Science:Math Exam Resources/Courses/MATH221/December 2007/Question 06 (c)

MATH221 December 2007

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Question 06 (c)
Suppose the standard matrix of a linear transformation $T:\mathbb {R} ^{2}\to \mathbb {R} ^{2}$ is ${\begin{pmatrix}2&-3\\0&2\end{pmatrix}}.$ Find the matrix of T with respect to the basis ${\mathcal {B}}=\{{\begin{pmatrix}1\\1\end{pmatrix}},{\begin{pmatrix}1\\-1\end{pmatrix}}\}$, i.e. find $[T]_{\mathcal {B}}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let ${\overline {\mathcal {B}}}=\{e_{1},e_{2}\}$ denote the standard basis of $\mathbb {R} ^{2}$, where $e_{1}={\begin{pmatrix}1\\0\end{pmatrix}}$ and $e_{2}={\begin{pmatrix}0\\1\end{pmatrix}},$ and denote the given basis by ${\mathcal {B}}=\{b_{1},b_{2}\}$, where $b_{1}={\begin{pmatrix}1\\1\end{pmatrix}}$ and $b_{2}={\begin{pmatrix}1\\-1\end{pmatrix}}.$ Let us briefly recall how given $[T]_{\mathcal {B}}$ (the matrix of T with respect to ${\mathcal {B}}$), we can find $[T]_{\overline {\mathcal {B}}}$ (the matrix of T with respect to ${\overline {\mathcal {B}}}.$). We first find the linear relation between the basis ${\mathcal {B}}$ and ${\overline {\mathcal {B}}}$: ${\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}}={\begin{pmatrix}1&1\\1&-1\end{pmatrix}}{\begin{pmatrix}e_{1}\\e_{2}\end{pmatrix}}.$ Letting $P={\begin{pmatrix}1&1\\1&-1\end{pmatrix}}$ the above equality gives us the following relation for the coordinates of a vector $v\in \mathbb {R} ^{2}$ with respect to the two bases: $[v]_{\mathcal {B}}=P[v]_{\overline {\mathcal {B}}}.$ Now, by definition, we have $[Tv]_{\mathcal {B}}=[T]_{\mathcal {B}}[v]_{\mathcal {B}}$ and $[Tv]_{\overline {\mathcal {B}}}=[T]_{\overline {\mathcal {B}}}[v]_{\overline {\mathcal {B}}}$. Therefore, we can write the following string of equalities: $P[T]_{\overline {\mathcal {B}}}[v]_{\overline {\mathcal {B}}}=P[Tv]_{\overline {\mathcal {B}}}=[Tv]_{\mathcal {B}}=[T]_{\mathcal {B}}[v]_{\mathcal {B}}=[T]_{\mathcal {B}}P[v]_{\overline {\mathcal {B}}}.$ Multiplying by $P^{-1}$ in the above gives us $[T]_{\overline {\mathcal {B}}}[v]_{\overline {\mathcal {B}}}=P^{-1}[T]_{\mathcal {B}}P[v]_{\overline {\mathcal {B}}}$. As this equality is true for every v, we conclude that $[T]_{\overline {\mathcal {B}}}=P^{-1}[T]_{\mathcal {B}}P.$ For this question, we are given $[T]_{\mathcal {B}}={\begin{pmatrix}2&-3\\0&2\end{pmatrix}},$ and we can compute $P^{-1}={\frac {1}{2}}{\begin{pmatrix}1&1\\1&-1\end{pmatrix}}.$ Putting these together, we find that $[T]_{\overline {\mathcal {B}}}={\frac {1}{2}}{\begin{pmatrix}1&1\\1&-1\end{pmatrix}}{\begin{pmatrix}2&-3\\0&2\end{pmatrix}}{\begin{pmatrix}1&1\\1&-1\end{pmatrix}}={\begin{pmatrix}1/2&3/2\\-3/2&7/2\end{pmatrix}}.$