(C_4\times C_2\), or \(C_2^3\), according as its exponent is \(8\), \(4\), or \(2\). In the exponent-\(4\) case, choose \(r\) of order \(4\) and an element outside \(\langle r\rangle\); multiplying the latter by \(r^{-1}\) when its square is \(r^2\) produces an outside involution and hence the direct product. A nonabelian group of order \(8\) has an element \(r\) of order \(4\), since a group in which every element squares to \(1\) is abelian. For \(s\notin\langle r\rangle\), conjugation must invert \(r\), and \(s^2\in\{1,r^2\}\). These give exactly the displayed presentations of \(D_8\) and \(Q_8\).
<2>3. A normal subgroup \(N\cong C_{15}\) is precisely the set of all odd-order elements of \(G\): the quotient is a \(2\)-group, and every element of \(N\) has odd order. Thus \(N\) is characteristic, and step <1>1 splits the extension. Two actions on the same \(P\) give isomorphic groups precisely when they differ by precomposition with an automorphism of \(P\). Indeed an isomorphism induces automorphisms of \(N\) and the quotient \(P\); the former conjugates the action, but \(\operatorname{Aut}(N)\) is abelian, so this conjugation does nothing. Conversely, a change of generators in \(P\) realizing the precomposition gives an isomorphism of semidirect products. Different types of \(P\) cannot give isomorphic quotients.
<2>4. Each row of the table lists one representative of each \(\operatorname{Aut}(P)\)-orbit of actions. For \(C_8\), changing a generator multiplies its image by an odd integer, hence identifies \(R\) with \(-R\); this gives its six listed images. For \(C_4\times C_2\), every automorphism has the form \[ r\longmapsto r^{\varepsilon}s^e,\qquad s\longmapsto r^{2f}s, \quad \varepsilon\in\{1,3\},\ e,f\in\{0,1\}. \] Consequently \((R,S)\) changes to \((\varepsilon R+eS,2fR+S)\). If \(2R=0\), \(S\) stays fixed and \(R\) is taken modulo \(\langle S\rangle\); these are the first ten tuples for this \(P\). If \(R\) has order \(4\), then \(2R=t\). When \(S\in\{0,t\}\), one can set \(S=0\) and reduce \(R\) to either \(v\) or \(u+v\). When \(S\in\{u,u+t\}\), the same operations reduce the pair to \((v,u)\). These are three further, distinct orbits.
For \(C_2^3\), all actions are linear maps to \(\mathcal A[2]\). Changing a basis of the domain identifies two maps exactly when they have the same image; there are the zero subspace, three lines, and the whole plane. For \(D_8\), the relations force \(R,S\in\mathcal A[2]\). Its automorphisms send \(r\) to \(r^{\pm1}\) and \(s\) to \(r^k s\), so \(R\) stays fixed and \(S\) is taken modulo \(\langle R\rangle\), giving the ten listed pairs. Finally, every homomorphism from \(Q_8\) factors through its abelianization \(C_2^2\). Automorphisms permuting the three cyclic subgroups of order \(4\) induce all of \(\operatorname{GL}_2(\mathbf F_2)\). Again the image subspace classifies the action, giving five pairs. This proves both exhaustiveness and irredundancy of the table.
<1>6. The remaining solvable groups of order \(120\) are exactly \[ C_5\times\mathrm{SL}_2(\mathbf F_3),\quad C_5\times S_4,\quad C_5\rtimes_{\mathrm{sgn}}S_4,\quad C_{10}\times A_4,\quad D_{10}\times A_4. \] In the third group, even permutations act trivially on \(C_5\) and odd permutations act by inversion. Here \(D_{10}\) has order \(10\).
Write \(G=C_5\rtimes H\) using step <1>4. If \(H\) has a normal Sylow \(3\)-subgroup, its inverse image in \(G\) has order \(15\) and is cyclic: the action of \(C_3\) on \(C_5\) is trivial because \(3\nmid4\). This is step <1>5. Otherwise step <1>3 gives the three possibilities for \(H\). Their abelianizations are respectively \(C_3\), \(C_2\), and \(C_3\times C_2\) for \(\mathrm{SL}_2(\mathbf F_3)\), \(S_4\), and \(A_4\times C_2\).
