# Linear Algebra (Similar Matrices)

1. Jul 20, 2009

### DanielFaraday

1. The problem statement, all variables and given/known data

Write the standard matrix representation TE for T.

2. Relevant equations

$$T\left(ax^3+bx^2+cx+d\right)=(a-b)x^2+(c-d)x+(a+b-c)$$

3. The attempt at a solution

$$[T]_E=\left( \begin{array}{c} a+b-c \\ c-d \\ a-b \\ 0 \end{array} \right)$$

I think this is right, but a subsequent problem asks me to show that this matrix is similar to another matrix. Similarity only makes sense for an n x n matrix, so this must be wrong.

Any ideas?

2. Jul 20, 2009

### Dick

The basis 'vectors' are x^3, x^2, x and 1. So ax^3+bx^2+cx+d=(a,b,c,d).This is exactly the same as other exercises you have already done, it's T(a,b,c,d)=(0,a-b,c-d,a+b-c). Sure, it's an 4x4 matrix.

3. Jul 20, 2009

### HallsofIvy

Is T here a function from the space of polynomials of degree 3 or less to itself or to the space of polynomials of degree 2 or less? If the former, the matrix is 4 by 4. If the latter, it is 4 by 3 (but the first row is all "0"s).
For example, taking x3 as the first "basis vector", T(x3)= ax2+ 0x+ a as a polynomial of degree 2 or less or 0x3+ ax2+ 0 x+ a as a polynomial of degree 3 or less. That is:
$$T\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ 0 \\ a\end{bmatrix}$$
in the first case or
$$T\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}0 \\ a \\ 0 \\ a\end{bmatrix}$$
in the second.

Do you see that those give you the first column of the matrix?

4. Jul 20, 2009

### DanielFaraday

Ah yes, I see. I was thinking of T as a vector instead of a transformation. Thanks!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook