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Let the linear map $ T\colon R^2 \to R^2$ has a matrix $$ A=\begin{pmatrix} -3 & 1 \\ 3 & -1 \end{pmatrix} $$

with respect to the basis $B=\{(1,-1), \ (-1,5) \}$ , then find the matrix $ \ A' \ $ with respect to the basis $B'=\{(-2,1), \ (-1,1) \}$?

Answer:

Let $A'$ be the matrix corresponding the basis $B'=\{(-2,1), \ (-1,1) \}$. Then, $A'=PAP^{-1}$, where $ \ P=\begin{pmatrix} -3 & 1 \\ 3 & -1 \end{pmatrix} $

Am I right ?

This is a basis change problem.

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See, it is very simple. How do we write the matrix of a transformation with choice of a certain basis? What we do is, find the images of the basis elements under the transformation, and write them as columns of the matrix. For example, if $A$ is the matrix with $B$ as given, then $T(1,-1) = (-3,3)$ and $T(-1,5) = (1,-1)$.

Now, to find $A'$, we have to find what $T(-2,1)$ is, and what $T(-1,1)$ is. Once we do this, we write down the answers as column vectors and put them together.

Of course, to find $T(-2,1)$ or the other, we need to use linearity!

For example, is there a linear combination of $(1,-1)$ and $(-1,5)$ which equals $(-2,1)$? Indeed, there must be, because these two form a basis, but we can explicitly find $a(1,-1) + b(-1,5) = (-2,1)$ : it is a pair of simultaneous equations which can be solved : $a -b = -2, 5b-a = 1$, gives $b = \frac{-1}4, a= \frac{-9}{4}$.

Therefore, $$ T(-2,1) = T(a(1,-1) + b(-1,5)) = aT(1,-1) + bT(-1,5) \\ = \frac{-1}{4}(-3,3) + \frac{-9}{4}(1,-1) = (-1.5,1.5) $$

We need not struggle for the other basis element , namely $(-1,1)$ : note that $(-1,1) = -1 \times (1,-1)$, so $T(-1,1) = -1 \times T(1,-1) = (3,-3)$.

Now, it is fairly clear what the answer is : putting these together as columns gives $\begin{pmatrix}-1.5 & 3 \\ 1.5 & -3\end{pmatrix}$.

Of course, we have found the change of basis matrix secretly, while computing which linear combination of the current basis gives elements of the other basis, but an approach by basics can be helpful in preventing you from going into blind computation, as can often be the case.

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