Definitions
A bilinear form \(b\) on a finite-dimensional vector space \(V\) has a Gram matrix \(G=(b(e_i,e_j))_{i,j}\) in each basis \((e_i)\) of \(V\), and \(b\) is nondegenerate if and only if \(G\) is invertible; for a nondegenerate symmetric form and a subspace \(W\subseteq V\), \(\dim W+\dim W^\perp=\dim V\). A complex matrix is normal if it commutes with its conjugate transpose, and normal matrices are exactly the unitarily diagonalizable ones. Over a perfect field, every linear operator \(T\) has a unique Jordan–Chevalley decomposition \(T=T_s+T_n\) with \(T_s\) semisimple, \(T_n\) nilpotent, and \(T_sT_n=T_nT_s\); for invertible \(T\), the multiplicative form is \(T=T_sT_u\) with \(T_u=I+\inverseof{T_s} T_n\) unipotent.
Nondegenerate symmetric and Hermitian forms (Tag D-5BR4D)
Let \(V\) be a real vector space with a symmetric bilinear form, or a complex vector space with a Hermitian form, and write \(\inner{\wait}{\wait}\) for the form. A vector \(v\in V\) is a null vector if \(\inner v w = 0\) for every \(w\in V\); the null vectors form the nullspace of the form. The form is nondegenerate if its nullspace is \(\theset 0\), that is, if for every \(v \neq 0\) there is some \(v'\in V\) with \(\inner{v}{v'} \neq 0\). Otherwise it is degenerate.
Let \(V\) and \(\inner{\wait}{\wait}\) be as in the definition, with \(V\) finite-dimensional.
Fix a basis of \(V\) and let \(A\) be the matrix of the form in that basis. Then \(v\) is a null vector if and only if its coordinate vector \(Y\) satisfies \(AY = 0\), so the form is nondegenerate if and only if \(A\) is invertible.
For a subspace \(W \subseteq V\) with \(W^\perp \coloneqq \theset{v\in V \st \inner w v = 0 \text{ for all } w\in W}\), the restriction of the form to \(W\) is nondegenerate if and only if \(W \cap W^\perp = \theset 0\), if and only if \(V = W \oplus W^\perp\).
Nondegeneracy of the form on \(V\) and nondegeneracy of its restriction to a subspace are independent. On \(V = \RR^2\) with standard basis \(e_1, e_2\), the form \(\inner x y = x_1y_1 - x_2y_2\) is nondegenerate, but its restriction to \(W = \RR(e_1 + e_2)\) is zero, hence degenerate. The form \(\inner x y = x_1y_1\) on \(\RR^2\) is degenerate, with nullspace \(\RR e_2\), but its restriction to \(\RR e_1\) is nondegenerate.
See [@Art11]. The formulation over an arbitrary field, through the adjoint \(V \to \dualof{V}\), is D-O4WWN.
Quadratic forms in coordinates (Tag D-SSGKC)
Let \(k\) be a field with \(\characteristic k\neq 2\) and \(n\geq 1\). A quadratic form in \(n\) variables over \(k\) is a homogeneous polynomial of degree \(2\) in \(x_1,\ldots,x_n\); it can be written uniquely as \[ q(x_1, \ldots, x_n) = \sum_{i} a_{ii}x_i^2 + \sum_{i < j} 2a_{ij}x_ix_j = X^t A X, \qquad X = (x_1,\ldots,x_n)^t, \] where \(A = (a_{ij})\in\Mat_{n\times n}(k)\) is symmetric, \(a_{ji}=a_{ij}\). The matrix \(A\) is the matrix of the form.
The coefficient of \(x_ix_j\) for \(i<j\) is written \(2a_{ij}\) so that \(X^tAX\) holds with \(A\) symmetric: the off-diagonal entries \(a_{ij}\) and \(a_{ji}\) each contribute \(a_{ij}x_ix_j\).
Let \(k=\RR\) and let \(q(X)=X^tAX\) with \(A\) real symmetric. There is an orthogonal matrix \(P\) such that the change of variables \(X = PX'\) gives \[ q = \sum_{i=1}^n \lambda_i {x_i'}^2, \] where \(\lambda_1,\ldots,\lambda_n\) are the eigenvalues of \(A\).
By the spectral theorem there is an orthogonal \(P\) with \(P^tAP=P^{-1}AP=\diag(\lambda_1,\ldots,\lambda_n)\), and \(q(PX')=X'^t(P^tAP)X'\).
