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I'm going through the various ways to construct a B-matrix of a linear transformation and I'm hitting a snag with one of the methods.

We have $A = \begin{pmatrix} -3 & 4 \\ 4 & 3 \\ \end{pmatrix}$ and $[ \vec{v_1}, \vec{v_2} ] = \bigg( \begin{bmatrix} 1 \\ 2 \end{bmatrix},\begin{bmatrix}-2\\1\end{bmatrix} \bigg)$

The question wants us to contruct the matrix $B$ "column by column" and so far, this is the best answer I've found:

$B = [[T(\vec{v_1}]_{\mathit{B}} [T(\vec{v_2}]_{\mathit{B}}]] = \bigg(\bigg[\begin{pmatrix} -3 & 4 \\ 4 & 3\end{pmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix}\bigg]_{\mathit{B}} \bigg[\begin{pmatrix} -3 & 4 \\ 4 & 3\end{pmatrix} \begin{bmatrix} -2 \\ 1 \end{bmatrix}\bigg]_{\mathit{B}}\bigg) \\ =\bigg[\begin{bmatrix} 5 \\ 10 \end{bmatrix}_{\mathit{B}} \begin{bmatrix}10 \\ -5 \end{bmatrix}_{\mathit{B}}\bigg] \\ =\begin{pmatrix} 5 & 0 \\ 0 & -5 \\ \end{pmatrix}$

I understand everything but how the solution gets from the second line to the third. Is the idea that the coordinates of the two vectors $ \bigg( \begin{bmatrix} 5 \\ 10 \end{bmatrix},\begin{bmatrix}-10\\5\end{bmatrix} \bigg)$ in this specific basis are given by the matrix $B = \begin{pmatrix} 5 & 0 \\ 0 & -5 \\ \end{pmatrix}$? I've been able to calculate the value of $B$ using the identity $B=S^{-1}AS$ and using a commutative diagram, but I feel like the conclusion is a bit tenuous using this third method. I'd like to be able to more fully explain the movement from the third to the fourth line. Thanks!

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  • $\begingroup$ I'm afraid I don't understand your notations. Are you speaking of the matrisof a linear map after a change of basis? $\endgroup$
    – Bernard
    Commented Feb 20, 2015 at 22:37

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It's pretty simple: the first column ($\begin{bmatrix} 5\\10\end{bmatrix}$) is $T(v_1)$ expressed in the canonical basis. To have the matrix of the linear map in the basis $(v_1,v_2)$, we need to express $T(v_1)$ and $T(v_2)$ in that basis. Now $$T(v_1)=\begin{bmatrix} 5\\10\end{bmatrix}=5\begin{bmatrix} 1\\2\end{bmatrix}=5v_1$$ Similarly for $T(v_2)$.

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