Question

Let$\epsilon =\left\{{\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3\right\}$ be the standard basis for ${\mathbb{R}}^3$,$B=\left\{{\mathbf{b}}_1,{\mathbf{b}}_2,{\mathbf{b}}_3\right\}$ be a basis for a vector space $V$ and$T:{\mathbb{R}}^3\to V$ be a linear transformation with the property that

$T\left(x_1,x_2,x_3\right)=\left(x_3-x_2\right){\mathbf{b}}_1-\left(x_1-x_3\right){\mathbf{b}}_2+\left(x_1-x_2\right){\mathbf{b}}_3$

1. Compute$T\left({\mathbf{e}}_1\right)$, $T\left({\mathbf{e}}_2\right)$ and $T\left({\mathbf{e}}_3\right)$.
2. Compute ${\left(T\left({\mathbf{e}}_1\right)\right)}_B$, ${\left(T\left({\mathbf{e}}_2\right)\right)}_B$ and ${\left(T\left({\mathbf{e}}_3\right)\right)}_B$.
3. Find the matrix for $T$ relative to $\epsilon$, and$B$.

Short answer

(a)$T\left({\mathbf{e}}_1\right)=-{\mathbf{b}}_2+{\mathbf{b}}_3$,$T\left({\mathbf{e}}_2\right)=-{\mathbf{b}}_1-{\mathbf{b}}_3$,and $T\left({\mathbf{e}}_3\right)={\mathbf{b}}_1-{\mathbf{b}}_2$.

(b)${\left(T\left({\mathbf{e}}_1\right)\right)}_B=\left(\begin{array}{r}0\\ -1\\ 1\end{array}\right)$,${\left(T\left({\mathbf{e}}_2\right)\right)}_B=\left(\begin{array}{r}-1\\ 0\\ -1\end{array}\right)$, and ${\left(T\left({\mathbf{e}}_3\right)\right)}_B=\left(\begin{array}{r}1\\ -1\\ 0\end{array}\right)$.

(c) The matrix for $T$ relative to $\epsilon$ and $B$ is $\left(\begin{array}{rlrlr}0& & -1& & 1\\ -1& & 0& & -1\\ 1& & -1& & 0\end{array}\right)$.

Step-by-step solution

The Standard basis

The standard basis for ${\mathbb{R}}^3$ are given by ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$.

Here, ${\mathbf{e}}_1=\left(\begin{array}{r}1\\ 0\\ 0\end{array}\right)$, ${\mathbf{e}}_2=\left(\begin{array}{r}0\\ 1\\ 0\end{array}\right)$, ${\mathbf{e}}_3=\left(\begin{array}{r}0\\ 0\\ 1\end{array}\right)$.

Find $T\left({\mathbf{e}}_1\right)$, $T\left({\mathbf{e}}_2\right)$, and $T\left({\mathbf{e}}_3\right)$ 

(a)

Find $T\left({\mathbf{e}}_1\right)$ by using ${\mathbf{e}}_1=\left(\begin{array}{r}1\\ 0\\ 0\end{array}\right)$ for $\left(\begin{array}{r}{x}_1\\ x_2\\ x_3\end{array}\right)$ into $T\left(x_1,x_2,x_3\right)=\left(x_3-x_2\right){\mathbf{b}}_1-\left(x_1+x_3\right){\mathbf{b}}_2+\left(x_1-x_2\right){\mathbf{b}}_3$.

$\begin{array}{rl}T\left({\mathbf{e}}_1\right)& =T\left(1,0,0\right)\\ & =\left(0-0\right){\mathbf{b}}_1-\left(1+0\right){\mathbf{b}}_2+\left(1-0\right){\mathbf{b}}_3\\ & =-{\mathbf{b}}_2+{\mathbf{b}}_3\end{array}$

Find $T\left({\mathbf{e}}_2\right)$ by using ${\mathbf{e}}_2=\left(\begin{array}{r}0\\ 1\\ 0\end{array}\right)$ for $\left(\begin{array}{r}{x}_1\\ x_2\\ x_3\end{array}\right)$.

