# Transformation matrix with respect to two bases?

1. Dec 11, 2013

### PhizKid

1. The problem statement, all variables and given/known data
Let $S = \{1, e^x, e^{-x}, e^{2x}, e^{-2x}\}$ and $B = \{1, sinh(x),cosh(x), sinh(2x), cosh(2x)\}$. S spans the vector space V, and a linear transformation T: V -> V is defined by T(y) = y'' - 3y' - 4y.

(a) Find the representation matrix of T with respect to the bases S and B.
(b) Use the change of basis matrix (transition matrix) P, from S to B, to find the representation matrix of T with respect to the bases B and B.

3. The attempt at a solution
(a) I don't understand what it means to find a single transformation matrix for two different bases.

(b) I already found the change of basis matrix P from S to B by representing the elements in S in terms of the ones in B and set them as columns of the matrix P. But again, I don't understand what it means to use this change of basis to find a transformation matrix for two different bases, but this time the two bases are the same bases?

2. Dec 11, 2013

### vela

Staff Emeritus
You have two vectors related by $\vec{y} = T(\vec{x})$. You're used to working with one basis to represent both $\vec{x}$ and $\vec{y}$ and find the matrix of T that relates the coordinate vectors, but you don't have to use the same basis for both. One vector could be represented relative to B while the other is relative to S. Since you'll have a different coordinate vector for one of the vectors, the matrix for T will be different.

3. Dec 11, 2013

### PhizKid

Then I really don't understand the questions because it's asking for a transformation matrix with respect to both S and B, and then another transformation matrix with respect to B and also B again?

4. Dec 11, 2013

### vela

Staff Emeritus
You want to find matrices TBB and TSB (or TBS — the question isn't clear). They're related by P.

TSB, for example, takes the coordinates of $\vec{x}$ wrt to basis B and gives you the coordinates of $\vec{y}$ wrt basis S.

TBB takes the coordinates of $\vec{x}$ wrt to basis B and gives you the coordinates of $\vec{y}$ wrt basis B.

Sorry...I have to run...I'll post a bit later if someone else already hasn't.

5. Dec 11, 2013

### PhizKid

Okay, for T(y) = y'' - 3y' - 4y, I found the matrix that takes S and transforms it into coordinates wrt S. I just took the derivatives of each element of S and plugged them into T(y) and found $\{-4, -6e^x, 0, -6e^{2x}, 6e^{-2x}\}$ so I just put each element in its respective positive in each column.

So for part (a), I want to take S, transform the basis into B by using P, and then somehow get a transformation for this new basis B, but I only have a transformation that takes me from S to S. What should I do here?

6. Dec 11, 2013

### vela

Staff Emeritus
Let me use a simple example with two basis vectors. The two bases are S={sinh x, cosh x} and B={ex, e-x}. Suppose we have the linear transformation T(y) = y'.

cosh x = T(sinh x) would correspond to
\begin{align*}
\begin{pmatrix} 0 \\ 1 \end{pmatrix}_S &= T_{SS}\begin{pmatrix} 1 \\ 0 \end{pmatrix}_S \\
\begin{pmatrix} 1/2 \\ 1/2 \end{pmatrix}_B &= T_{BS}\begin{pmatrix} 1 \\ 0 \end{pmatrix}_S \\
\begin{pmatrix} 0 \\ 1 \end{pmatrix}_S &= T_{SB}\begin{pmatrix} 1/2 \\ -1/2 \end{pmatrix}_B \\
\begin{pmatrix} 1/2 \\ 1/2 \end{pmatrix}_B &= T_{BB}\begin{pmatrix} 1/2 \\ -1/2 \end{pmatrix}_B
\end{align*} depending on which basis you use for the domain and which basis you use for the range.

7. Dec 11, 2013

### PhizKid

For the 2nd transformation you listed, how do I obtain that $T_{BS}$? Is it the same thing as changing the basis S into basis B, then applying the transformation with respect to B to obtain the same answer?

I think I got part (b) by doing this: $T_{BB} = PT_{SS}P^{-1}$ where $T_{SS}$ is the transformation I listed above

8. Dec 11, 2013

### vela

Staff Emeritus
You have
$$\sinh x = (1) \sinh x + (0) \cosh x = \left(\frac{1}{2}\right) e^x + \left(-\frac{1}{2}\right) e^{-x}$$ so with respect to the two bases, the representation for sinh x is $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 1/2 \\ -1/2 \end{pmatrix}$. You can do the same thing for cosh x. I just plugged in the various representations for sinh x and cosh x to get the four equations.

9. Dec 11, 2013

### vela

Staff Emeritus
If you're asking how to get the matrix itself, the second equation tells you the first column of TBS is $\begin{pmatrix} 1/2 \\ 1/2 \end{pmatrix}$. If you don't see this, just multiply in the vector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ into a 2x2 matrix, and you'll see you get the first column. You can get the second column similarly, by multiplying by $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$.

10. Dec 11, 2013

### PhizKid

For the second equation, you have $(1,0)T_{BS} = (1/2, 1/2)$. I was asking how to obtain the general matrix $T_{BS}$ to directly transform something in S directly to some coordinates in B.

11. Dec 11, 2013

### PhizKid

When something says a matrix is respect to 'B' how can you tell if it means with respect to 'S to B' or 'B to B'? It says that when T is a transformation with respect to 'S', $PTP^{-1}$ is = to the transformation with respect to B.

12. Dec 11, 2013

### vela

Staff Emeritus
If only one basis is specified, you can assume both the domain and range are expressed in this basis.

In your example, matrix $P$ maps the representation of a vector wrt to S to its representation wrt B. $P^{-1}$ goes in the opposite direction. So you have
\begin{align*}
\vec{x}_S &= P^{-1}\vec{x}_B \\
\vec{y}_S &= T\vec{x}_S = TP^{-1}\vec{x}_B \\
\vec{y}_B &= P\vec{y}_S = PTP^{-1}\vec{x}_B
\end{align*} You can therefore identify $PTP^{-1}$ as the matrix of T with respect to B.

13. Dec 11, 2013

### PhizKid

I am still having trouble understanding how to get the transformations. From your example here: https://www.physicsforums.com/showpost.php?p=4599971&postcount=6

The 3rd equation from the top. It means that some 2x2 transformation matrix multiplied on the left side of some arbitrary vector in B will equal to some arbitrary vector in S, correct? So my goal here is to find that 2x2 transformation matrix. However, in order to solve for this 2x2 transformation matrix, I need the inverse of a 2x1 matrix which is impossible. So how can I solve for the 2x2 matrix $T_{SB}$ in your example?

14. Dec 12, 2013

### vela

Staff Emeritus
Do you know how to find the matrix for T(y)=y' if you're just dealing with one basis, say S={sinh x, cosh x}?