Computing [T]_B and determine whether Beta is a basis consisting of eigenvectors

**alice8675309** · November 16th 2010, 07:30 PM

For the following linear operators T on a vector space V and ordered basis B (for beta), compute [T]_B, and determine whether B is a basis consisting of eigenvectors of T.

V=P_2(R), T(a+bx+cx^2)=(-4a+2b-2c)-(7a+3b+7c)x+(7a+b+5c)x^2, and B={x-x^2, -1+x^2, -1-x+x^2}

Can someone lead me in the right direction on getting this problem started? I'm pretty decent at the problem when it is given in matrix form and in column vector form (the basis) however, I don't know how to get this one started and/or set it up.

**topspin1617** · November 16th 2010, 08:16 PM

The first thing to do is to compute $\left[ T\right]_B$ , the matrix for the linear transformation $T$ with respect to the basis $B$ .

It will be a 3 by 3 matrix, as 3 is the dimension of our vector space. Each row and column will stand for a basis vector; lets name them $b_1=x-x^2,b_2=-1+x^2,b_3=-1-x+x^2$ . The columns and rows of the matrix will correspond to our basis vectors; say the columns and rows correspond to $b_1,b_2,b_3$ respectively. We need to do the following:

Find the image of each basis vector under $T$ . Write these images out in terms of our basis. Then we have $\left[ T\right]_B=(t_{ij})$ ; that is, the entry in column j, row i of $\left[ T\right]_B$ is the coefficient of $b_i$ in the expression for $T(b_j)$ .

Let's try it for our first basis vector $b_1=x-x^2$ :

$T(x-x^2)=4-4x-4x^2=-8(-1+x^2)+4(-1-x+x^2)=0b_1-8b_2+4b_3$

Thus, the first column in the matrix should be $\left( \begin{array}{c}0\\-8\\4\end{array}\right)$

**alice8675309** · November 16th 2010, 08:23 PM

wait, i'm confused. I thought i was going to have to put this information into a matrix and then go through the steps to get [T]_B with the eigenvectors. but continuing your process: actually, i seem to be completely confused..

**HallsofIvy** · November 17th 2010, 02:42 AM

That is what topspin1617 is telling you to do- put the information into a matrix. And he has shown you how to get one column. Use the same idea to get the other columns. I don't know what you mean by "go through the steps to get [T]_B". What he is getting is [T]_B.

**alice8675309** · November 17th 2010, 11:12 AM

Originally Posted by topspin1617

The first thing to do is to compute $\left[ T\right]_B$ , the matrix for the linear transformation $T$ with respect to the basis $B$ .

It will be a 3 by 3 matrix, as 3 is the dimension of our vector space. Each row and column will stand for a basis vector; lets name them $b_1=x-x^2,b_2=-1+x^2,b_3=-1-x+x^2$ . The columns and rows of the matrix will correspond to our basis vectors; say the columns and rows correspond to $b_1,b_2,b_3$ respectively. We need to do the following:

Find the image of each basis vector under $T$ . Write these images out in terms of our basis. Then we have $\left[ T\right]_B=(t_{ij})$ ; that is, the entry in column j, row i of $\left[ T\right]_B$ is the coefficient of $b_i$ in the expression for $T(b_j)$ .

Let's try it for our first basis vector $b_1=x-x^2$ :

$T(x-x^2)=4-4x-4x^2=-8(-1+x^2)+4(-1-x+x^2)=0b_1-8b_2+4b_3$

Thus, the first column in the matrix should be $\left( \begin{array}{c}0\\-8\\4\end{array}\right)$

I'm sorry, I was confusing it with another problem that I did that was given in matrix form. However, I'm a little confused as to how you got 4-4x-4x^2. I think I'm missing something that your doing.

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