Smith Normal Form. Matrix Equivalence. Part 1

Published 8/23/2025

\text{Let }A=\begin{bmatrix}1&0&1\\[2pt]0&1&1\\[2pt]0&0&0\end{bmatrix}\in M_{3\times 3}(\mathbb{R}).\\[8pt] \text{(a) Find invertible }P\in GL_3(\mathbb{R}),\,Q\in GL_3(\mathbb{R})\text{ such that }PAQ=\operatorname{diag}(1,1,0).\\ \text{(b) Give bases of }\ker A\text{ and }\operatorname{im}A.\\[8pt] \textbf{Answer: }Q=\begin{bmatrix}1&0&-1\\[2pt]0&1&-1\\[2pt]0&0&1\end{bmatrix},\quad P=I_3,\quad AQ=\begin{bmatrix}1&0&0\\[2pt]0&1&0\\[2pt]0&0&0\end{bmatrix}.\\ \ker A=\operatorname{span}\{(1,1,-1)\},\qquad \operatorname{im}A=\operatorname{span}\{e_1,e_2\}.\\[10pt] \textbf{Explanation: }c_3=c_1+c_2\ \Rightarrow\ c_3\leftarrow c_3-c_1-c_2. \text{ The right multiplication by } Q=I-E_{1,3}-E_{2,3}\text{ realizes this, so }AQ=\operatorname{diag}(1,1,0).\\ \text{No row operations are needed, hence }P=I_3.\ \text{ Then }\ker A=\{(x,y,z):x+z=0,\ y+z=0\}=\operatorname{span}\{(1,1,-1)\},\\ \text{and the first two columns of }AQ\text{ show }\operatorname{im}A=\operatorname{span}\{e_1,e_2\}.

\text{Let } A\in \mathbb{R}^{m\times n}\ \text{and suppose there exist invertible matrices } P\in \mathrm{GL}_m(\mathbb{R}),\ Q\in \mathrm{GL}_n(\mathbb{R})\ \text{such that } PAQ=D,\\ \text{where } D=\begin{bmatrix} I_r & 0 \\ 0 & 0 \end{bmatrix}\ \text{is the Smith normal form of }A.\ \text{Prove that }\ker A=Q(\ker D).\\[10pt] \text{Proof outline:}\\ \quad (1)\ \text{If }\mathbf{x}\in \ker D,\ \text{then }Q\mathbf{x}\in \ker A.\\ \quad (2)\ \text{If }\mathbf{y}\in \ker A,\ \text{then }\exists\,\mathbf{x}\in \ker D\ \text{with }\mathbf{y}=Q\mathbf{x}.\\ \quad (3)\ \text{Therefore }\ker A=Q(\ker D).

\text{Let }A,D\in \mathbb{R}^{m\times n}\ \text{and suppose there exist invertible matrices }P\in \mathrm{GL}_m(\mathbb{R}),\ Q\in \mathrm{GL}_n(\mathbb{R})\ \text{such that }PAQ=D.\ \text{Prove that }\ker A \cong \ker D.\\[10pt] \textbf{Proof Outline:}\\ \quad (1)\ \text{Define }\varphi:\ker D\to \ker A\ \text{by }\varphi(\mathbf{x})=Q\mathbf{x}.\\ \quad (2)\ \text{Well-defined: if }\mathbf{x}\in \ker D,\ D\mathbf{x}=0\Rightarrow PAQ\mathbf{x}=0\Rightarrow AQ\mathbf{x}=0\Rightarrow Q\mathbf{x}\in \ker A.\\ \quad (3)\ \text{Injective: }Q\mathbf{x}=0\Rightarrow \mathbf{x}=0\ \text{(since }Q\text{ is invertible).}\\ \quad (4)\ \text{Surjective: for }\mathbf{y}\in \ker A,\ A\mathbf{y}=0\Rightarrow PA\mathbf{y}=0\Rightarrow DQ^{-1}\mathbf{y}=0\Rightarrow Q^{-1}\mathbf{y}\in \ker D,\ \mathbf{y}=Q(Q^{-1}\mathbf{y}).\\ \quad (5)\ \text{Therefore }\varphi\text{ is an isomorphism, hence }\ker A\cong \ker D.