Let \(q\) be a real quadratic form with symmetric matrix \(A\), nondegenerate in the sense that \(\det A\neq 0\). The signature of \(q\) is the pair \((p,m)\), where \(p\) and \(m\) are the numbers of positive and negative coefficients \(\lambda_i\) in a diagonalization \(q=\sum_i\lambda_i{x_i'}^2\) obtained by an invertible change of variables.
(Sylvester's law of inertia)
For a nondegenerate real quadratic form, the numbers \(p\) and \(m\) do not depend on the invertible change of variables that diagonalizes it.
For a nondegenerate real quadratic form \(q\) and \(c>0\), the signature determines the level set \(\theset{q = c}\) up to an invertible linear change of variables. In two variables, signature \((2,0)\) gives an ellipse, \((1,1)\) a hyperbola, and \((0,2)\) the empty set. In three variables, signature \((3,0)\) gives an ellipsoid, \((2,1)\) a hyperboloid of one sheet, \((1,2)\) a hyperboloid of two sheets, and \((0,3)\) the empty set.
See [@Art11]. A basis-free quadratic form is a function \(q\) with \(q(\lambda x) = \lambda^2 q(x)\) whose polarization is bilinear.
Gram matrix of a bilinear form (Tag D-RG5FO)
Let \(k\) be a field, \(V\) a finite-dimensional \(k\)-vector space, \(\inner{\wait}{\wait}\) a bilinear form on \(V\), and \(\mathbf B = (v_1, \ldots, v_n)\) an ordered basis of \(V\). The Gram matrix of the form with respect to \(\mathbf B\), also called the matrix of the form, is \[ A = (a_{ij}), \qquad a_{ij} \coloneqq \inner{v_i}{v_j}. \]
If \(X,Y\in k^n\) are the coordinate vectors of \(v,w\in V\) with respect to \(\mathbf B\), then \[ \inner{v}{w} = X^t A Y. \] Consequently, for a fixed basis \(\mathbf B\), the assignment of the Gram matrix is a bijection from bilinear forms on \(V\) to \(\Mat_{n\times n}(k)\), and a form is symmetric if and only if its Gram matrix is symmetric.
Writing \(v=\sum_i x_iv_i\) and \(w=\sum_j y_jv_j\), bilinearity gives \(\inner{v}{w}=\sum_{i,j}x_iy_j\inner{v_i}{v_j}=X^tAY\). Conversely, every \(A\in\Mat_{n\times n}(k)\) defines the bilinear form \((v,w)\mapsto X^tAY\), whose Gram matrix is \(A\). The form is symmetric if and only if \(a_{ij}=a_{ji}\) for all \(i,j\), by the displayed formula.
If \(\mathbf B'\) is another ordered basis and \(P\in\GL_n(k)\) is the change-of-basis matrix, so that coordinates satisfy \(X=PX'\), then the Gram matrix with respect to \(\mathbf B'\) is \(P^t A P\). This preserves the rank of \(A\) and, for \(k=\RR\) and a symmetric form, its signature, but not its eigenvalues: for \(P=2I_n\) the new Gram matrix is \(4A\).
See [@Art11].
Normal matrix (Tag D-BSUV4)
Let \(n\geq 1\). A matrix \(A\in \Mat(n\times n; \CC)\) is normal if \(A^{\dagger} A = AA^{\dagger}\), where \(A^{\dagger} = \bar A^t\) is the conjugate transpose of \(A\).
Semisimple matrix (Tag D-B4VTH)
Let \(k\) be a field with algebraic closure \(k^\alg\). A matrix \(A \in \Mat_{n\times n}(k)\) is semisimple if \(A\) is diagonalizable over \(k^\alg\), that is, if there exist \(P\in\GL_n(k^\alg)\) and a diagonal \(D\in\Mat_{n\times n}(k^\alg)\) with \(A = PDP^{-1}\).
Nilpotent matrix (Tag D-HGMOW)
Let \(k\) be a field and \(n\geq 1\). A matrix \(A\in\Mat_{n\times n}(k)\) is nilpotent if there exists an integer \(m\geq 1\) such that \(A^m = 0\).
