Finding T(u) given eigenvalues and eigenvectors.

[peripatein]

Hello,

Homework Statement

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.

Homework Equations

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.

 

[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!

 

[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.
I'd appreciate your help.