\text{Let }A=\begin{bmatrix}1&0&2&1\\[2pt]0&1&3&1\end{bmatrix}\in M_{2\times 4}(\mathbb{R}).\ \text{Find }P\in GL_2,\ Q\in GL_4\text{ with }PAQ=\begin{bmatrix}I_2&0\end{bmatrix}.\\[6pt] \textbf{Answer: }Q=\begin{bmatrix} 1&0&-2&-1\\[2pt] 0&1&-3&-1\\[2pt] 0&0& 1& 0\\[2pt] 0&0& 0& 1 \end{bmatrix},\quad P=I_2,\quad AQ=\begin{bmatrix}1&0&0&0\\[2pt]0&1&0&0\end{bmatrix}.

\text{Let }A,B\in M_{m\times n}(\mathbb{F}).\ \text{Prove that }A\text{ and }B\text{ are equivalent }(\exists P\in GL_m,\ Q\in GL_n:\ B=PAQ)\\ \text{iff }\operatorname{rank}(A)=\operatorname{rank}(B).\\[8pt] \textbf{Answer (outline): }(\Rightarrow)\ \text{Left/right multiplication by invertible matrices preserves rank, so }\operatorname{rank}(B)=\operatorname{rank}(A).\\ (\Leftarrow)\ \text{Reduce each matrix by row/column operations to }\operatorname{diag}(I_r,0)\ \text{with }r=\operatorname{rank}(A)=\operatorname{rank}(B).\\ \text{If }A=P_1^{-1}\operatorname{diag}(I_r,0)Q_1^{-1}\ \text{and }B=P_2^{-1}\operatorname{diag}(I_r,0)Q_2^{-1},\ \text{then }B=(P_2^{-1}P_1)\,A\,(Q_1Q_2^{-1}).

\text{Let }A=\begin{bmatrix}1&2\\[2pt]2&4\\[2pt]3&6\end{bmatrix}\in M_{3\times 2}(\mathbb{R}).\ \text{Find }P\in GL_3,\ Q\in GL_2\text{ such that }PAQ=\begin{bmatrix}1&0\\[2pt]0&0\\[2pt]0&0\end{bmatrix}.\\[6pt] \textbf{Answer: }Q=\begin{bmatrix}1&-2\\[2pt]0&1\end{bmatrix},\ AQ=\begin{bmatrix}1&0\\[2pt]2&0\\[2pt]3&0\end{bmatrix},\ P=\begin{bmatrix}1&0&0\\[2pt]-2&1&0\\[2pt]-3&0&1\end{bmatrix},\ PAQ=\begin{bmatrix}1&0\\[2pt]0&0\\[2pt]0&0\end{bmatrix}.\\ \ker A=\operatorname{span}\{(-2,1)\},\ \operatorname{rank}(A)=1.

\text{Let }A=\begin{bmatrix}0&1&0\\[2pt]1&1&1\\[2pt]0&0&0\end{bmatrix}\in M_{3\times 3}(\mathbb{R}).\quad \text{Find }P\in GL_3,\ Q\in GL_3\text{ such that }PAQ=\operatorname{diag}(1,1,0).\\[6pt] \textbf{Answer: }Q=\begin{bmatrix}0&1&0\\[2pt]1&0&0\\[2pt]0&0&1\end{bmatrix}\quad(\text{swap }C_1\leftrightarrow C_2),\quad AQ=\begin{bmatrix}1&0&0\\[2pt]1&1&1\\[2pt]0&0&0\end{bmatrix},\quad P=\begin{bmatrix}1&0&0\\[2pt]-1&1&0\\[2pt]0&0&1\end{bmatrix},\quad PAQ=\begin{bmatrix}1&0&0\\[2pt]0&1&0\\[2pt]0&0&0\end{bmatrix}.