# Matrix of. Linear operator question

I am trying to figure out what the matrix of this linear operator would be:
T:M →AMB where A, M, B are all 2X2 matrices with respect to the standard bases of a 2x2 matrix viz. e11, e12 e21 and e22. Any ideas? Il know it should be 2X8 matrix. I am trying to teach myself Abstract algebra using Artin's book and this is listed in the problem section in one of the chapters.

Thanks,
Frowdow

## Answers and Replies

chiro
Science Advisor
Hey frowdow and welcome to the forums.

Just to clarify what are the domain and range (or codomain) of the actual mappings?

What are the images of the standard basis elements under T?

HallsofIvy
Science Advisor
Homework Helper
In general, one can find the matrix representing a given linear transformation, in a given basis, by applying the linear transformation to each basis "vector" in turn, writing the result as a linear combination of the basis vectors. The coefficients of the linear combination form the columns of the matrix.

$$e_{11}= \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}$$
If, say,
$$A= \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}$$
and
$$B= \begin{bmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{bmatrix}$$
then
$$Ae_{11}B= \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix} \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\begin{bmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{bmatrix}$$
$$= \begin{bmatrix}a_{11} & 0 \\ a_{21} & 0\end{bmatrix}\begin{bmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{bmatrix}= \begin{bmatrix}a_{11}b_{11} & a_{11}b_{21} \\ a_{21}b_{12} & a_{21}b_{22}\end{bmatrix}$$
$$= a_{11}b_{11}e_{11}+ a_{11}b_{21}e_{12}+ a_{21}b{21}e_{21}+ a_{21}b_{22}e_{22}$$
so the first 'column' consists of those four matrices. That is, the matrix is 2 by 2 but each entry is a 2 by 2 matrix so, expanded, it has 4 rows and 4 columns.