# Homework Help: Transition matrix question.

1. Jan 13, 2013

### peripatein

Hello,
1. The problem statement, all variables and given/known data
D:R3[x]->R3[x] is defined thus for any p(x)=(a0)+(a1)x+(a2)x2+(a3)x3:
D(p(x)) = a1 + (2a2)x + (33)2
I am asked to find [D]B where B is the standard basis {1,x,x2,x3}
I am then asked to find the transition matrix from B to C, where C={1,1+x,x+x2,x2+x3}.
Based on these two I am then asked to find [D]C.

2. Relevant equations

3. The attempt at a solution
I have found [D]B to be (0 1 0 0)T (0 0 2 0)T (0 0 0 3)T (0 0 0 0)T
I have found the transition matrix to be (1 -1 1 -1)T (0 1 -1 1)T (0 0 1 -1)T (0 0 0 1)T
But then, the multiplication of (0 1 0 0)T (0 0 2 0)T (0 0 0 3)T (0 0 0 0)T by (1 -1 1 -1)T (0 1 -1 1)T (0 0 1 -1)T (0 0 0 1)T does not yield the expected (0 1 -1 1)T (0 0 2 -1)T (0 0 0 3)T (0 0 0 0)T
Any ideas where I might be wrong? I have gone over the algebra several times, and have tried multiplying in the opposite order too.

2. Jan 13, 2013

### stripes

Hi there,

The linear transformation D(p(x)) = a1 + (2a2)x + (33)2; is this correct, or should it read D(p(x)) = a1 + (2a2)x + (3a3)x2? If my assumption is correct, then I can help you with your question. If you are not mistaken, then I can't really help because I am unfamiliar with that notation.

Anyways, I'll chip in assuming D(p(x)) = a1 + (2a2)x + (3a3)x2. I started by finding the transition matrix from the B basis to the C basis. When you say: (1 -1 1 -1)T (0 1 -1 1)T (0 0 1 -1)T (0 0 0 1)T, I guess you mean the transpose of (a b c d) where each (a b c d) is the column. I think you got the columns in opposite order. For the transition matrix, I got:

\begin{bmatrix}
1 & -1 & 1 & -1\\
0 & 1 & -1 & 1 \\
0 & 0 & 1 & -1 \\
0 & 0 & 0 & 1
\end{bmatrix}.

When asked for [D]B, you're trying to find the matrix of the linear transformation, right? Again, this is assuming the linear transformation I wrote above: D(p(x)) = a1 + (2a2)x + (3a3)x2, but I get

\begin{bmatrix}
0 & 1 & -2 & -3\\
0 & 0 & 2 & -3 \\
0 & 0 & 0 & 3 \\
0 & 0 & 0 & 0
\end{bmatrix},

since the coordinate vectors of the linear transformation of each member of the B basis with respect to the C basis are the above matrice's column vectors; i.e.,

[D(1)]C =

\begin{bmatrix}
0\\
0 \\
0 \\
0
\end{bmatrix},

[D(x)]C =

\begin{bmatrix}
1\\
0 \\
0 \\
0
\end{bmatrix},

[D(x2)]C =

\begin{bmatrix}
-2\\
2 \\
0 \\
0
\end{bmatrix}, and

[D(x3)]C =

\begin{bmatrix}
3\\
-3 \\
3 \\
0
\end{bmatrix}.

I only have enough time to compute those two matrices. Either way let me know if you wrote the linear transformation correctly so I can determine if what I said was pointless or possibly helpful!

3. Jan 14, 2013

### peripatein

Hi,
I have managed to sort it out myself. Regardless, thank you!
However, I have by now come across the following difficulty: finding T(u) where u = [-7 2 9]T, given T:R3->R3 a linear transformation satisfying:
v1=[-2 1 4]T eignevector of T whose eigenvalue is p1=2
AND
v2=[3 0 -1]T eignevector of T whose eigenvalue is p2=5

Attempt at solution:
Clearly, T[-2 1 4]T = 2[-2 1 4]T = [-4 2 8]T and T[3 0 -1]T = 5[3 0 -1]T = [15 0 -5]T
But I am not sure how to continue and derive T(u) based on the above.