Linear algebra · Qual Corpus

A vector space is a module over a field: every vector space has a basis, every short exact sequence of vector spaces splits, and $\dim\ker T+\dim\im T=\dim V$ for a linear map $T$ on a finite-dimensional space $V$. A bilinear form $b$ on $V$ with ordered basis $e=(e_1,\ldots,e_n)$ has Gram matrix $G_e=(b(e_i,e_j))_{i,j}$; for $e'=eP$ with $P\in\GL_n(F)$, $G_{e'}=P^tG_eP$, so $b$ is nondegenerate if and only if $\det G_e\ne0$, a condition independent of $e$. In characteristic different from $2$, $q(v)=b(v,v)$ and $b(v,w)=\frac12\big(q(v+w)-q(v)-q(w)\big)$ identify symmetric bilinear forms with quadratic forms. A real symmetric matrix is orthogonally diagonalizable (the spectral theorem), and by Sylvester’s law of inertia a real symmetric bilinear form is determined up to change of basis by its rank and signature. The groups $\GL_n(F)$, $\operatorname{O}(q)$, and $\operatorname{Sp}(b)$ are the automorphism groups of $V$, of a quadratic form $q$, and of an alternating form $b$; counting ordered bases one vector at a time gives $\abs{\GL_n(\FF_q)}=\prod_{i=0}^{n-1}(q^n-q^i)$. On a Hilbert space, a bounded operator $T$ factors as $T=U\abs{T}$ with $\abs{T}=(T^*T)^{1/2}$ and $U$ a partial isometry (the polar decomposition), and a compact self-adjoint operator has an orthonormal basis of eigenvectors with real eigenvalues that accumulate only at $0$ (the spectral theorem). For $\varphi\in L^\infty(S^1)$, the Toeplitz operator $T_\varphi$ on the Hardy space $H^2=\{f\in L^2(S^1):\hat f(n)=0\text{ for }n<0\}$ is $f\mapsto P(\varphi f)$, with $P$ the orthogonal projection of $L^2(S^1)$ onto $H^2$; its matrix in the basis $(z^n)_{n\ge0}$ is $(\hat\varphi(j-k))_{j,k\ge0}$.

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 $AB=BA=I_n$.

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.

Special linear group $\SL_n(\CC)$ (Tag D-P5D3T)

Let $n\geq 1$. The special linear group is \[ \SL_n(\CC) \coloneqq \theset{ A\in\Mat_{n\times n}(\CC) \st \det A = 1 }. \]

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 $n(n-1)/2$.

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 \[ \U_n(\CC) \coloneqq \theset{ A\in\Mat_{n\times n}(\CC) \st A^\dagger A = AA^\dagger = I_n }. \]

Unitary group $\U_n(\CC)$ (Tag D-QZ2LQ)

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.

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 $x^t\begin{bmatrix} 0 & I_n \\ I_n & 0 \end{bmatrix}y$.

Symplectic group $\operatorname{Sp}_{2n}(\CC)$ (Tag D-3V3SP)

Order of $\GL_n(\FF_q)$ (Tag PR-UQ3XJ)

Count invertible matrices by choosing an ordered basis one independent column at a time.Suggest a slogan

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.

The trace pairing identifies $\Hom(V,W)$ with $\dualof{\Hom(W,V)}$ (Tag PR-GXII2)

Trace of composition gives a perfect pairing between the two opposite Hom spaces.Suggest a slogan

Let $k$ be a field and $V, W$ finite-dimensional $k$-vector spaces. The map \[ \begin{aligned} \Hom_k(V, W) &\to \dualof{\Hom_k(W, V)} \\ T &\mapsto \big(S \mapsto \Tr(T \circ S)\big) \end{aligned} \] is an isomorphism of $k$-vector spaces, where $T \circ S\colon W \to W$.

The trace pairing identifies $\Hom(V,W)$ with $\dualof{\Hom(W,V)}$ (Tag PR-TGFTL)

Trace of composition gives a perfect pairing between opposite Hom spaces.Suggest a slogan

Let $k$ be a field and let $V$ and $W$ be finite-dimensional $k$-vector spaces. The map \[ \begin{aligned} \Hom_k(V, W) &\to \dualof{\Hom_k(W, V)},\\ T &\mapsto \big(S\mapsto\Tr(T \circ S)\big), \end{aligned} \] is an isomorphism of $k$-vector spaces, where $T\circ S\colon W\to W$.

The map is $k$-linear because the trace is, and both sides have dimension $\dim_k V\cdot\dim_k W$. Choose bases, so that $T$ is an $m\times n$ matrix and $S$ an $n\times m$ matrix; then $\Tr(TS)=\sum_{a,b}T_{ab}S_{ba}$. If $\Tr(TS)=0$ for all $S$, taking $S$ to be the matrix unit $E_{ba}$ gives $T_{ab}=0$ for all $a,b$, so $T=0$. Thus the map is injective, hence an isomorphism.

