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Q65E

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Found in: Page 185

### Linear Algebra With Applications

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

# (a) Show that T is a linear transformation.(b) Find the kernel of T.(c) Show that the image of T is a space ${\mathbf{L}}{\mathbf{\left(}}{{\mathbf{R}}}^{{\mathbf{m}}}{\mathbf{,}}{{\mathbf{R}}}^{{\mathbf{n}}}{\mathbf{\right)}}$of all linear transformation ${{\mathbf{R}}}^{{\mathbf{m}}}$ to role="math" localid="1659420398933" ${{\mathbf{R}}}^{\mathbit{n}}$.(d) Find the dimension of $\mathbf{L}\mathbf{\left(}{{\mathbf{R}}}^{{\mathbf{m}}}{\mathbf{,}}{{\mathbf{R}}}^{{\mathbf{n}}}{\mathbf{\right)}}$.

(a) The solution is a linear transformation.

(b) The solution is the kernel of T contains only zero matrix.

(c) The solution is the image of T is the space $\mathrm{L}\left({\mathrm{R}}^{\mathrm{m}},{\mathrm{R}}^{\mathrm{n}}\right)$ of all linear transformation from ${\mathrm{R}}^{\mathrm{m}}$ to role="math" localid="1659420484059" ${\mathrm{R}}^{\mathrm{n}}$.

(d) The solution is the dimension of $\mathrm{L}\left({\mathrm{R}}^{\mathrm{m}},{\mathrm{R}}^{\mathrm{n}}\right)$is mn.

See the step by step solution

## (a) 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.

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

For all elements f,g of V and k is scalar.

## Step2: Explanation of the solution

Consider the transformation as follows.

$\mathrm{T}:{\mathrm{R}}^{\mathrm{n}×\mathrm{m}}\to \mathrm{F}\left({\mathrm{R}}^{\mathrm{n}},{\mathrm{R}}^{\mathrm{m}}\right)$ is defined by $\left(T\left(A\right)\right)\left(\stackrel{\to }{v}\right)=A\stackrel{\to }{v}$.

Simplify for the linear transformation first condition as follows.

$\left(\mathrm{T}\left(\mathrm{A}\right)\right)\left(\stackrel{\to }{\mathrm{v}}+\stackrel{\to }{\mathrm{w}}\right)=\mathrm{A}\left(\stackrel{\to }{\mathrm{v}}+\stackrel{\to }{\mathrm{w}}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{A}\stackrel{\to }{\mathrm{v}}+\mathrm{A}\stackrel{\to }{\mathrm{w}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{T}\left(\mathrm{A}\right)\right)\left(\stackrel{\to }{\mathrm{v}}+\stackrel{\to }{\mathrm{w}}\right)=\left(\mathrm{T}\left(\mathrm{A}\right)\right)\stackrel{\to }{\mathrm{v}}+\left(\mathrm{T}\left(\mathrm{A}\right)\right)\stackrel{\to }{\mathrm{w}}$

Similarly, simplify for the second condition as follows.

$\left(T\left(A\right)\right)\left(k\stackrel{\to }{v}\right)=A\left(k\stackrel{\to }{v}\right)\phantom{\rule{0ex}{0ex}}=k\left(A\stackrel{\to }{v}\right)\phantom{\rule{0ex}{0ex}}=K\left(T\left(A\right)\right)\left(\stackrel{\to }{v}\right)\phantom{\rule{0ex}{0ex}}\left(T\left(A\right)\right)\left(k\stackrel{\to }{v}\right)=k\left(T\left(A\right)\right)\left(\stackrel{\to }{v}\right)$

Thus, T is a linear transformation.

## (b) Step3: Definition of image and kernel

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

## Step4: Explanation of the solution

Consider for a non-zero vector $\stackrel{\to }{v}$ as follows.

$\left(T\left(A\right)\right)\left(\stackrel{\to }{v}\right)=\stackrel{\to }{0}\phantom{\rule{0ex}{0ex}}A\stackrel{\to }{v}=0$

Then by equality of two matrix as follows.

$\stackrel{\to }{v}=\stackrel{\to }{0}$

Thus, the kernel of T contains only zero matrix.

## (c) Step5: Definition of image and kernel

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

## Step6: Explanation of the solution

Since, the transformation is linear.

Therefore, the image of T is the linear space of all the linear transformation from ${\mathrm{R}}^{\mathrm{m}}$ to ${\mathrm{R}}^{\mathrm{n}}$.

Hence, the image of T is the space $\mathrm{L}\left({\mathrm{R}}^{\mathrm{m}},{\mathrm{R}}^{\mathrm{n}}\right)$ of all linear transformation from ${\mathrm{R}}^{\mathrm{m}}$ to ${\mathrm{R}}^{\mathrm{n}}$.

## (d)Step7: Definition of Linear Transformation

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

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

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 ${\mathbf{ker}}{\mathbf{\left(}}{\mathbf{T}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{\left\{}}{\mathbf{0}}{\mathbf{\right\}}}$ and ${\mathbf{im}}{\mathbf{\left(}}{\mathbf{T}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{W}}$ or ${\mathbf{dim}}{\mathbf{\left(}}{\mathbf{V}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{dim}}{\mathbf{\left(}}{\mathbf{W}}{\mathbf{\right)}}$.

## Step8: Explanation of the solution

Consider the transformation as follows.

$\mathrm{T}:{\mathrm{R}}^{\mathrm{n}×\mathrm{m}}\to \mathrm{F}\left({\mathrm{R}}^{\mathrm{n}},{\mathrm{R}}^{\mathrm{m}}\right)$ is defined by $\left(\mathrm{T}\left(\mathrm{A}\right)\right)\left(\stackrel{\to }{\mathrm{v}}\right)=\mathrm{A}\stackrel{\to }{\mathrm{v}}$.

The dimension of the space is as follows.

$\mathrm{dim}\left(\mathrm{L}\left({\mathrm{R}}^{\mathrm{m}},{\mathrm{R}}^{\mathrm{n}}\right)\right)=\mathrm{dim}\left({\mathrm{R}}^{\mathrm{n}×\mathrm{m}}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{mn}\phantom{\rule{0ex}{0ex}}\mathrm{dim}\left(\mathrm{L}\left({\mathrm{R}}^{\mathrm{m}},{\mathrm{R}}^{\mathrm{n}}\right)\right)=\mathrm{mn}$

Thus, the dimension of $\mathrm{L}\left({\mathrm{R}}^{\mathrm{m}},{\mathrm{R}}^{\mathrm{n}}\right)$ is mn.