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 $T_B$.

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

 

[Redbelly98]

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[vela]

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

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

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

$$\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1=P\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}_2$$

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

$$\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}_2=P^{-1}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1$$

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

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

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

$$\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1=P\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}_2$$

$$\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}_1=A\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1=AP\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}_2$$

$$\begin{pmatrix}y_1'\\y_2'\\y_3'\end{pmatrix}_2=P^{-1}\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}_1=P^{-1}AP\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}_2$$

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.