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Q43E

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Found in: Page 402

### Linear Algebra With Applications

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

# Consider the transformation ${\mathbf{T}}{\mathbf{\left(}}{\mathbf{q}}{\mathbf{\left(}}{{\mathbf{x}}}_{{\mathbf{1}}}{\mathbf{,}}{{\mathbf{x}}}_{{\mathbf{2}}}{\mathbf{\right)}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{q}}{\mathbf{\left(}}{{\mathbf{x}}}_{{\mathbf{1}}}{\mathbf{,}}{\mathbf{0}}{\mathbf{\right)}}$ from ${{\mathbit{Q}}}_{{\mathbf{2}}}$ to ${{\mathbf{P}}}_{{\mathbf{2}}}$. Is T a linear transformation? If so, find the image, rank, kernel, and nullity of T.

the solution is

Yes,T is a linear transformation.

$\begin{array}{r}\mathrm{ker}\left(\mathrm{T}\right)=\left\{\mathrm{q}\left({\mathrm{x}}_{1},{\mathrm{x}}_{2}\right)={\mathrm{bx}}_{1}{\mathrm{x}}_{2}+{\mathrm{cx}}_{2}^{2}\in {\mathrm{Q}}_{2}:\mathrm{b},\mathrm{c}\in \mathrm{R}\right\}\\ \mathrm{Im}\left(\mathrm{T}\right)=\left\{\mathrm{p}\left(\mathrm{x}\right)={\mathrm{ax}}^{2}\in {\mathrm{P}}_{2}:\mathrm{a}\in \mathrm{R}\right\}\\ \mathrm{rank}\left(\mathrm{T}\right)=1\mathrm{and}\mathrm{nullity}\left(\mathrm{T}\right)=2\end{array}$

See the step by step solution

## Step 1: given information

$\mathrm{T}\left(\mathrm{q}\left({\mathrm{x}}_{1},{\mathrm{x}}_{2}\right)\right)=\mathrm{q}\left({\mathrm{x}}_{1},0\right)$

## Step 2: linear transformation

Consider ${\mathrm{q}}_{1}\left({\mathrm{x}}_{1},{\mathrm{x}}_{2}\right)={\mathrm{a}}_{1}{\mathrm{x}}_{1}^{2}+{\mathrm{b}}_{1}{\mathrm{x}}_{1}{\mathrm{x}}_{2}+{\mathrm{c}}_{1}{\mathrm{x}}_{2}^{2}$

${q}_{2}\left({x}_{1},{x}_{2}\right)={a}_{2}{x}_{1}^{2}+{b}_{2}{x}_{1}{x}_{2}+{c}_{2}{x}_{2}^{2}\in {Q}_{2}\mathrm{and}\mathrm{\alpha }\in \mathrm{R}$ then9

$T\left({a}_{1}\left({x}_{1},{x}_{2}\right)+{a}_{2}\left({x}_{1},{x}_{2}\right)\right)=T\left(a\left({a}_{1}{x}_{1}^{2}+{b}_{1}{x}_{1}{x}_{2}+{c}_{1}{x}_{2}^{2}\right)+{a}_{2}{x}_{1}^{2}+{b}_{2}{x}_{1}{x}_{2}+{c}_{2}{x}_{2}^{2}\right)\phantom{\rule{0ex}{0ex}}=T\left(\left(a{a}_{1}+{a}_{2}\right){x}_{1}^{2}+\left(a{b}_{1}+{b}_{2}\right){x}_{1}{x}_{2}+\left({a}_{1}+{c}_{2}\right){x}_{2}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(a{a}_{1}+{a}_{2}\right){x}_{1}^{2}\phantom{\rule{0ex}{0ex}}=\left(a{a}_{1}+{a}_{2}\right){x}_{1}^{2}\phantom{\rule{0ex}{0ex}}=a{q}_{1}\left({x}_{1},0\right)+{a}_{2}\left({x}_{1},0\right)\phantom{\rule{0ex}{0ex}}=\alpha T\left({a}_{1}\left({x}_{1},{x}_{2}\right)\right)+T\left({a}_{2}\left({x}_{1},{x}_{2}\right)\right)$

