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Found in: Page 185

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# For which constant k is a linear transformation ${\mathbf{T}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{=}}\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]{\mathbf{M}}{\mathbf{-}}{\mathbf{M}}\left[\begin{array}{cc}3& 0\\ 0& k\end{array}\right]$is an isomorphism form ${{\mathbf{R}}}^{\mathbf{2}\mathbf{×}\mathbf{2}}$ to ${{\mathbf{R}}}^{\mathbf{2}\mathbf{×}\mathbf{2}}$.

The solution is an isomorphism when $\mathrm{\left\{}k\in \mathrm{R}-\left\{2,4\right\}$.

See the step by step solution

## Step1: Definition of Linear Transformation

Consider two linear spaces V and W. A function T is said to be linear transformation if the following holds.

${\mathbf{T}}\mathbf{\left(}\mathbf{f}\mathbf{+}\mathbf{g}\mathbf{\right)}{\mathbf{=}}{\mathbf{T}}\mathbf{\left(}\mathbf{f}\mathbf{\right)}{\mathbf{+}}{\mathbf{T}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{T}}\mathbf{\left(}\mathbf{kf}\mathbf{\right)}{\mathbf{=}}{\mathbf{kT}}\mathbf{\left(}\mathbf{f}\mathbf{\right)}$

For all elements ${\mathbf{f}}{\mathbf{,}}{\mathbf{g}}$ of V and k is scalar.

A linear transformation ${\mathbf{T}}{\mathbf{:}}{\mathbf{V}}{\mathbf{\to }}{\mathbf{W}}$ is said to be an isomorphism if and only if ${\mathbf{ker}}{\mathbf{\left(}}{\mathbf{T}}{\mathbf{\right)}}{\mathbf{=}}\left\{0\right\}$ and ${\mathbf{im}}{\mathbf{\left(}}{\mathbf{T}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{W}}$ or ${\mathbf{dim}}{\mathbf{\left(}}{\mathbf{V}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{dim}}{\mathbf{\left(}}{\mathbf{W}}{\mathbf{\right)}}$.

## Step2: Explanation of the solution

The given transformation as follows.

$\mathrm{T}\left(\mathrm{M}\right)=\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\mathrm{M}-\mathrm{M}\left[\begin{array}{cc}3& 0\\ 0& \mathrm{k}\end{array}\right]$, from ${\mathrm{R}}^{2×2}$ to ${\mathrm{R}}^{2×2}$.

By using the definition of linear transformation as follows.

$T\left(A+B\right)=T\left(A\right)+T\left(B\right)\phantom{\rule{0ex}{0ex}}T\left(kA\right)=kT\left(A\right)$

Now, to check the first condition as follows.

Let A and B be arbitrary matrices from ${\mathrm{R}}^{2×2}$ and as follows.

$\mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\left(\mathrm{A}+\mathrm{B}\right)\left(\mathrm{A}-\mathrm{B}\right)\left[\begin{array}{cc}3& 0\\ 0& \mathrm{k}\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\mathrm{A}+\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\mathrm{B}-\mathrm{A}\left[\begin{array}{cc}3& 0\\ 0& \mathrm{k}\end{array}\right]-\mathrm{B}\left[\begin{array}{cc}3& 0\\ 0& \mathrm{k}\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cc}2& 3\\ 0& \mathrm{k}\end{array}\right]\mathrm{A}-\mathrm{A}\left[\begin{array}{cc}3& 0\\ 0& \mathrm{k}\end{array}\right]+\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\mathrm{B}-\mathrm{B}\left[\begin{array}{cc}3& 0\\ 0& \mathrm{k}\end{array}\right]\phantom{\rule{0ex}{0ex}}\mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right)$

Similarly, to check the second condition as follows.

Let $\alpha$ be an arbitrary scalar, and $\mathrm{A}\in {\mathrm{R}}^{2×2}$as follows.

$T\left(\alpha A\right)=\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\alpha A-\alpha A\left[\begin{array}{cc}3& 0\\ 0& k\end{array}\right]\phantom{\rule{0ex}{0ex}}=\alpha \left(\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]A-A\left[\begin{array}{cc}3& 0\\ 0& k\end{array}\right]\right)\phantom{\rule{0ex}{0ex}}\mathrm{T}\left(\mathrm{\alpha A}\right)=\mathrm{\alpha T}\left(\mathrm{A}\right)$

Thus,T is a linear transformation.

## Step3: Properties of isomorphism

A linear transformation ${\mathbf{T}}{\mathbf{:}}{\mathbf{V}}{\mathbf{\to }}{\mathbf{W}}$ is isomorphism if and only if ${\mathbf{ker}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{\right)}}{\mathbf{=}}\left\{0\right\}$and localid="1659426664071" ${\mathbf{Im}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{W}}{\mathbf{}}$

Now, check if $\mathrm{ker}\left(\mathrm{t}\right)=\left\{0\right\}$ as follows.

$ker\left(T\right)=\left\{A\in {R}^{2×2}|T\left(A\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right\}$

Consider a matrix A as follows.

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

The next equation as follows.

$T\left(A\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}2& 3\\ 0& 4\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]-\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}3& 0\\ 0& k\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}2a+3c& 2b+3d\\ 0a+4c& 0d+4d\end{array}\right]-\left[\begin{array}{cc}3a+0b& 0a+kb\\ 3c+0d& 0c+kd\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}2a+3c& -3a-0d\\ 0a+4c& -3c-3d\end{array}\begin{array}{cc}2b+3d& -0a-kd\\ 0b+4d& -0c-kd\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Simplify further as follows.

$\left[\begin{array}{cc}2a+3c-3a-0d& 2b+3d-0a-kb\\ 0a+4c-3c-0d& 0d+4d-0c-kd\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}a+3c& 2b-kb+3d\\ c& 4d-kd\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}-a+3c& \left(2-k\right)b+3d\\ c& \left(4-k\right)d\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Equating the corresponding entries as follows.

$c=0$ and $a+3c=0$

Solve and find the values as follows.

$a=0,c=0$

Substitute the value 0 for and 0 for in $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$as follows.

$\begin{array}{l}A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\\ =\left[\begin{array}{cc}0& b\\ 0& d\end{array}\right]\end{array}$

For $k=4$ the solution as follows.

$T\left(\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\phantom{\rule{0ex}{0ex}}ker\left(T\right)\ne \left\{\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\right\}\phantom{\rule{0ex}{0ex}}ker\left(T\right)\ne \left\{0\right\}$

Therefore,T to be an isomorphism for $k\ne 4$.

Similarly, for T to be an isomorphism for $\mathrm{k}\ne 2$.

Thus, is a linear transformation and is an isomorphism when $k\in R-\left\{2,4\right\}$