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Let $$B=\left\{\begin{bmatrix}1&0\\0&2\end{bmatrix},\begin{bmatrix}0&1\\1&0 \end{bmatrix},\begin{bmatrix}2&0\\0&3\end{bmatrix}\right\}$$ a base of $W=\{A\in \mathrm{Mat}_{2 \times 2}(\mathbb{R}) \mid A^t=A\}$

Given a linear transformation $T:W\rightarrow W$ which is defined by $[T]_B=\begin{bmatrix}1&2&0\\-1&0&2\\1&3&1\end{bmatrix}$.

Calculate $T\begin{bmatrix}-1&2\\2&3\end{bmatrix}$.

So I found out that $$\begin{bmatrix}-1&2\\2&3\end{bmatrix}=9\begin{bmatrix}1&0\\0&2\end{bmatrix}+2\begin{bmatrix}0&1\\1&0\end{bmatrix}-5\begin{bmatrix}2&0\\0&3\end{bmatrix}$$ and I'm not quite sure how to continue.

Any help would be appreciated.

2 Answers 2

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Let denote with $B_1$, $B_2$ and $B_3$ the first, the second and the third matrix from $B$.

From the first column of $[T]_B$ it results:

$T(B_1)= B_1 - B_2 + B_3$.

From the second column of $[T]_B$ it results

$T(B_2)= 2\cdot B_1 + 3\cdot B_3$

and from the third column it results that

$T(B_3)= 2\cdot B_2 + B_3$.

Now, you have to compute $T(9\cdot B_1 + 2\cdot B_2 - 5\cdot B_3)$.

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    Yes, that's what I did at first but thought it wasn't right. Could you explain the logic? – 2017-02-22
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    @Itay4 The matrix of an endomorphism in a basis is obtained in this way: first you have to compute $T$(element of basis) then you have to express it as a combination of basis elements, and after that you have to write the coefficents on column. – 2017-02-22
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You have found that $(9,2,-5) $ is the coordinate of $\left[\begin{array}{rrr} -1&2\\ 2&3\\ \end{array}\right]$ with respect to the ordered basis $B$.

Notice that $[T]_B$ sends all coordinates $\vec w$ (that each coordinate represents a unique matrix with respect to the ordered basis $B$, here just consider $\vec v$ is the coordinate representing the matrix $A$) to some coordinates(which is exactly the coordinate of $T(A)$ with respect to $B$).

Now what does $\left[\begin{array}{rrr} 1&2&0\\ -1&0&2\\ 1&3&1\\ \end{array}\right] \left[\begin{array}{rrr} 9\\ 2\\ -5\\ \end{array}\right]$ mean?

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    That would be the coordinate of $T\begin{bmatrix}-1&2\\2&3\end{bmatrix}$ with respect to $B$? – 2017-02-22
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    @Itay4 Exactly! – 2017-02-22
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    Ok, so I get that $\left[\begin{array}{rrr} 1&2&0\\ -1&0&2\\ 1&3&1\\ \end{array}\right] \left[\begin{array}{rrr} 9\\ 2\\ -5\\ \end{array}\right]$$=$$\left[\begin{array}{rrr} 13\\ -19\\ 10\\ \end{array}\right]$ Now, let's denote with $b1$, $b2$ and $b3$ the first, the second and the third matrix from B, $T\begin{bmatrix}-1&2\\2&3\end{bmatrix}$$=13b1-19b2+10b3$ and the result is my answer? – 2017-02-22