A strictly upper triangular matrix \(A\in\Mat_{n\times n}(k)\) is nilpotent, with \(A^n=0\): if \(e_1,\ldots,e_n\) is the standard basis, then \(A\) maps \(\spanof\theset{e_1,\ldots,e_j}\) into \(\spanof\theset{e_1,\ldots,e_{j-1}}\) for each \(j\). Conversely, every nilpotent \(A\in\Mat_{n\times n}(k)\) is similar to a strictly upper triangular matrix: every eigenvalue \(\lambda\) of \(A\) in an algebraic closure of \(k\) satisfies \(\lambda^m=0\), so the characteristic polynomial of \(A\) is \(\pm x^n\), which splits over \(k\), and a matrix whose characteristic polynomial splits is similar to an upper triangular matrix whose diagonal entries are its eigenvalues.
Unipotent element (Tag D-23FX7)
Let \(R\) be a ring. An element \(a\in R\) is unipotent if \(a-1\) is nilpotent.
Let \(k\) be a field and let \(A\in\Mat_{n\times n}(k)\) be upper triangular with every diagonal entry equal to \(1\). Then \(A - I\) is strictly upper triangular, so \((A-I)^n = 0\) and \(A\) is unipotent.
Jordan--Chevalley decomposition (Tag PR-24CPI)
Let \(k\) be a perfect field, let \(V\) be a finite-dimensional \(k\)-vector space, and let \(T\colon V\to V\) be \(k\)-linear. Then there are unique \(k\)-linear maps \(S,N\colon V\to V\) with \[ T=S+N,\qquad SN=NS, \] where \(S\) is semisimple, that is, diagonalizable over an algebraic closure \(\kbar\), and \(N\) is nilpotent. If \(T\) is invertible, there are unique \(k\)-linear maps \(S,U\colon V\to V\) with \(T=SU\) and \(SU=US\), where \(S\) is semisimple and \(U\) is unipotent.
Orthogonal complement of a sum of subspaces (Tag PR-WDPF7)
Let \(V\) be an inner product space and \((W_i)_{i\in I}\) a family of subspaces of \(V\), where \(W^\perp\coloneqq\theset{v\in V\suchthat \inner{w}{v}=0\ \forall w\in W}\). Then \[ \qty{ \sum_{i\in I} W_i}^\perp = \bigcap_{i\in I} W_i^\perp. \]
Each \(W_i\subseteq\sum_i W_i\), so \(\qty{\sum_i W_i}^\perp\subseteq W_i^\perp\) for every \(i\). Conversely, if \(v\in\bigcap_i W_i^\perp\) and \(w=w_{i_1}+\cdots+w_{i_r}\) with \(w_{i_j}\in W_{i_j}\), then \(\inner{w}{v}=\sum_j\inner{w_{i_j}}{v}=0\).
Square matrices \(A\) and \(B\) are similar if \(B=PA\inverseof{P}\) for an invertible \(P\), and \(m\times n\) matrices are equivalent if \(B=PAQ\) for invertible \(P\) and \(Q\). Similar matrices have the same characteristic polynomial, minimal polynomial, and Jordan form; two matrices over a field are equivalent if and only if they have the same size and rank.
Similar matrices (Tag D-JIGMN)
Let \(k\) be a field and \(n\geq 1\). Matrices \(A,B\in\Mat_{n\times n}(k)\) are similar if there exists \(P\in\GL_n(k)\) such that \(A = PBP^{-1}\).
Let \(k\) be an algebraically closed field and \(A,B\in\Mat_{n\times n}(k)\). Then \(A\) and \(B\) are similar if and only if they have the same Jordan canonical form, up to the order of the Jordan blocks.
Every matrix over \(k\) is similar to a Jordan canonical form, and similarity is an equivalence relation, so matrices with the same Jordan form are similar. Conversely, for \(\lambda\in k\) and \(j\geq 1\), the number of Jordan blocks of \(A\) with eigenvalue \(\lambda\) and size at least \(j\) is \(\dim\ker(A-\lambda I)^j-\dim\ker(A-\lambda I)^{j-1}\). If \(A=PBP^{-1}\), then \((A-\lambda I)^j=P(B-\lambda I)^jP^{-1}\), so these dimensions agree for \(A\) and \(B\), and hence so do their Jordan blocks.
Equivalent matrices (Tag D-JRPTK)
Let \(R\) be a principal ideal domain and \(m,n\geq 1\). Matrices \(A, B\in\Mat_{m\times n}(R)\) are equivalent if there exist \(P\in\GL_m(R)\) and \(Q\in\GL_n(R)\) such that \(A = PBQ\).
Let \(R\) be a principal ideal domain and \(A,B\in\Mat_{m\times n}(R)\). Then \(A\) and \(B\) are equivalent if and only if they have the same invariant factors, that is, the same Smith normal form. In particular, equivalent matrices have the same rank, namely the number of nonzero invariant factors.
