Matrix Representation & Linear Mappings - Part 3

Published 8/26/2025

Change of Bases

Problem 1.1

\text{Let }A=\begin{bmatrix}2&-1&3\\ 1&0&4\end{bmatrix}. \text{ Recall that }A \text{ determines }F:\mathbb{R}^3\to\mathbb{R}^2 \text{ by }F(v)=Av \text{ (vectors are columns).}\\[8pt] \text{Find the matrix }[F]\text{ relative to the bases below.}\\[6pt] \text{(a) The usual bases of }\mathbb{R}^3\text{ and }\mathbb{R}^2.\\ \text{(b) }S=\{w_1,w_2,w_3\}=\{(1,1,0),(1,0,1),(0,1,1)\},\quad S'=\{u_1,u_2\}=\{(1,1),(1,2)\}.\ \text{ Compute }[F]_{S'\leftarrow S}\ \text{without forming change-of-basis matrices (solve by linear combinations).}\\ \text{(c) Confirm (b) using }Q=[I]_{\mathbf e\leftarrow S},\ P=[I]_{S'\leftarrow \mathbf e}\text{ and verify }[F]_{S'\leftarrow S}=P\,A\,Q.\\[10pt] \textbf{Solution:}\\ \text{(a) }[F]_{\mathbf e}=A=\begin{bmatrix}2&-1&3\\ 1&0&4\end{bmatrix}.\\[6pt] \text{(b) }F(w_1)=A\!\begin{bmatrix}1\\[1pt]1\\[1pt]0\end{bmatrix}=\begin{bmatrix}1\\[1pt]1\end{bmatrix},\quad F(w_2)=A\!\begin{bmatrix}1\\[1pt]0\\[1pt]1\end{bmatrix}=\begin{bmatrix}5\\[1pt]5\end{bmatrix},\quad F(w_3)=A\!\begin{bmatrix}0\\[1pt]1\\[1pt]1\end{bmatrix}=\begin{bmatrix}2\\[1pt]4\end{bmatrix}.\\ \text{Write }F(w_j)=a_j u_1+b_j u_2\ (u_1=(1,1),\ u_2=(1,2)).\ \text{Solve } a_j+b_j=(F(w_j))_1,\ a_j+2b_j=(F(w_j))_2.\\ \Rightarrow\ b_j=(F(w_j))_2-(F(w_j))_1,\quad a_j=2(F(w_j))_1-(F(w_j))_2.\\ \text{Hence }(a_1,b_1)=(1,0),\ (a_2,b_2)=(5,0),\ (a_3,b_3)=(0,2).\\ \Rightarrow\ [F]_{S'\leftarrow S}=\begin{bmatrix}1&5&0\\ 0&0&2\end{bmatrix}.\\[8pt] \text{(c) }Q=[I]_{\mathbf e\leftarrow S}=\begin{bmatrix}1&1&0\\ 1&0&1\\ 0&1&1\end{bmatrix},\quad [I]_{\mathbf e\leftarrow S'}=\begin{bmatrix}1&1\\ 1&2\end{bmatrix},\ \ P=[I]_{S'\leftarrow \mathbf e}=\begin{bmatrix}2&-1\\ -1&1\end{bmatrix}.\\ P\,A\,Q=\begin{bmatrix}2&-1\\ -1&1\end{bmatrix}\begin{bmatrix}2&-1&3\\ 1&0&4\end{bmatrix}\begin{bmatrix}1&1&0\\ 1&0&1\\ 0&1&1\end{bmatrix} =\begin{bmatrix}1&5&0\\ 0&0&2\end{bmatrix}=[F]_{S'\leftarrow S}.