For the first assertion, abelianize the presentation in step <1>2 with \(m=3\): \(a^4=b^3=(ab)^3=1\) implies \(a^3=a^4=1\), hence \(a=1\); the quotient onto \(A_4/V_4=C_3\) shows that the abelianization is exactly \(C_3\). Also \([S_4,S_4]=A_4\) and \([A_4,A_4]=V_4\): the respective quotients are abelian, commutators of transpositions give \(3\)-cycles, and a commutator of two \(3\)-cycles in \(A_4\) is a nonidentity double transposition whose conjugates generate \(V_4\).
A homomorphism from these abelianizations to \(\operatorname{Aut}(C_5)=C_4\) is trivial in the first case and is either trivial or the unique nontrivial map from the \(C_2\) factor in each of the other two cases. This gives exactly the five displayed groups. The quotient by the unique Sylow \(5\)-subgroup distinguishes the three types of \(H\); within each type, a trivial action cannot become nontrivial under an automorphism. None has a normal subgroup of order \(15\), since its image in \(H\) would be a normal subgroup of order \(3\).
<1>7. The nonsolvable groups of order \(120\) are exactly \[ S_5,\qquad C_2\times A_5,\qquad \mathrm{SL}_2(\mathbf F_5). \] Thus part (a) has \(39+5+3=47\) isomorphism classes. Its abelian members are \(C_{120}\), \(C_{60}\times C_2\), and \(C_{30}\times C_2^2\).
<2>1. The only nonabelian simple group whose order divides \(120\) is \(A_5\). Here are the order exclusions. Prime-power orders have nontrivial center. Orders \(6,10,15,20,40\) have a normal Sylow \(3\)- or \(5\)-subgroup. For orders \(12\) and \(24\), a nonnormal Sylow \(3\)-subgroup gives an action on four such subgroups. A simple group would embed in \(A_4\), which is impossible for order \(24\) and would identify the order-\(12\) group with the nonsimple group \(A_4\). For order \(30\), absence of normal Sylow \(3\)- and \(5\)-subgroups would require \(10\cdot2+6\cdot4=44\) nonidentity elements of orders \(3\) and \(5\).
For a simple group of order \(60\), Sylow theory gives \(n_5=6\) and \(n_3\in\{4,10\}\). The value \(n_3=4\) would embed it in \(A_4\), so \(n_3=10\). These account for \(24+20=44\) nonidentity elements. The possibilities for \(n_2\) are \(3,5,15\); \(3\) is excluded by the degree-\(3\) conjugation action, and \(5\) embeds the group in \(A_5\), proving the required isomorphism. If \(n_2=15\) and the Sylow \(2\)-subgroups were cyclic, their \(30\) distinct generators of order \(4\) would exceed the \(15\) remaining nonidentity elements. Thus they are Klein four groups. For an involution \(t\), its centralizer has order \(4,12\), or \(20\); order \(60\) is excluded by the trivial center. Order \(20\) gives an impossible faithful action of degree \(3\), while order \(12\) gives an embedding in \(A_5\). If all these centralizers have order \(4\), each involution belongs to a unique Sylow \(2\)-subgroup, so \(15\) Klein four subgroups require \(45\) distinct involutions, again impossible. This proves the order-\(60\) assertion.
Finally, a simple group of order \(120\) would have six Sylow \(5\)-subgroups and hence embed in \(A_6\). Its image would have index \(3\). The action of the simple group \(A_6\) on those three cosets is nontrivial but cannot be injective, since \(360>6\), a contradiction. These cases exhaust all divisors of \(120\). In each use of simplicity, the nontrivial permutation action is faithful and its composite with the sign homomorphism is trivial, so the image lies in the alternating group.
<2>2. A composition series of a nonsolvable group of order \(120\) must therefore have factors \(A_5\) and \(C_2\). The group has either a normal subgroup \(C_2\) with quotient \(A_5\), or a normal subgroup \(A_5\) with quotient \(C_2\). In the first case the kernel is central, so step <1>2 applies.
In the second case, conjugation gives \(G\to\operatorname{Aut}(A_5)\cong S_5\). To justify the last isomorphism, the five Sylow \(2\)-subgroups of \(A_5\) are the Klein four groups fixing each of the five letters. An automorphism permutes these groups. If it fixes all five, then for each \(g\in A_5\), \(g\) and its image induce the same permutation of them; the faithful conjugation action implies that the automorphism fixes \(g\). Hence \(\operatorname{Aut}(A_5)\) embeds in \(S_5\). Conversely conjugation by \(S_5\) induces all natural permutations of these five groups, so embeds \(S_5\) in the automorphism group. This proves equality.