Every matrix over \(R\) is equivalent to its Smith normal form \(\diag(d_1,\ldots,d_s,0,\ldots,0)\) with \(d_1\divides d_2\divides\cdots\divides d_s\) nonzero, so matrices with the same Smith normal form are equivalent. Conversely, for each \(i\) let \(\Delta_i(A)\) be a greatest common divisor of the \(i\times i\) minors of \(A\). By the Cauchy–Binet formula, every \(i\times i\) minor of \(PBQ\) is an \(R\)-linear combination of \(i\times i\) minors of \(B\), so \(\Delta_i(B)\divides\Delta_i(PBQ)\); applying this also to \(B=P^{-1}AQ^{-1}\) shows that \(\Delta_i(A)\) and \(\Delta_i(B)\) are associates. For the Smith normal form, \(\Delta_i=d_1\cdots d_i\) for \(i\leq s\) and \(\Delta_i=0\) for \(i>s\), so \(s\) and the invariant factors \(d_i=\Delta_i/\Delta_{i-1}\) are determined up to units by the equivalence class.
Over a field \(k\), the invariant factors of a matrix of rank \(s\) are \(1,\ldots,1\) (\(s\) times), so \(A,B\in\Mat_{m\times n}(k)\) are equivalent if and only if they have the same rank. Equivalence is coarser than similarity: if \(A=PBP^{-1}\) then \(A\) and \(B\) are equivalent, but over \(\QQ\) the matrices \(I_2\) and \(\diag(2,1)\) both have rank \(2\), so they are equivalent, while they are not similar, since their eigenvalues differ.
Determinants
(Determinant formulas)
For \(A=(a_{ij})\in\Mat(n\times n;R)\), \[ \det A = \sum_{\sigma \in S_n} \sgn(\sigma) \prod_{i=1}^n a_{i, \sigma(i)}, \] which for \(n=3\) reads \[ \det\left(\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right)=\begin{gathered} a_{11} a_{22} a_{33}+a_{12} a_{23} a_{31}+a_{13} a_{21} a_{32} \\ -a_{13} a_{22} a_{31}-a_{12} a_{21} a_{33}-a_{11} a_{23} a_{32} \end{gathered}. \]
Writing \(\minor_A(i, j)\) for the matrix obtained from \(A\) by deleting row \(i\) and column \(j\), expansion along row \(i\) is \[ \det(A) = \sum_{j=1}^n (-1)^{i+j} a_{ij} \det \minor_A(i, j), \] and if \(\det A\) is invertible, the adjugate gives the inverse: \[ \inverseof{A} = {1\over \det A} \operatorname{adj}(A), \qquad \operatorname{adj}(A)_{ij} \da (-1)^{i+j} \det \minor_A(j, i). \]
(Block multiplication)
For block matrices whose block sizes make the products \(AE\), \(BG\), \(AF\), \(BH\), \(CE\), \(DG\), \(CF\), \(DH\) defined, \[ \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} E & F \\ G & H \end{bmatrix} = \matt{AE + BG}{AF + BH}{CE + DG}{ CF + DH}. \]
Eigenvalues and the characteristic polynomial
(Trace and determinant from the eigenvalues)
Let \(A\) be an \(n\times n\) matrix over a field whose characteristic polynomial \(\chi_A\) splits, with eigenvalues \(\lambda_1,\ldots,\lambda_n\) listed with algebraic multiplicity. The coefficients of \(\chi_A\) are, up to sign, the elementary symmetric functions of the eigenvalues: \[ \chi_A(t) = t^n - \qty{\sum_i \lambda_i }t^{n-1} + \qty{\sum_{i < j} \lambda_i \lambda_j }t^{n-2} - \cdots + (-1)^n\qty{\prod_i \lambda_i}. \] In particular \[ \tr(A) = \sum_i \lambda_i, \qquad \det(A) = \prod_i \lambda_i. \] The eigenvalues of a triangular matrix are its diagonal entries, so its determinant is the product of its diagonal entries.
(Similarity invariants)
\(\tr(AB) = \tr(BA)\), so similar matrices have the same trace: \(\tr(PJ\inverseof{P}) = \tr(\inverseof{P} P J) = \tr J\). The determinant, characteristic polynomial, and minimal polynomial are also similarity invariants.
