Matrix of. Linear operator question

[frowdow]

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

 

[chiro]

Hey frowdow and welcome to the forums.

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

 

[micromass]

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

 

[HallsofIvy]

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.