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Let $T: R^2 \to R^2$ be represented by $\begin{bmatrix}5 & -3\\2 & -2\end{bmatrix}$ with respect to the standard basis. Find the matrix T with respect to the basis B = { $\begin{bmatrix}3 \\1\end{bmatrix}$ , $\begin{bmatrix}1\\2 \end{bmatrix}$ }.

I found T$\begin{bmatrix}3 \\1\end{bmatrix}$ and T$\begin{bmatrix}1\\2 \end{bmatrix}$ by:

$\begin{bmatrix}5 & -3\\2 & -2\end{bmatrix}$$\begin{bmatrix}3 \\1\end{bmatrix}$ = $\begin{bmatrix}12 \\4\end{bmatrix}$ and

$\begin{bmatrix}5 & -3\\2 & -2\end{bmatrix}$$\begin{bmatrix}1 \\2\end{bmatrix}$ = $\begin{bmatrix}-1 \\-2\end{bmatrix}$

so that $[T]_B$ = $\begin{bmatrix}12 & -1\\4 & -2\end{bmatrix}$

but I'm not sure if this is correct. Am I doing the right thing or are my steps wrong?

user127273
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1 Answers1

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Note: I have assumed that the basis was changed in both the domain and the range in this answer.

If $y$ is a point expressed in the new basis, then it corresponds to a point $x=By$ in the old basis.

So, to compute the new form of the operator $T$, first map from the new coordinates to the old (multiply by $B$), then perform the mapping to get the mapped point in the old coordinates (multiply by $T$) and then map into the new coordinates (multiply by $B^{-1}$).

Putting these together gives $\tilde{T} = B^{-1} T B$.

Note that in this particular example, $T$ behaves as multiplication on the rows of $B$ (that is, $B$ is a matrix of eigenvectors), this should help considerably with the computations. In fact, if you think carefully, little computation will be needed (other than multiplying the columns of $B$ by $T$)...

Note that $B^{-1} T B = \operatorname{diag} (4, -1)$.

copper.hat
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  • I like your explanation, but I came across this lecture: https://math.dartmouth.edu/archive/m24w07/public_html/Lecture12.pdf . One page 5, it seems they are doing what the OP did. That is, to calculate T wrt to new basis $B$, they just did $T*B$. Isn't this wrong? Am I missing something? – larry Apr 16 '15 at 16:07
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    If the basis is only changed for the domain then this is correct. I presumed the basis was changed in both the domain and range. – copper.hat Apr 16 '15 at 23:45
  • "If the basis is only changed for the domain then this is correct" ... does the "this" refer to your solution or to the work referred to by the link? – larry Apr 17 '15 at 16:48
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    My solution presumed the basis was changed in both the domain and range, the solution in the link only changes the domain basis. – copper.hat Apr 17 '15 at 16:53
  • In the link that you gave out. We are changing the basis of $R^3$ which is the domain of the linear transformation. We are not changing the basis of $A$, the representation of the linear transformation as a matrix. Watch out for that, and read the question carefully!. In your question, If basis of the $R^2$ was changed, than your answer would have been correct. But we are asked to changed to basis of the transformation matrix. –  Jul 20 '16 at 15:53
  • What link are you referring to? – copper.hat Jul 20 '16 at 15:58