(Powers of a triangular matrix)
\[ A\da\left(\begin{array}{ccc} a_1 & & * \\ & \ddots & \\ 0 & & a_n \end{array}\right) \implies A^k = \left(\begin{array}{ccc} a_1^k & & * \\ & \ddots & \\ 0 & & a_n^k \end{array}\right). \]
Matrix groups
The general linear group \(\GL_n(F)\) is the group of invertible \(n\times n\) matrices over \(F\), and the special linear group \(\SL_n(F)\) is its subgroup of matrices of determinant \(1\). The orthogonal group \(\ts{A \st A^tA=I}\) is the group of matrices preserving the standard symmetric bilinear form, the unitary group \(\ts{A\st A^\dagger A=I}\) preserves the standard Hermitian form, and the symplectic group \(\ts{A\st A^tJA=J}\) preserves the standard alternating form with Gram matrix \(J=\matt{0}{I_n}{-I_n}{0}\). The special orthogonal and special unitary groups are the subgroups of matrices of determinant \(1\).
General linear group $\GL_n(\RR)$ (Tag D-J5AAX)
Let \(n\geq 1\). The general linear group is \[ \GL_n(\RR) \coloneqq \theset{ A \in \Mat_{n\times n}(\RR) \st \det A \neq 0 }, \] a group under matrix multiplication.
A matrix \(A\in\Mat_{n\times n}(\RR)\) lies in \(\GL_n(\RR)\) if and only if there exists \(B\in\Mat_{n\times n}(\RR)\) with \(AB=BA=I_n\). If \(AB=I_n\), then \(\det A\det B=1\), so \(\det A\neq 0\); conversely, if \(\det A\neq0\), then \(B=(\det A)^{-1}\operatorname{adj}(A)\) satisfies \(AB=BA=I_n\).
Special linear group $\SL_n(\CC)$ (Tag D-P5D3T)
Let \(n\geq 1\). The special linear group is \[ \SL_n(\CC) \coloneqq \theset{ A\in\Mat_{n\times n}(\CC) \st \det A = 1 }. \]
Orthogonal group $\Orth_n(\CC)$ (Tag D-3ZPR7)
Let \(n\geq 1\). The orthogonal group is \[ \Orth_n(\CC) \coloneqq \theset{ A\in\Mat_{n\times n}(\CC) \st A^tA = AA^t = I }. \]
\(\Orth_n(\CC)\) is a complex Lie group of dimension \(n(n-1)/2\).
Special orthogonal group $\SO_n(\RR)$ (Tag D-GY7ZN)
Let \(n\geq 1\). The special orthogonal group is \[ \SO_n(\RR) \coloneqq \theset{ A\in\Mat_{n\times n}(\RR) \st AA^t = I \text{ and } \det A = 1} = \Orth_n(\RR) \cap \SL_n(\RR). \]
Every \(A\in\Orth_n(\RR)\) satisfies \(\det A = \pm 1\), and \(\SO_n(\RR)\) is the kernel of \(\det\colon\Orth_n(\RR)\to\theset{\pm1}\), of index \(2\) in \(\Orth_n(\RR)\). For \(n\geq 2\), \(\SO_n(\RR)\) is a proper subgroup of \(\SL_n(\RR)\): the matrix \(\operatorname{diag}(2, 1/2, 1, \ldots, 1)\) lies in \(\SL_n(\RR)\) but not in \(\Orth_n(\RR)\).
Unitary group $\U_n(\CC)$ (Tag D-QZ2LQ)
Let \(n\geq 1\), and for \(A\in\Mat_{n\times n}(\CC)\) let \(A^\dagger\) be its conjugate transpose. The unitary group is \[ \U_n(\CC) \coloneqq \theset{ A\in\Mat_{n\times n}(\CC) \st A^\dagger A = AA^\dagger = I_n }. \]
Special unitary group $\SU_n(\CC)$ (Tag D-MCUTE)
Let \(n\geq 1\). The special unitary group is \[ \SU_n(\CC) \coloneqq \theset{ A \in \U_n(\CC) \st \det A = 1 }, \] where \(\U_n(\CC)\) is the unitary group.