$\begin{array}{rl}cT\left({\mathbf{e}}_2\right)& =T\left(0,1,0\right)\\ & =\left(0-1\right){\mathbf{b}}_1-\left(0+0\right){\mathbf{b}}_2+\left(0-1\right){\mathbf{b}}_3\\ & =-{\mathbf{b}}_1-{\mathbf{b}}_3\end{array}$

Find $T\left({\mathbf{e}}_3\right)$ by using ${\mathbf{e}}_3=\left(\begin{array}{r}0\\ 0\\ 1\end{array}\right)$ for $\left(\begin{array}{r}{x}_1\\ x_2\\ x_3\end{array}\right)$.

$\begin{array}{rl}T\left({\mathbf{e}}_3\right)& =T\left(0,0,1\right)\\ & =\left(1-0\right){\mathbf{b}}_1-\left(0+1\right){\mathbf{b}}_2+\left(0-0\right){\mathbf{b}}_3\\ & ={\mathbf{b}}_1-{\mathbf{b}}_2\end{array}$

Hence, $T\left({\mathbf{e}}_1\right)=-{\mathbf{b}}_2+{\mathbf{b}}_3$, $T\left({\mathbf{e}}_2\right)=-{\mathbf{b}}_1-{\mathbf{b}}_3$ and $T\left({\mathbf{e}}_3\right)={\mathbf{b}}_1-{\mathbf{b}}_2$.

Find ${\left(T\left({\mathbf{e}}_1\right)\right)}_B$, ${\left(T\left({\mathbf{e}}_2\right)\right)}_B$, and ${\left(T\left({\mathbf{e}}_3\right)\right)}_B$

(b)

Write the $B$-coordinate vectors of the images of ${\mathbf{e}}_1$, ${\mathbf{e}}_2$, and ${\mathbf{e}}_3$.

${\left(T\left({\mathbf{e}}_1\right)\right)}_B=\left(\begin{array}{r}0\\ -1\\ 1\end{array}\right)$

${\left(T\left({\mathbf{e}}_2\right)\right)}_B=\left(\begin{array}{r}-1\\ 0\\ -1\end{array}\right)$

${\left(T\left({\mathbf{e}}_3\right)\right)}_B=\left(\begin{array}{r}1\\ -1\\ 0\end{array}\right)$

The matrix for a linear transformation 

A matrix associated with a linear transformation $T$ for $V$ and $W$ is given by ${\left(T\left(\mathbf{x}\right)\right)}_C=\left(\begin{array}{rlrl}{\left(T\left({\mathbf{b}}_1\right)\right)}_C& {\left(T\left({\mathbf{b}}_2\right)\right)}_C& \cdots & {\left(T\left({\mathbf{b}}_n\right)\right)}_C\end{array}\right)$, where $V$ and $W$ are $n$ and $m$-dimensional subspaces respectively, and $B$, and $C$ are bases for $V$, and $W$.

Find the matrix for a linear transformation 

(c)

Form a matrix $T$for the obtained vectors in step 3, by using the formula ${\left(T\left(\mathbf{x}\right)\right)}_C=\left(\begin{array}{rlrl}{\left(T\left({\mathbf{b}}_1\right)\right)}_C& {\left(T\left({\mathbf{b}}_2\right)\right)}_C& \cdots & {\left(T\left({\mathbf{b}}_n\right)\right)}_C\end{array}\right)$, where $n=3$.

$\begin{array}{rl}c{\left(T\left(\mathbf{x}\right)\right)}_B& =\left(\begin{array}{rlr}{\left(T\left({\mathbf{e}}_1\right)\right)}_B& {\left(T\left({\mathbf{e}}_2\right)\right)}_B& {\left(T\left({\mathbf{e}}_3\right)\right)}_B\end{array}\right)\\ & =\left(\begin{array}{rlrlr}0& & -1& & 1\\ -1& & 0& & -1\\ 1& & -1& & 0\end{array}\right)\end{array}$

So, the required matrix is $\left(\begin{array}{rlrlr}0& & -1& & 1\\ -1& & 0& & -1\\ 1& & -1& & 0\end{array}\right)$.