Nondegenerate bilinear form (Tag D-O4WWN)

Let $k$ be a field, $V$ a $k$-vector space, and $b\colon V\times V \to k$ a bilinear form. The form $b$ is nondegenerate if its adjoint map \[ \begin{aligned} V &\to V^{\vee} \\ x &\mapsto b(x, \wait) \end{aligned} \] is injective.

The kernel of the adjoint map is $\theset{ x \in V \st b(x,y) = 0 \text{ for all } y \in V }$, so $b$ is nondegenerate if and only if this subspace is zero. If $\dim_k V = n< \infty$, then $\dim_k V^{\vee}=n$, so the adjoint map is injective if and only if it is an isomorphism; with respect to a basis $e_1,\ldots,e_n$ of $V$ and its dual basis, the adjoint map has matrix the transpose of the Gram matrix $\bigl(b(e_i,e_j)\bigr)_{i,j}$, so $b$ is nondegenerate if and only if a Gram matrix of $b$ is invertible.

See [@Art11].

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.

Gram matrix of a bilinear form (Tag D-H4TDM)

Let $k$ be a field, let $V$ be a finite-dimensional $k$-vector space with ordered basis $\mathcal B = (e_1, \ldots, e_n)$, and let $b\colon V\times V\to k$ be a bilinear form. The Gram matrix of $b$ with respect to $\mathcal B$ is \[ G \coloneqq \qty{ b(e_i, e_j) }_{1\leq i,j\leq n}. \]

Let $k$, $V$, $\mathcal B$, $b$, and $G$ be as in the definition, and write $[x]_{\mathcal B}\in k^n$ for the coordinate vector of $x\in V$.

For all $x, y\in V$, $b(x, y) = [x]_{\mathcal B}^t\, G\, [y]_{\mathcal B}$.
If $\mathcal B'$ is another ordered basis and $P\in\GL_n(k)$ satisfies $[x]_{\mathcal B} = P[x]_{\mathcal B'}$ for all $x\in V$, then the Gram matrix of $b$ with respect to $\mathcal B'$ is $P^t G P$. In particular, $\det G$ is well defined up to multiplication by the square of a nonzero element of $k$.
$b$ is nondegenerate if and only if $G$ is invertible.

See [@Art11].

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].

Orthogonal complement of a sum of subspaces (Tag PR-WDPF7)

Orthogonal complement turns sums into intersections.Suggest a slogan

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 $\inner{w}{v}=\sum_j\inner{w_{i_j}}{v}=0$.

Quadratic form (Tag D-NRRIT)

Let $k$ be a field and $V$ a $k$-vector space. A quadratic form on $V$ is a function $q\colon V\to k$ such that $q(\lambda x) = \lambda^2 q(x)$ for all $\lambda\in k$ and $x\in V$, and such that \[ b_q(x, y) \coloneqq q(x+y) - q(x) - q(y) \] is a bilinear form $V\times V\to k$.

The form $b_q$ is symmetric and satisfies $b_q(x,x)=q(2x)-2q(x)=2q(x)$. If $\characteristic k \neq 2$, then $q\mapsto \frac12 b_q$ is a bijection from quadratic forms on $V$ to symmetric bilinear forms on $V$, with inverse $b\mapsto \bigl(x\mapsto b(x,x)\bigr)$. If $V$ has basis $e_1,\ldots,e_n$, then in the coordinates $x=\sum_i x_ie_i$ the form $q$ is a homogeneous polynomial of degree $2$, $q(x) = \sum_{i \leq j} a_{ij} x_i x_j$ with $a_{ii}=q(e_i)$ and $a_{ij}=b_q(e_i,e_j)$ for $i<j$.

See [@Art11].

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.

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 $A$.

Spectral theorem for Hermitian and real symmetric matrices (Tag T-WQHMA)

Hermitian matrices have unitary real diagonalizations; real symmetric matrices have orthogonal diagonalizations.Suggest a slogan

If $A\in\Mat_n(\CC)$ is Hermitian, that is, $A^{\dagger} = A$, then $A$ is diagonalizable over $\CC$; more precisely, there is a unitary matrix $U$ such that $U^{\dagger}AU$ is diagonal with real entries.
If $A\in\Mat_n(\RR)$ is symmetric, that is, $A^t = A$, then $A$ is diagonalizable over $\RR$; more precisely, there is an orthogonal matrix $Q$ such that $Q^tAQ$ is diagonal.

Diagonalizable matrix (Tag FD-K6FVX)

Let $R$ be a commutative ring and $A$ an $n\times n$ matrix over $R$. $A$ is diagonalizable if it is similar to a diagonal matrix: there exist a diagonal $n\times n$ matrix $D$ over $R$ and $P\in \GL_n(R)$ such that $A = PDP^{-1}$.

What does it mean for a matrix to be diagonalizable?