Since T satisfy

$\mathrm{T}\left({\mathrm{\alpha q}}_{1}+{\mathrm{q}}_{2}\right)=\mathrm{\alpha T}\left({\mathrm{q}}_{1}\right)+\mathrm{T}\left({\mathrm{q}}_{2}\right)\mathrm{for\alpha }\in \mathrm{R}\mathrm{and}{\mathrm{q}}_{1},{\mathrm{q}}_{2}\in {\mathrm{Q}}_{2}$

$\mathrm{T}:{\mathrm{Q}}_{2}\to {\mathrm{P}}_{2},\mathrm{T}\left(\mathrm{q}\left({\mathrm{x}}_{1},{\mathrm{x}}_{2}\right)\right)=\mathrm{q}\left({\mathrm{x}}_{1},0\right)$is a linear transformation.

## Step 3: find kernel of T and image of T

Kernel of T:

$\mathrm{ker}\left(T\right)=\left\{q\left({x}_{1},{x}_{2}\right)\in {Q}_{2}:T\left(q\left({x}_{1},{x}_{2}\right)\right)=0\in {P}_{2}\right\}\phantom{\rule{0ex}{0ex}}=\left\{q\left({x}_{1},{x}_{2}\right)\in {Q}_{2}:q\left({x}_{1},0\right)=0\in {P}_{2}\right\}\phantom{\rule{0ex}{0ex}}=\left\{q\left({x}_{1},{x}_{2}\right)=a{x}_{1}^{2}+b{x}_{1}{x}_{2}+c{x}_{2}^{2}\in {Q}_{2}:a{x}_{1}^{2}=0\right\}\phantom{\rule{0ex}{0ex}}=\left\{q\left({x}_{1},{x}_{2}\right)=a{x}_{1}^{2}+b{x}_{1}{x}_{2}+c{x}_{2}^{2}\in {Q}_{2}:a=0\right\}\phantom{\rule{0ex}{0ex}}=\left\{q\left({x}_{1},{x}_{2}\right)=b{x}_{1}{x}_{2}+c{x}_{2}^{2}\in {Q}_{2}:b,c\in R\right\}$

Image of T:

$\mathrm{Im}\left(T\right)=\left\{T\left(q\left({x}_{1},{x}_{2}\right)\right)\in {P}_{2}:q\left({x}_{1},{x}_{2}\right)\in {Q}_{2}\right\}\phantom{\rule{0ex}{0ex}}=\left\{q\left({x}_{1},0\right)\in {P}_{2}:q\left({x}_{1},{x}_{2}\right)=a{x}_{1}^{2}+b{x}_{1}{x}_{2}+c{x}_{2}^{2}\in {Q}_{2}\right\}\phantom{\rule{0ex}{0ex}}=\left\{q\left({x}_{1},0\right)=a{x}_{1}^{2}\in {P}_{2}:a\in R\right\}\phantom{\rule{0ex}{0ex}}=\left\{p\left(x\right)=a{x}^{2}\in {P}_{2}:a\in R\right\}$

$\mathrm{rank}\left(\mathrm{T}\right)=\mathrm{dim}\left(\mathrm{Img}\left(\mathrm{T}\right)\right)=1\phantom{\rule{0ex}{0ex}}\mathrm{nullity}\left(\mathrm{T}\right)=\mathrm{dim}\left(\mathrm{ker}\left(\mathrm{T}\right)\right)=2$

## Step 4: conclusion

Yes, is a linear transformation.

$\mathrm{ker}\left(\mathrm{T}\right)=\mathrm{q}\left({\mathrm{x}}_{1},{\mathrm{x}}_{2}\right)={\mathrm{bx}}_{1}{\mathrm{x}}_{2}+{\mathrm{cx}}_{2}^{2}\in {\mathrm{Q}}_{2}:\mathrm{b},\mathrm{c}\in \mathrm{R}\phantom{\rule{0ex}{0ex}}\mathrm{Im}\left(\mathrm{T}\right)=\mathrm{p}\left(\mathrm{x}\right)={\mathrm{ax}}^{2}\in {\mathrm{P}}_{2}:\mathrm{a}\in \mathrm{R}\phantom{\rule{0ex}{0ex}}\mathrm{rank}\left(\mathrm{T}\right)=1\mathrm{and}\mathrm{nullity}\left(\mathrm{T}\right)=2$