Given a matrix block diagonal matrix as follows
$${\bf A} =
\left[ \begin{array}{cccc}
{\bf a}_1 & 0 & 0 & 0 & \\
0 &{\bf a}_2 & 0 & 0 \\
0 & 0 & {\bf a}_3& 0 \\
0 & 0 & 0& {\bf a}_4
\end{array} \right]$$
where each non zero entry ${\bf a_i}$is $M\times 1$ vector and the zero are also $M\times 1$ vectors. Is it correct to say that the product of this matrix and any other matrix for example ${\bf D}$
$${\bf P = AD} =
\left[ \begin{array}{cccc}
{\bf a}_1 & 0 & 0 & 0 & \\
0 &{\bf a}_2 & 0 & 0 \\
0 & 0 & {\bf a}_3& 0 \\
0 & 0 & 0& {\bf a}_4 \\
\end{array} \right]\left[ \begin{array}{cccc}
d_1 & d_2 & d_3 & d_4 & \\
d_5 &d_6 &d_7 &d_8\\
d_9 &d_{10} &d_{11} &d_{12} \\
d_{13} & d_{14} &d_{15} & d_{16} &
\end{array} \right]=\\
\stackrel{=}{?????} \left[ \begin{array}{cccc}
{\bf a}_1 & 0 & 0 & 0 & \\
0 &{\bf a}_2 & 0 & 0 \\
0 & 0 & {\bf a}_3& 0 \\
0 & 0 & 0& {\bf a}_4
\end{array} \right] \times \operatorname{diag} ({\bf D})$$
where $d_i$ are scalars, where $\operatorname{diag} ({\bf D})$ is a matrix with only the diagonal entries of matrix ${\bf D}$?
Thanks
Best Answer
Suppose $M=1$ (so that $A$ is $4\times4$) and the $a_i$'s are all nonzero. Then $A$ is invertible. Do you think the equality $AD=A\operatorname{diag}(D)$ always holds?