Diagonalizability via the minimal polynomial (Tag T-6ABNR)

A matrix is diagonalizable exactly when its minimal polynomial splits with no repeated root.Suggest a slogan

Let $F$ be a field and $M\in M_n(F)$ with minimal polynomial $m_M\in F[x]$. Then $M$ is diagonalizable over $F$ if and only if $m_M$ splits over $F$ into distinct linear factors.

If $M=PDP^{-1}$ with $D$ diagonal whose distinct diagonal entries are $\lambda_1,\ldots,\lambda_r$, then $\prod_i(x-\lambda_i)$ annihilates $D$ and hence $M$, so $m_M$ divides a product of distinct linear factors.

Conversely, suppose $m_M=\prod_{i=1}^r(x-\lambda_i)$ with the $\lambda_i\in F$ distinct. The factors are pairwise coprime, so by the kernel decomposition $F^n=\ker m_M(M)=\bigoplus_{i=1}^r\ker(M-\lambda_i I)$. Choosing a basis of each eigenspace $\ker(M-\lambda_i I)$ gives a basis of $F^n$ of eigenvectors of $M$.

A matrix whose minimal polynomial splits over $F$ need not be diagonalizable over $F$. Over any field $F$, $M=\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}$ has $m_M = (x-1)^2$, which splits over $F$, but $\ker(M-I)$ is one-dimensional, so $M$ is not diagonalizable.

Characterizations of diagonalizability (Tag FT-BC6S2)

Diagonalizable means an eigenbasis, equivalently a squarefree split minimal polynomial.Suggest a slogan

Let $F$ be a field and let $M\in\Mat_n(F)$. The following are equivalent:

$M$ is diagonalizable over $F$.
The minimal polynomial $\min_M(x)$ is a product of distinct monic linear factors in $F[x]$.
There is a basis of $F^n$ consisting of eigenvectors of $M$.
Every elementary divisor of $M$ has degree $1$.

In particular, if $M$ has $n$ distinct eigenvalues in $F$, equivalently if $\min_M(x)$ has $n$ distinct roots in $F$, then $M$ is diagonalizable over $F$.

A matrix whose minimal polynomial splits over $F$ need not be diagonalizable over $F$. For $M=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ over any field $F$, $\min_M(x)=(x-1)^2$ splits over $F$, and $M$ is not diagonalizable, because its only eigenvalue is $1$ and $M\ne I$.

What conditions each characterise diagonalizability of a square matrix?

Simultaneous diagonalizability of commuting diagonalizable operators (Tag PR-GV5CF)

Commuting diagonalizable operators share an eigenbasis.Suggest a slogan

Let $V$ be a finite-dimensional vector space over a field $k$, and let $\theset{A_i}_{i \in I}$ be a family of diagonalizable linear operators on $V$. The $A_i$ pairwise commute if and only if they are simultaneously diagonalizable: there is a basis of $V$ in which every $A_i$ is diagonal.

The implication from simultaneous diagonalizability to commuting holds for any family of operators, since diagonal matrices commute.

Commuting operators need not be simultaneously diagonalizable without the diagonalizability hypothesis: $\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}$ commutes with itself and is not diagonalizable.

A linear operator over an algebraically closed field has an eigenvector in every nonzero invariant subspace (Tag L-C5DDK)

Every nonzero invariant subspace over an algebraically closed field contains an eigenvector.Suggest a slogan

Let $k$ be an algebraically closed field, let $V$ be a finite-dimensional $k$-vector space, and let $A\colon V\to V$ be $k$-linear. If $W\subseteq V$ is a nonzero subspace with $A(W)\subseteq W$, then $W$ contains an eigenvector of $A$.

The restriction $A|_W\colon W\to W$ is a linear operator on a nonzero finite-dimensional space, so its characteristic polynomial has positive degree and has a root $\lambda\in k$. Then $A|_W-\lambda\id_W$ is not injective, and a nonzero $w\in W$ with $Aw=\lambda w$ is an eigenvector of $A$.

A linear operator on an infinite-dimensional vector space over an algebraically closed field need not have an eigenvector: multiplication by $x$ on $V=W=k[x]$ has no eigenvector, since $xf=\lambda f$ has no nonzero solution $f\in k[x]$ by comparing degrees.

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 $A = PDP^{-1}$.

Jordan--Chevalley decomposition (Tag PR-24CPI)

Over a perfect field, every linear map splits uniquely into commuting semisimple and nilpotent parts.Suggest a slogan

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.

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.

$A^n = B^n$ does not imply $A = B$ for matrices (Tag FF-VKYM3)

Matrix powers are not injective: $A^2=(-A)^2$ can hold with $A\ne -A$.Suggest a slogan

For square matrices $A,B$ and $n\ge2$, $A^n=B^n$ does not imply $A=B$. In $\Mat_2(\RR)$, let \[ A=\begin{bmatrix}0&1\\-1&0\end{bmatrix}. \] Then \[ A^2=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}=(-A)^2, \] but $A\ne-A$.

For matrices, does $A^n=B^n\implies A=B$?

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