I have a block diagonal matrix $\mathbf{A}$ of size $ NK\times K$, whose elements are copies of a column vector $\mathbf{a}=[a_1, \dots, a_n]^T$ with $n=1,\dots,N$ and of size $N \times 1$. Such a block diagonal matrix reads as follows:
$\mathbf{A}=\begin{bmatrix}
\mathbf{a} & 0 & \dots & 0\\
0 & \mathbf{a} & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \dots & \mathbf{a} \\
\end{bmatrix}=
\begin{bmatrix}
a_1 & 0 & \dots & 0\\
a_2 & 0 & \dots & 0\\
\vdots & 0 & \dots & 0\\
a_N & 0 & \dots & 0\\
0 & a_1 & \dots & 0\\
0 & a_2 & \dots & 0\\
0 & \vdots & \dots & 0\\
0 & a_N & \dots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \dots & a_1 \\
0 & 0 & \dots & a_2 \\
0 & 0 & \dots & \vdots \\
0 & 0 & \dots & a_N \\
\end{bmatrix}.
$
Is it possible to factor out the vector $\mathbf{a}$, or $\mathbf{a}^T$ from $\mathbf{A}$ ?
I would like to have either $\mathbf{a}$, or $\mathbf{a}^T$ as the rigthmost or leftmost term.
Best Answer
If I understand the question correctly, you would like to write $\mathbf A = \mathbf a \cdot {?}$, $\mathbf A = \mathbf a^T \cdot {?}$, $\mathbf A = {?} \cdot \mathbf a$ or $\mathbf A = {?} \cdot \mathbf a^T$.
Since $\mathbf A$ is a $NK\times K$ matrix and $\mathbf a$ is $N\times 1$, neither of these factorizations can work with a usual matrix product, where “$(k\times l)\cdot(l\times m)=k\times m$”.
However, your matrix $\mathbf A$ can be written as a Kronecker product: $$ \mathbf A = \mathbf I_k \otimes \mathbf a. $$ Here $\mathbf I_k$ denotes the $k\times k$ identity matrix.