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

MATH221 December 2007

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[hide] 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 $[T]_{\mathcal {B}}$.

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[show] Hint

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[show] 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 $[T]_{\overline {\mathcal {B}}}={\frac {1}{2}}{\begin{pmatrix}1&1\\1&-1\end{pmatrix}}{\begin{pmatrix}2&-3\\0&2\end{pmatrix}}{\begin{pmatrix}1&1\\1&-1\end{pmatrix}}={\begin{pmatrix}1/2&3/2\\-3/2&7/2\end{pmatrix}}.$