# Transition Matrix

1. Feb 16, 2010

### 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)}

-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.

2. Feb 16, 2010

### Fredrik

Staff Emeritus
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.

3. Feb 16, 2010

### 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 $T_B$.

$$B^{-1}T = T_B B^{-1}[/itex] [tex]B^{-1}TB = T_B.[/itex] 4. Feb 16, 2010 ### Redbelly98 Staff Emeritus Moderator's note: Homework assignments or textbook style exercises for which you are seeking assistance are to be posted in the appropriate forum in our https://www.physicsforums.com/forumdisplay.php?f=152" area. This should be done whether the problem is part of one's assigned coursework or just independent study. Last edited by a moderator: Apr 24, 2017 5. Feb 16, 2010 ### vela Staff Emeritus Say vectors $\vec{x}$ and $\vec{y}$ have the representations [tex] \vec{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}_1=\begin{pmatrix}x_1'\\x_2'\\x_3'\end{pmatrix}_2 \hspace{0.5in} \vec{y}=\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}_1=\begin{pmatrix}y_1'\\y_2'\\y_3'\end{pmatrix}_2$$

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

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

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.