Symplectic group $\operatorname{Sp}_{2n}(\CC)$ (Tag D-3V3SP)
Let \(n\geq 1\), and let \[ J \coloneqq \begin{bmatrix} 0 & I_n \\ -I_n & 0 \end{bmatrix} \in \GL_{2n}(\CC). \] The symplectic group is \[ \operatorname{Sp}_{2n}(\CC) \coloneqq \theset{ A \in \GL_{2n}(\CC) \st A^tJA = J }. \]
The matrix \(J\) is skew-symmetric, \(J^t = -J\), so \(\operatorname{Sp}_{2n}(\CC)\) is the group of invertible matrices preserving the alternating bilinear form \(\omega(x,y) = x^tJy\) on \(\CC^{2n}\). Replacing \(J\) by the symmetric matrix \(\begin{bmatrix} 0 & I_n \\ I_n & 0 \end{bmatrix}\) gives instead the orthogonal group of the symmetric bilinear form \(x^t\begin{bmatrix} 0 & I_n \\ I_n & 0 \end{bmatrix}y\).
Counting over a finite field
An element of \(\GL_n(\FF_q)\) is determined by its columns, which form an ordered basis of \(\FF_q^n\). The first column is any of the \(q^n-1\) nonzero vectors, and the \((i+1)\)st column is any of the \(q^n-q^i\) vectors outside the span of the first \(i\) columns, so \[ \size{\GL_n(\FF_q)} = \prod_{i=0}^{n-1}(q^n-q^i). \] Since \(\det\colon\GL_n(\FF_q)\to\FF_q^\times\) is surjective with kernel \(\SL_n(\FF_q)\), \(\size{\SL_n(\FF_q)} = \size{\GL_n(\FF_q)}/(q-1)\).
Order of $\GL_n(\FF_q)$ (Tag PR-UQ3XJ)
Let \(q\) be a prime power and \(n\geq1\). Then \[ \abs{\GL_n(\FF_q)} = \prod_{k=0}^{n-1}\qty{q^n - q^k} = (q^n-1)(q^n-q)\cdots(q^n - q^{n-1}). \]
A matrix in \(M_n(\FF_q)\) is invertible if and only if its columns form a basis of \(\FF_q^n\) [@DF04], so we count ordered bases. If the first \(k\) columns are linearly independent, the \((k+1)\)-st column can be any vector outside their span, a subspace with \(q^k\) elements, so there are \(q^n-q^k\) choices. Multiplying over \(k=0,\ldots,n-1\) gives the formula.
Exercises
Complex normal matrices are unitarily diagonalizable (Tag E-LJ7PF)
Let \(A\in M_n(\CC)\) be normal, so \(AA^*=A^*A\). Show that \(A\) is unitarily diagonalizable.
Solution
<1>1. By Schur triangularization, there is a unitary matrix \(U\) such that \[ T:=U^*AU \] is upper triangular.
Schur’s theorem applies to every complex square matrix. Thus \(A=UTU^*\) for some unitary \(U\) and upper-triangular \(T\).
<1>2. The triangular matrix \(T\) is normal.
Since \(A\) is normal, \[ AA^*=A^*A. \] Conjugating by \(U\) gives \[ TT^*=T^*T. \] Thus \(T\) is normal.
<1>3. Every upper-triangular normal complex matrix is diagonal.
We argue by induction on \(n\). The result is immediate for \(n=1\).
Write \(T=(t_{ij})\). Comparing the \((1,1)\) entries of \(TT^*\) and \(T^*T\) gives \[ |t_{11}|^2+|t_{12}|^2+\cdots+|t_{1n}|^2=|t_{11}|^2. \] Hence \[ t_{12}=\cdots=t_{1n}=0. \] Since \(T\) is upper triangular, its first column also has no entries below \(t_{11}\), so \[ T=[t_{11}]\oplus T_1 \] for an upper-triangular \((n-1)\times(n-1)\) matrix \(T_1\). The equality \(TT^*=T^*T\) then implies \[ T_1T_1^*=T_1^*T_1, \] so \(T_1\) is normal. By induction \(T_1\) is diagonal. Therefore \(T\) is diagonal.
<1>4. Therefore \(A\) is unitarily diagonalizable.
By <1>1 and <1>3, \(T=D\) is diagonal. Hence \[ A=UDU^*, \] which is a unitary diagonalization of \(A\).
Vandermonde determinant $\det A=\prod_{i<j}(\lambda_j-\lambda_i)$ (Tag E-D62SD)
Consider the Vandermonde matrix: \[ A \da \left(\begin{array}{ccc} 1 & \cdots & 1 \\ \lambda_{1} & \cdots & \lambda_{k} \\ \vdots & & \vdots \\ \lambda_{1}^{k-1} & \cdots & \lambda_{k}^{k-1} \end{array}\right) .\]
Show that \[ \det A = \prod_{i < j} (\lambda_j - \lambda_i) .\]
Solution
Regard \(D(\lambda_1,\ldots,\lambda_k)=\det A\) as a polynomial in the \(\lambda_i\).
