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Matrix Representation & Linear Operators Part 1 - nerjik.com

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Matrix Representation & Linear Operators Part 1

Matrix Representation & Linear Operators Part 1

Almaz Khusnutdinov

Almaz Khusnutdinov•Math enthusiast, love coding and mathematics

Published 8/15/2025

Problem 1 — Two different bases for domain and codomain (worked).

\text{Let } B=\{(1,1,0),(0,1,1),(2,0,1)\},\quad C=\{(1,0,1),(0,2,1),(1,1,0)\},\quad [F]_{C\leftarrow B}= \begin{bmatrix} 2&-1&0\\ 3&1&2\\ -2&0&1 \end{bmatrix},\quad v=(3,-1,4). \\[18pt] \textit{Solution.}\\ \text{1) Change } v \text{ from standard to } B\text{-coordinates.}\\[10pt] P_B=\begin{bmatrix} 1&0&2\\ 1&1&0\\ 0&1&1 \end{bmatrix},\qquad P_B^{-1}= \begin{bmatrix} \frac{1}{3}&\frac{2}{3}&-\frac{2}{3}\\[4pt] -\frac{1}{3}&\frac{1}{3}&\frac{2}{3}\\[4pt] \frac{1}{3}&-\frac{1}{3}&\frac{1}{3} \end{bmatrix}. \\[10pt] [v]_B = P_B^{-1}\,[v]_{\varepsilon} = \begin{bmatrix} \frac{1}{3}&\frac{2}{3}&-\frac{2}{3}\\ -\frac{1}{3}&\frac{1}{3}&\frac{2}{3}\\ \frac{1}{3}&-\frac{1}{3}&\frac{1}{3} \end{bmatrix} \begin{bmatrix}3\\ -1\\ 4\end{bmatrix} = \begin{bmatrix}-\tfrac{7}{3}\\ \tfrac{4}{3}\\ \tfrac{8}{3}\end{bmatrix}. \\[18pt] \text{2) Apply the given matrix to get }[F(v)]_C.\\[10pt] [F(v)]_C =[F]_{C\leftarrow B}\,[v]_B = \begin{bmatrix} 2&-1&0\\ 3&1&2\\ -2&0&1 \end{bmatrix} \begin{bmatrix} -\tfrac{7}{3}\\[4pt] \tfrac{4}{3}\\[4pt] \tfrac{8}{3} \end{bmatrix} = \begin{bmatrix} -6\\[4pt] -\tfrac{1}{3}\\[4pt] \tfrac{22}{3} \end{bmatrix}. \\[18pt] \text{3) Convert }[F(v)]_C\text{ to standard coordinates.}\\[10pt] P_C=\begin{bmatrix} 1&0&1\\ 0&2&1\\ 1&1&0 \end{bmatrix},\qquad F(v)=P_C\,[F(v)]_C. \\[10pt] F(v) = \begin{bmatrix} 1&0&1\\ 0&2&1\\ 1&1&0 \end{bmatrix} \begin{bmatrix} -6\\[4pt] -\tfrac{1}{3}\\[4pt] \tfrac{22}{3} \end{bmatrix} = \begin{bmatrix} \tfrac{4}{3}\\[4pt] \tfrac{20}{3}\\[4pt] -\tfrac{19}{3} \end{bmatrix}. \\[18pt] \boxed{\,F(v)=\left(\tfrac{4}{3},\,\tfrac{20}{3},\,-\tfrac{19}{3}\right)\,}

Problem 2 — Composition with different domain/codomain bases

\text{Let } B=\{(1,1),(2,0)\},\quad C=\{(1,0),(1,1)\},\quad D=\{(2,1),(0,1)\}\ \text{be bases of }\mathbb{R}^2.\\[12pt] \text{Suppose}\\[6pt] [F]_{C\leftarrow B}=\begin{bmatrix}1&2\\ 0&1\end{bmatrix},\quad [G]_{D\leftarrow C}=\begin{bmatrix}2&-1\\ 1&3\end{bmatrix}.\\[12pt] \text{(a) Compute }[G\circ F]_{D\leftarrow B}.\\ \text{(b) For }v=(5,-2)\text{ in standard coordinates, find }(G\circ F)(v)\text{ in standard coordinates.}\\[12pt] \textbf{Answer: }[G\circ F]_{D\leftarrow B}= \begin{bmatrix}2&3\\ 1&5\end{bmatrix},\quad (G\circ F)(5,-2)=(13,\,22).\\ \\

Problem 3 — Recover the standard matrix and evaluate F

\text{Let }B=\{(1,2),(1,0)\},\quad C=\{(0,1),(1,1)\}\text{ be bases of }\mathbb{R}^2.\\[12pt] \text{Given }[F]_{C\leftarrow B}=\begin{bmatrix}0&1\\ -2&3\end{bmatrix},\\ \text{compute the standard matrix }A\text{ of }F\ (\text{so that }[F(v)]_{\varepsilon}=A[v]_{\varepsilon}\ \forall v),\\ \text{and then compute }F(w)\text{ for }w=(1,-1).\\[12pt] \textbf{Answer: }A=\begin{bmatrix}3&-\tfrac{5}{2}\\ 4&-3\end{bmatrix},\quad F(1,-1)=\left(\tfrac{11}{2},\,7\right).\\

Problem 4 — Preimage with mixed bases

\text{Let }B=\{(1,0,1),(0,1,1),(1,1,0)\},\quad C=\{(2,0,0),(0,1,1),(1,0,1)\}\ \text{be bases of }\mathbb{R}^3.\\[12pt] \text{Suppose }[F]_{C\leftarrow B}= \begin{bmatrix} 1&0&2\\ 0&2&1\\ 1&-1&1 \end{bmatrix}.\\[12pt] \text{Given }y=(4,1,3)\text{ in standard coordinates, find }u\text{ in standard coordinates such that }F(u)=y.\\[12pt] \textbf{Answer: }u=(4,\,-1,\,9).\\ \\

Problem 5 — Invertibility and image of a basis vector

\text{Let }B=\{(2,1),(1,1)\},\quad C=\{(1,0),(1,2)\}\ \text{be bases of }\mathbb{R}^2.\\[12pt] \text{With }[F]_{C\leftarrow B}=\begin{bmatrix}3&1\\ 0&2\end{bmatrix}:\\[12pt] \text{(a) Find the standard matrix }A=P_C[F]_{C\leftarrow B}P_B^{-1}.\\ \text{(b) Compute }\det A,\ \text{decide if }F\text{ is invertible}.\\ \text{(c) Compute }F(e_1)\ \text{where }e_1=(1,0).\\[12pt] \textbf{Answer: }A=\begin{bmatrix}0&3\\ -4&8\end{bmatrix},\quad \det A=12\ (\neq 0)\Rightarrow F\ \text{invertible},\quad F(e_1)=(0,\,-4).\\