The kernel of \(G\to\operatorname{Aut}(A_5)\) is the centralizer of its normal \(A_5\) and has trivial intersection with that subgroup. Its order is therefore \(1\) or \(2\). In the first case \(G\cong S_5\); in the second the commuting subgroups give \(G\cong A_5\times C_2\).
<2>3. The three groups exist and have a section isomorphic to \(A_5\), so are nonsolvable. They are distinct: \(S_5\) has trivial center, whereas the other two have a central involution, and \(\mathrm{SL}_2(\mathbf F_5)\) is perfect while \(C_2\times A_5\) has abelianization \(C_2\). For perfectness, abelianize its presentation from step <1>2: \(4a=0\), \(3b=0\), and \(5a+5b=0\) imply \(a+2b=0\), then \(b=0\) and \(a=0\); also \(z=2a=0\).
Finally an abelian group of order \(120\) is the product of its Sylow subgroups; its odd part is \(C_{15}\) and its \(2\)-part is one of the three abelian groups of order \(8\) from step <1>5. This gives the stated three abelian groups.
<1>8. For part (b), put \(R=\mathbf Q[x]\), \(p=x-3\), and \(q=x+5\). The isomorphism classes are \[ V_{a,b,c}=(R/(p^2))^a\oplus(R/(p))^b\oplus(R/(q))^c, \qquad a,b,c\geq0,\quad 2a+b+c=6. \] There are \(7+5+3+1=16\) classes, corresponding to \(a=0,1,2,3\).
Let \(T\) be multiplication by \(x\). Since \(p^2\) and \(q\) are relatively prime and \(p(T)^2q(T)=0\), a Bezout identity decomposes \[ V=\ker(T-3I)^2\oplus\ker(T+5I). \] Indeed, if \(h p^2+kq=1\), then \(w=h(T)p(T)^2w+k(T)q(T)w\), with the first term in \(\ker q(T)\) and the second in \(\ker p(T)^2\). Their intersection is zero by the same identity. On the first summand put \(U=T-3I\), so \(U^2=0\). Choose a basis \(e_1,\ldots,e_a\) of \(\operatorname{im}U\), extend it by \(f_1,\ldots,f_b\) to a basis of \(\ker U\), and choose \(w_i\) with \(Uw_i=e_i\). The vectors \(w_i,e_i,f_j\) form a basis: applying \(U\) proves independence of the \(w_i\) modulo the kernel, and subtracting the appropriate \(w_i\) from any vector puts it in the kernel. Each pair \(w_i,e_i\) gives \(R/(p^2)\) and each \(f_j\) gives \(R/(p)\). On the other summand \(T=-5I\), giving \(c\) copies of \(R/(q)\).
The dimension condition is exactly \(2a+b+c=6\). Conversely every displayed module has that dimension and is annihilated by \(p^2q\). The dimensions of the two primary summands and the rank of \(U\) recover \(a,b,c\), so distinct triples give nonisomorphic modules.
<1>9. For part (c), let \(C(f)\) denote the companion matrix of a monic polynomial \(f\): it is the matrix of multiplication by \(x\) on \(R/(f)\) in the ordered basis \(1,x,\ldots,x^{\deg f-1}\). The six possible rational canonical forms are the block diagonal matrices with the following ordered companion blocks.
| \((a,b,c)\) | Invariant factors, in divisibility order |
|---|---|
| \((1,0,4)\) | \(q,\ q,\ q,\ p^2q\) |
| \((1,1,3)\) | \(q,\ pq,\ p^2q\) |
| \((1,2,2)\) | \(p,\ pq,\ p^2q\) |
| \((1,3,1)\) | \(p,\ p,\ p,\ p^2q\) |
| \((2,0,2)\) | \(p^2q,\ p^2q\) |
| \((2,1,1)\) | \(p,\ p^2,\ p^2q\) |
In step <1>8 the minimal polynomial is exactly \(p^2q\) precisely when \(a\geq1\) and \(c\geq1\). The nonnegative solutions of \(2a+b+c=6\) with these inequalities are exactly the six triples displayed. For each, combine the primary cyclic factors by \(R/(p^i)\oplus R/(q)\cong R/(p^iq)\), arranging the \(p\)-powers in increasing order and pairing the \(q\)-factors with the rightmost ones. This gives each listed divisibility chain. Its degrees sum to \(6\) and its largest factor is \(p^2q\), so its companion blocks have the required size and minimal polynomial. Distinct triples were proved nonisomorphic in step <1>8, and therefore give distinct similarity classes. Every matrix in question yields one of those modules, proving completeness.