<1>1. For every \(i<j\), the polynomial \(\lambda_j-\lambda_i\) divides \(D\).
Interchanging \(\lambda_i\) and \(\lambda_j\) interchanges columns \(i\) and \(j\) of \(A\), so it changes the sign of \(D\). In particular, after setting \(\lambda_i=\lambda_j\), the two columns coincide and \(D=0\). Viewing \(D\) as a polynomial in \(\lambda_j\) over the coefficient ring generated by the other variables, the factor theorem gives \[ \lambda_j-\lambda_i\mid D. \]
<1>2. Hence \[ \prod_{i<j}(\lambda_j-\lambda_i)\mid D. \]
The factors \(\lambda_j-\lambda_i\) are pairwise nonassociate irreducible polynomials in the UFD generated by the \(\lambda_i\), hence are pairwise relatively prime. By <1>1 each divides \(D\), so their product divides \(D\).
<1>3. Both \(D\) and the product in <1>2 have total degree \(k(k-1)/2\).
Expanding the determinant, \[ D=\sum_{\sigma\in S_k}\operatorname{sgn}(\sigma) \prod_{r=1}^k \lambda_{\sigma(r)}^{r-1}. \] Every monomial in this sum has total degree \[ 0+1+\cdots+(k-1)=\frac{k(k-1)}2. \] The product \(\prod_{i<j}(\lambda_j-\lambda_i)\) has one linear factor for each of the \(\binom{k}{2}\) pairs, so it has the same total degree.
<1>4. The quotient in <1>2 is the constant \(1\).
By <1>2 and <1>3, there is a constant \(c\) such that \[ D=c\prod_{i<j}(\lambda_j-\lambda_i). \] Consider the monomial \[ \lambda_2\lambda_3^2\cdots\lambda_k^{k-1}. \] In the determinant expansion of <1>3, this monomial occurs only for the identity permutation and has coefficient \(1\). In the product, it is obtained uniquely by choosing the term \(\lambda_j\) from every factor with second index \(j\), and again has coefficient \(1\). Hence \(c=1\).
<1>5. Therefore \[ \det A=\prod_{i<j}(\lambda_j-\lambda_i). \]
This is <1>4 with \(c=1\).
Nonzero nilpotent matrices are not diagonalizable (Tag E-NUJ7W)
Show that a nonzero nilpotent matrix \(A\) is not diagonalizable over any field. Some useful facts:
\(\spec A = \ts{0}\): if \(Av=\lambda v\) and \(A^N=0\), then \(0=A^Nv=\lambda^N v\), so \(\lambda=0\). Thus every Jordan block is nilpotent.
If \(r\) is the nilpotency index of \(A\), then \(\min_A(x)=x^r\).
If \(A\) were diagonalizable, its diagonal form would be \(0\).
Solution
<1>1. The only eigenvalue of a nilpotent matrix is \(0\).
Choose \(N\ge1\) with \(A^N=0\). If \(Av=\lambda v\) for some nonzero vector \(v\), then \[ 0=A^Nv=\lambda^N v. \] Since \(v\neq0\) and the scalars form a field, \(\lambda^N=0\), hence \(\lambda=0\).
<1>2. If a nilpotent matrix \(A\) is diagonalizable, then \(A=0\).
If \(A\) is diagonalizable, there is an invertible matrix \(P\) and a diagonal matrix \(D\) such that \[ A=PDP^{-1}. \] The diagonal entries of \(D\) are the eigenvalues of \(A\). By <1>1 every eigenvalue is \(0\), so \(D=0\). Therefore \[ A=P0P^{-1}=0. \]
<1>3. Hence every nonzero nilpotent matrix is not diagonalizable.
If a nonzero nilpotent matrix were diagonalizable, <1>2 would force it to equal \(0\), a contradiction.
<1>4. Equivalently, if \(r\) is the nilpotency index of a nonzero nilpotent matrix, then its minimal polynomial is \(x^r\) with \(r\ge2\), so it cannot have the distinct linear factors required for diagonalizability.
By definition of the nilpotency index, \(A^r=0\) and \(A^{r-1}\neq0\), so the monic polynomial of least degree annihilating \(A\) is \(x^r\). Since \(A\neq0\), one has \(r\ge2\), and \(x^r\) has a repeated root. A matrix is diagonalizable over a field containing its eigenvalues exactly when its minimal polynomial splits into distinct linear factors; here the only factor is \(x\), repeated \(r\) times.
