Find Matrix of Linear Transformation T w/ Respect to Basis B

[kiwifruit]

I came across this problem in one of my linear algebra books.
A linear transformation T:R^3 ->R^3 has matrix

2 3 0
-1 1 2
2 0 1
with respect to the standard basis for R^3. Find the matrix of T with respect to the basis
B={(1,2,1),(0,1,-1),(2,3,2)}

The answer given is
-28 -19 -43
5 4 7
18 11 28
but i have no idea how to get to that answer as the book does not provide workings/steps. Any help would be appreciated thanks.

 

[Fredrik]

I recommend that you read this post (the part above the quote) to make sure that you understand the relationship between linear operators and matrices.

 

[Rasalhague]

I'll use subscript B to indicate a vector or linear transformation expressed as a matrix in the new basis, thus

\left ( Tx \right )_B = T_B x_B

Let B be a matrix whose columns are the basis vectors of the new basis, expressed in the standard basis. Then the inverse of B will convert the components of a general vector from the standard basis to the new basis:

B^{-1}Tx = T_B B^{-1}x.

So B^{-1}T has the same effect on x as T_B B^{-1}. Now all we have to do is solve for T_B.

B^{-1}T = T_B B^{-1}[/itex]<br /> <br /> B^{-1}TB = T_B.[/itex]

 

[Redbelly98]

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[vela]

Say vectors \vec{x} and \vec{y} have the representations

<br /> \vec{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1=\begin{pmatrix}x_1&#039;\\x_2&#039;\\x_3&#039;\end{pmatrix}_2<br /> \hspace{0.5in}<br /> \vec{y}=\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}_1=\begin{pmatrix}y_1&#039;\\y_2&#039;\\y_3&#039;\end{pmatrix}_2

with respect to basis 1 and basis 2. You can construct a matrix P that will convert between the two representations:

\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1=P\begin{pmatrix}x_1&#039;\\x_2&#039;\\x_3&#039;\end{pmatrix}_2

and its inverse P^-1 will take you in the other direction:

\begin{pmatrix}x_1&#039;\\x_2&#039;\\x_3&#039;\end{pmatrix}_2=P^{-1}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1

If \vec{y}=T(\vec{x}), there are matrices A and B such that

\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}_1=A\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1<br /> \hspace{0.5in}<br /> \begin{pmatrix}y_1&#039;\\y_2&#039;\\y_3&#039;\end{pmatrix}_2=B\begin{pmatrix}x_1&#039;\\x_2&#039;\\x_3&#039;\end{pmatrix}_2<br />.

It turns out that A and B are related by B=P^{-1}AP because

\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1=P\begin{pmatrix}x_1&#039;\\x_2&#039;\\x_3&#039;\end{pmatrix}_2

\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}_1=A\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1=AP\begin{pmatrix}x_1&#039;\\x_2&#039;\\x_3&#039;\end{pmatrix}_2

\begin{pmatrix}y_1&#039;\\y_2&#039;\\y_3&#039;\end{pmatrix}_2=P^{-1}\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}_1=P^{-1}AP\begin{pmatrix}x_1&#039;\\x_2&#039;\\x_3&#039;\end{pmatrix}_2

In your problem, you're given A, and you want to find B. So the problem boils down to finding P given the information you have about the two bases.