Problem 1.2

\text{Let }A=\begin{bmatrix}3&1&0\\ -1&2&1\end{bmatrix}. \text{ Recall that }A\text{ determines }F:\mathbb{R}^3\to\mathbb{R}^2\text{ by }F(v)=Av.\\[6pt] \text{Find the matrix }[F]\text{ relative to the bases below:}\\ \text{(a) The usual bases of }\mathbb{R}^3\text{ and }\mathbb{R}^2.\\ \text{(b) }S=\{w_1,w_2,w_3\}=\{(1,0,1),(1,1,0),(0,1,1)\},\quad S'=\{u_1,u_2\}=\{(1,0),(1,1)\}.\ \text{ Compute }[F]_{S'\leftarrow S}\ \textbf{without} change-of-basis matrices (solve by linear combinations).\\ \text{(c) Confirm (b) using }Q=[I]_{\mathbf e\leftarrow S},\ P=[I]_{S'\leftarrow \mathbf e}\text{ and verify }[F]_{S'\leftarrow S}=P\,A\,Q.\\[8pt] \textbf{Solution:}\\ \text{(a) }[F]_{\mathbf e}=A=\begin{bmatrix}3&1&0\\ -1&2&1\end{bmatrix}.\\[6pt] \text{(b) }F(w_1)=\begin{bmatrix}3\\ 0\end{bmatrix},\ F(w_2)=\begin{bmatrix}4\\ 1\end{bmatrix},\ F(w_3)=\begin{bmatrix}1\\ 3\end{bmatrix}.\\ \text{Write }F(w_j)=a_j u_1+b_j u_2\ \ (u_1=(1,0),u_2=(1,1)).\ \text{Then }(a_j+b_j,\ b_j)=F(w_j)\Rightarrow b_j=(F(w_j))_2,\ a_j=(F(w_j))_1-(F(w_j))_2.\\ \Rightarrow (a_1,b_1)=(3,0),\ (a_2,b_2)=(3,1),\ (a_3,b_3)=(-2,3).\\ \therefore\ [F]_{S'\leftarrow S}=\begin{bmatrix}3&3&-2\\ 0&1&3\end{bmatrix}.\\[8pt] \text{(c) }Q=[I]_{\mathbf e\leftarrow S}=\begin{bmatrix}1&1&0\\ 0&1&1\\ 1&0&1\end{bmatrix},\quad [I]_{\mathbf e\leftarrow S'}=\begin{bmatrix}1&1\\ 0&1\end{bmatrix}\Rightarrow P=[I]_{S'\leftarrow \mathbf e}=\begin{bmatrix}1&-1\\ 0&1\end{bmatrix}.\\ P\,A\,Q=\begin{bmatrix}1&-1\\ 0&1\end{bmatrix} \begin{bmatrix}3&1&0\\ -1&2&1\end{bmatrix} \begin{bmatrix}1&1&0\\ 0&1&1\\ 1&0&1\end{bmatrix} =\begin{bmatrix}3&3&-2\\ 0&1&3\end{bmatrix}=[F]_{S'\leftarrow S}.

Problem 1.3

\text{Let }A=\begin{bmatrix}1&2\\ 0&1\\ 1&0\end{bmatrix}. \text{ Recall that }A\text{ determines }F:\mathbb{R}^2\to\mathbb{R}^3\text{ by }F(v)=Av.\\[6pt] \text{Find the matrix }[F]\text{ relative to the bases below:}\\ \text{(a) The usual bases of }\mathbb{R}^2\text{ and }\mathbb{R}^3.\\ \text{(b) }S=\{b_1,b_2\}=\{(1,1),(1,-1)\},\quad S'=\{u_1,u_2,u_3\}=\{(1,1,0),(0,1,1),(1,0,1)\}.\ \text{ Compute }[F]_{S'\leftarrow S}\ \textbf{without} change-of-basis matrices.\\ \text{(c) Confirm (b) using }Q=[I]_{\mathbf e\leftarrow S},\ P=[I]_{S'\leftarrow \mathbf e}\text{ and verify }[F]_{S'\leftarrow S}=P\,A\,Q.\\[8pt] \textbf{Solution:}\\ \text{(a) }[F]_{\mathbf e}=A=\begin{bmatrix}1&2\\ 0&1\\ 1&0\end{bmatrix}.\\[6pt] \text{(b) }F(b_1)=A\!\begin{bmatrix}1\\[1pt]1\end{bmatrix}=\begin{bmatrix}3\\ 1\\ 1\end{bmatrix},\quad F(b_2)=A\!\begin{bmatrix}1\\[1pt]-1\end{bmatrix}=\begin{bmatrix}-1\\ -1\\ 1\end{bmatrix}.\\ \text{Write }F(b_j)=a_j u_1+b_j u_2+c_j u_3.\ \text{Since }u_1=(1,1,0),u_2=(0,1,1),u_3=(1,0,1),\\ (a_j+c_j,\ a_j+b_j,\ b_j+c_j)=F(b_j).\ \text{Solve } a_j=\tfrac{(F(b_j))_1+(F(b_j))_2-(F(b_j))_3}{2},\ b_j=\tfrac{(F(b_j))_2+(F(b_j))_3-(F(b_j))_1}{2},\ c_j=\tfrac{(F(b_j))_1+(F(b_j))_3-(F(b_j))_2}{2}.\\ \Rightarrow (a_1,b_1,c_1)=\bigl(\tfrac{3}{2},-\tfrac{1}{2},\tfrac{3}{2}\bigr),\quad (a_2,b_2,c_2)=\bigl(-\tfrac{3}{2},\tfrac{1}{2},\tfrac{1}{2}\bigr).\\ \therefore\ [F]_{S'\leftarrow S}= \begin{bmatrix} \tfrac{3}{2}&-\tfrac{3}{2}\\[2pt] -\tfrac{1}{2}&\ \tfrac{1}{2}\\[2pt] \ \tfrac{3}{2}&\ \tfrac{1}{2} \end{bmatrix}.\\[8pt] \text{(c) }Q=[I]_{\mathbf e\leftarrow S}=\begin{bmatrix}1&1\\ 1&-1\end{bmatrix},\quad [I]_{\mathbf e\leftarrow S'}=\begin{bmatrix}1&0&1\\ 1&1&0\\ 0&1&1\end{bmatrix}\Rightarrow P=[I]_{S'\leftarrow \mathbf e}= \begin{bmatrix} \tfrac12&\tfrac12&-\tfrac12\\[2pt] -\tfrac12&\tfrac12&\ \tfrac12\\[2pt] \tfrac12&-\tfrac12&\ \tfrac12 \end{bmatrix}.\\ P\,A\,Q= \begin{bmatrix} \tfrac12&\tfrac12&-\tfrac12\\ -\tfrac12&\tfrac12&\tfrac12\\ \tfrac12&-\tfrac12&\tfrac12 \end{bmatrix} \begin{bmatrix}1&2\\ 0&1\\ 1&0\end{bmatrix} \begin{bmatrix}1&1\\ 1&-1\end{bmatrix} = \begin{bmatrix} \tfrac{3}{2}&-\tfrac{3}{2}\\[2pt] -\tfrac{1}{2}&\ \tfrac{1}{2}\\[2pt] \ \tfrac{3}{2}&\ \tfrac{1}{2} \end{bmatrix} =[F]_{S'\leftarrow S}.

