# Math Help - Matrix of a Linear Transformation

1. ## Matrix of a Linear Transformation

L: P2 --> M22 given by

L(ax^3 + bx^2 + cx + d) = Matrix
 -3a - 2c -b + 4d 4b - c + 3d -6a - b + 2d

For this linear transformation, what is the matrix for L with respect to the standard bases?

Is it the 4x4
 -3 0 -2 0 0 -1 0 4 0 4 -1 3 -6 -1 0 2
?

Thanks,

2. ## Re: Matrix of a Linear Transformation

Assume you mean P3 since that is your example. The standard basis for P3 is $\{x^3, x^2, x, 1\}$ and, applying L to each of those,
$L(x^3)= \begin{bmatrix}-3(1)- 2(0) & -0+ 4(0)\\ 4(0)- 0+ 3(0) & -6(1)- (0)+ 2(0)\end{bmatrix}= \begin{bmatrix}-3 & 0 \\ 0 - 6\end{bmatrix}$
$L(x^2)= \begin{bmatrix}-3(0)- 2(0) & -(1)+ 4(0) \\ 4(1)- 0+ 3(0) & -6(0)- 1+ 2(0)\end{bmatrix}= \begin{bmatrix}0 & -1 \\ 4 & -1\end{bmatrix}$
$L(x)= \begin{bmatrix}-3(0)- 2(1) & -(0)+ 4(0) \\ 4(0)- 1+ 3(0) & -6(0)- 0+ 2(0)\end{bmatrix}= \begin{bmatrix}-2 & 0 \\ -1 &U 0\end{bmatrix}$
$L(1)= \begin{bmatrix}-3(0)- 2(0)& -0+ 4(1) \\ 4(0)- 0+ 3(1) & -6(0)- 0+ 2(1)\end{bmatrix}= \begin{bmatrix}0 & 4 \\ 3 & 2\end{bmatrix}$

A basis for $M_2$ is
[tex]\{M1= \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix},M2= \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}, M3=\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}, M4=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}
s $L(x^3)= -3M1_+ 0M2+ 0M3- 6M4$
so the first column is $\begin{bmatrix}-3 \\ 0 \\ 0 \\ -6\end{bmatrix}$.

Yes, I get the same thing you do.