# Some interesting linear algebra proofs

Below are some cute linear algebra results and proofs cherrypicked from various sources. All the standard hypotheses (on the base field, the size of the matrices, etc.) that make the claims valid are assumed. The list will likely be updated.

Fact 1. Let $A,B,X$ be matrices. If $AX=XB$, then $p(A)X=Xp(B)$ for any polynomial $p$.

Proof. We have $A^2X=A(AX)=A(XB)=(AX)B=(XB)B=XB^2$. By induction, $A^nX=XB^n$ for any $n\in\mathbb N$. Hence the result follows. $\square$

Remark. Note that $X$ need not be square, let alone invertible.

Fact 2. Let $x_0,\dots,x_n$ be distinct. Then the Vandermonde matrix

$\displaystyle V=\begin{pmatrix} 1 & x_0 & \cdots & x_0^n\\ 1 & x_1 & \cdots & x_1^n\\ \vdots & \vdots & \ddots & \vdots\\ 1 & x_n & \cdots & x_n^n\end{pmatrix}$

is invertible.

Proof. It suffices to show that the kernel of the linear transformation $f(x)=Vx$ is trivial. If $a=(a_0,\dots,a_n)$ is in the kernel, then $p(X)=a_0+a_1X+\cdots+a_nX^n=0$ for each $X\in\{x_0,\dots,x_n\}$. Since $\deg(p)=n$, this forces $p(X)$ to be identically zero. Thus $a=0$. $\square$

Fact 3. A matrix is diagonalisable iff its minimal polynomial decomposes into distinct linear factors.

Proof. A matrix is diagonalisable iff every Jordan block has size $1$. Since the multiplicity of an eigenvalue in the minimal polynomial corresponds to the size of the largest Jordan block, the result follows. $\square$

Corollary. Idempotent matrices are diagonalisable. Moreover, the rank of an idempotent matrix is equal to the algebraic multiplicity of the eigenvalue $1$.

Fact 4. If $A^nx=0$ and $A^{n-1}x\neq 0$, then $x,Ax,\dots,A^{n-1}x$ are linearly independent.

Proof. Note that $\ker(A^n)\supseteq\ker(A^{n-1})$. Let $x\in\ker(A^n)\setminus\ker(A^{n-1})$ and suppose that $a_0x+a_1Ax+\cdots+a_{n-1}A^{n-1}x=0$. Multiplying both sides by $A^i$ for $i=n-1,\dots,1$ shows that $a_i=0$ for all $i$, as desired. $\square$

Corollary 1. If $\ker(A^n)\neq\ker(A^{n-1})$, then $\dim(\ker(A^j))\ge j$ for $j=1,\dots,n$.

Proof. If $x\in\ker(A^n)\setminus\ker(A^{n-1})$, then $A^{n-1}x,\dots,A^{n-j}x\in\ker(A^j)$. $\square$

Corollary 2. If $A$ is $n\times n$, and $\ker(A^n)\neq\ker(A^{n-1})$, then $\dim(\ker(A^j))=j$ for each $0\le j\le n$. In particular, $A$ is similar to the nilpotent Jordan block of size $n$.

Fact 5. If $f$ is linear on $V$, then $V\cong\ker(f)\oplus f(V)$.

Proof. The short exact sequence

$0\to\ker(f)\hookrightarrow V\twoheadrightarrow f(V)\to 0$

is split. So the result follows by the splitting lemma. $\square$

Corollary. If $W\subseteq V$ is a subspace, then $V\cong W\oplus W^\perp$.

Fact 6. If $A$ is $n\times n$, then $r(A)\ge n-k$, where $k$ is the algebraic multiplicity of the eigenvalue $0$ of $A$.

Proof. Since the nullity of $A$ is the geometric multiplicity of the eigenvalue $0$, and the geometric multiplicity of an eigenvalue is at most its algebraic multiplicity, we get $r(A)=n-n(A)\ge n-k$. $\square$

Fact 7. The number of distinct eigenvalues of $A$ is at most $r(A)+1$.

Proof. The rank of a matrix is the number of non-zero eigenvalues and the nullity is the number of zero eigenvalues, both counted with multiplicity. $\square$