Smith Normal Form

Problem 2.1

\text{Let }A=\begin{bmatrix}1&0&1\\[2pt] 0&1&1\end{bmatrix}\ \text{be the (standard-basis) }2\times3\text{ matrix of }T:\mathbb{R}^3\to\mathbb{R}^2.\\[8pt] \text{(a) Construct a basis }\mathcal B\text{ of }\mathbb{R}^3\text{ whose last vector spans }\ker A,\ \text{and a basis }\mathcal C\text{ of }\mathbb{R}^2\text{ whose first two vectors form a basis of }\operatorname{im}A.\\ \text{Let }Q=[I]_{\mathcal B}\in GL_3(\mathbb{R}),\ P=[I]_{\mathcal C}^{-1}\in GL_2(\mathbb{R}).\ \text{Show that }PAQ=\begin{bmatrix}1&0&0\\[2pt] 0&1&0\end{bmatrix}.\\[6pt] \text{(b) Compute }\operatorname{rank}(A),\ \dim\ker A,\ \dim\ker(A^\top).\\[10pt] \textbf{Answer: } \ker A=\operatorname{span}\{(-1,-1,1)^\top\}.\ \text{Choose }\mathcal B=\{e_1,\ e_2,\ \underbrace{(-1,-1,1)^\top}_{b_3}\},\quad \mathcal C=\{Ae_1,\ Ae_2\}=\{(1,0)^\top,\ (0,1)^\top\}.\\[6pt] Q=[I]_{\mathcal B}=\begin{bmatrix} 1&0&-1\\ 0&1&-1\\ 0&0&\ \ 1 \end{bmatrix},\qquad A Q=\begin{bmatrix} 1&0&0\\ 0&1&0 \end{bmatrix}.\\[6pt] [I]_{\mathcal C}=\begin{bmatrix} 1&0\\ 0&1 \end{bmatrix},\quad P=[I]_{\mathcal C}^{-1}=I_2.\\[4pt] \Rightarrow\ PAQ=\begin{bmatrix}1&0&0\\[2pt] 0&1&0\end{bmatrix}.\\[8pt] \operatorname{rank}(A)=2,\quad \dim\ker A=3-2=1,\quad \dim\ker(A^\top)=2-2=0.

Matrix Representation & Linear Mappings - Part 3

Change of Bases

Problem 1.1

Problem 1.2

Problem 1.3

Smith Normal Form

Problem 2.1

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