Cayley-Hamilton via invariant flags of an upper-triangular basis (Tag E-GNYRR)
Prove Cayley-Hamilton in the following way. Let \(V=\spanof\ts{\vector v_1,\ldots,\vector v_n}\) and define \[ \Fil_iV=\spanof\ts{\vector v_1,\ldots,\vector v_i}, \qquad \Fil_0V=\ts{0}. \] Suppose the matrix of \(A\) in this basis is upper triangular, with diagonal entries \(\lambda_1,\ldots,\lambda_n\). Show that \[ A(\Fil_iV)\subseteq\Fil_iV, \qquad (A-\lambda_iI)\Fil_iV\subseteq\Fil_{i-1}V. \] Deduce that \[ (A-\lambda_1I)\cdots(A-\lambda_nI)V=0 \] and hence \(\chi_A(A)=0\). Finally, deduce Cayley-Hamilton for an arbitrary operator over an arbitrary field by extending scalars to an algebraic closure and triangularizing there.
Solution
<1>1. For every \(i\), the flag subspace \(\Fil_iV\) is \(A\)-invariant.
Because the matrix of \(A\) in the ordered basis \(v_1,\ldots,v_n\) is upper triangular, for each \(j\) one has \[ Av_j\in\spanof\ts{v_1,\ldots,v_j}=\Fil_jV. \] Thus if \(j\le i\), then \(Av_j\in\Fil_iV\). Since \(\Fil_iV\) is spanned by \(v_1,\ldots,v_i\), it follows that \[ A(\Fil_iV)\subseteq\Fil_iV. \]
<1>2. For every \(i\), one has \[ (A-\lambda_iI)\Fil_iV\subseteq\Fil_{i-1}V. \]
For \(j<i\), both \(Av_j\) and \(\lambda_i v_j\) lie in \(\Fil_{i-1}V\), hence so does \((A-\lambda_iI)v_j\). For \(j=i\), upper triangularity gives \[ Av_i=\lambda_i v_i+w \] for some \(w\in\Fil_{i-1}V\). Therefore \[ (A-\lambda_iI)v_i=w\in\Fil_{i-1}V. \] Since \(v_1,\ldots,v_i\) span \(\Fil_iV\), the inclusion follows.
<1>3. The product of the linear factors annihilates \(V\): \[ (A-\lambda_1I)\cdots(A-\lambda_nI)V=0. \]
By <1>2, \[ (A-\lambda_nI)\Fil_nV\subseteq\Fil_{n-1}V. \] Applying successively the factors with indices \(n-1,n-2,\ldots,1\) gives \[ (A-\lambda_1I)\cdots(A-\lambda_nI)\Fil_nV \subseteq\Fil_0V=0. \] Since \(\Fil_nV=V\), the displayed product is the zero operator.
<1>4. For an upper-triangular matrix, \[ \chi_A(t)=\prod_{i=1}^n(t-\lambda_i), \] so \(\chi_A(A)=0\).
The determinant of \(tI-A\) is the product of its diagonal entries, namely \[ \det(tI-A)=\prod_{i=1}^n(t-\lambda_i). \] Evaluating this polynomial at \(A\) yields exactly the product in <1>3. Hence \(\chi_A(A)=0\).
<1>5. Cayley-Hamilton holds for every endomorphism of a finite-dimensional vector space over an arbitrary field.
Let \(A\in\Endo_k(V)\) and let \(\overline{k}\) be an algebraic closure of \(k\). Extend scalars: \[ \overline{V}=V\otimes_k\overline{k}, \qquad \overline{A}=A\otimes1. \] Over \(\overline{k}\), the characteristic polynomial of \(A\) splits, so \(\overline{A}\) admits an upper-triangular matrix in some basis. By <1>4, \[ \chi_{\overline{A}}(\overline{A})=0. \] Scalar extension does not change the characteristic polynomial coefficients, so \[ \chi_{\overline{A}}(t)=\chi_A(t) \] viewed in \(\overline{k}[t]\). Thus \[ \chi_A(A)\otimes1=0 \] in \(\Endo_{\overline{k}}(\overline{V})\). The natural map \[ \Endo_k(V)\longrightarrow\Endo_{\overline{k}}(V\otimes_k\overline{k}), \qquad T\longmapsto T\otimes1, \] is injective because \(k\to\overline{k}\) is a field extension. Therefore \(\chi_A(A)=0\) already over \(k\).