Emerson Barnes

Answered

2022-01-20

Find matrix of linear transformation

A linear transformation

$T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$

is given by

$T\left(i\right)=i+j$

$T\left(j\right)=2i-j$

A linear transformation

is given by

Answer & Explanation

egowaffle26ic

Expert

2022-01-21Added 7 answers

Step 1

If$T\left(i\right)=(1,\text{}1)$ and $T\left(j\right)=(2,\text{}-1)$ and

${e}_{1}=i-j=(1,\text{}-1)$ and ${e}_{2}=3i+j=(3,\text{}1)$ , then

$T\left({e}_{1}\right)=T(i-j)$

$=T\left(i\right)-T\left(j\right)$

$=(1,\text{}2)$

$={a}_{1}{e}_{1}+{b}_{1}{e}_{2}$ (say)

and

$T\left({e}_{2}\right)=T(3i+j)$

$=3(1,\text{}1)+(2,\text{}-1)$

$=(5,\text{}2)$

$={a}_{2}{e}_{1}+{b}_{2}{e}_{2}$ (say)

Thus the matrix of T w.r.t. the new basis$\{{e}_{1},\text{}{e}_{2}\}$ is

$\left(\begin{array}{cc}{a}_{1}& {a}_{2}\\ {b}_{1}& {b}_{2}\end{array}\right)$

and you need to find the values of$a}_{1},\text{}{a}_{2},\text{}{b}_{1$ and $b}_{2$ . The above systems of equations reduces to

${a}_{1}+2{b}_{1}=-1$

$-{a}_{1}+{b}_{1}=2$

and

${a}_{2}+3{b}_{2}=5$

$-{a}_{2}+{b}_{2}=2$

Solve these equations to obtain

$a}_{1}=-\frac{7}{4$

$b}_{1}=\frac{1}{4$

$a}_{2}=-\frac{1}{4$

and

$b}_{2}=\frac{7}{4$

If

and

Thus the matrix of T w.r.t. the new basis

and you need to find the values of

and

Solve these equations to obtain

and

suzzzlesv7

Expert

2022-01-22Added 7 answers

Step 1

The matrix of the transformation of basis is:

$P=\left[\begin{array}{cc}1& 3\\ -1& 1\end{array}\right]$

and

${P}^{-1}=\frac{1}{4}\left[\begin{array}{cc}1& -3\\ 1& 1\end{array}\right]$

Your transformed matrix is:

${P}^{-1}TP=\frac{1}{4}\left[\begin{array}{cc}1& -3\\ 1& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 1& -1\end{array}\right]\left[\begin{array}{cc}1& 3\\ -1& 1\end{array}\right]$

$=\frac{1}{4}\left[\begin{array}{cc}-7& -1\\ 1& 7\end{array}\right]$

Step 2

The figure illustrate how operate the given transformation in the two basis.

Indices$\{i,\text{}j\}$ refer to the canonical basis, indices $\{1,\text{}2\}$ to the new basis $e}_{1},\text{}{e}_{2$ .

In the figure we have the vector$v={v}_{i,j}={[1,2]}^{T}$ (in canonical basis), that is transformed in $v}^{\prime}={v}_{i,j}^{\prime}={[5,-1]}^{T$

In the new basis we have:

${v}_{1,2}={P}^{-1}{v}_{i,j}=\frac{1}{4}\left[\begin{array}{c}-5\\ 3\end{array}\right]$

and, for the tranformed vector:

${v}_{1,2}^{\prime}={P}^{-1}{v}_{i,j}=\frac{1}{4}\left[\begin{array}{c}2\\ 1\end{array}\right]$

$={P}^{-1}{T}_{i.j}{v}_{i,j}$

$={P}^{-1}{T}_{i,j}P{v}_{1,2}$

$={T}_{1,2}{v}_{1,2}$

ao we have:

${T}_{1,2}={P}^{-1}{T}_{i,j}P=\frac{1}{4}\left[\begin{array}{cc}-7& -1\\ 1& 7\end{array}\right]$

I hope this can be helpful.

The matrix of the transformation of basis is:

and

Your transformed matrix is:

Step 2

The figure illustrate how operate the given transformation in the two basis.

Indices

In the figure we have the vector

In the new basis we have:

and, for the tranformed vector:

ao we have:

I hope this can be helpful.

RizerMix

Expert

2022-01-27Added 437 answers

Step 1
${e}_{1}+{e}_{2}=4i$
$i=\frac{{e}_{1}+{e}_{2}}{4}=P(i|j{)}^{T}$
${e}_{2}-3{e}_{1}=4j$
$j=\frac{{e}_{2}-3{e}_{1}}{4}=P(i|j{)}^{T}$
Now compute the Matrix ${P}^{-1}$ and your T in the new Basis is given by ${P}^{-1}TP$

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