The matrix multiplication of square matrices of different order is often claimed to be impossible. Yet, if the order of one matrix is divisible by the order of the other, a natural multiplication rule is visible. The bigger matrix simply should be considered a "matrix of matrices" or, alternatively, in small matrix all elements should be replaced with equivalent diagonal $m\times m$ (in this case, $2\times2$) square matrices:
$\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right).\left( \begin{array}{cc} \left( \begin{array}{cc} {a_1} & {b_1} \\ {c_1} & {d_1} \\ \end{array} \right) & \left( \begin{array}{cc} {a_2} & {b_2} \\ {c_2} & {d_2} \\ \end{array} \right) \\ \left( \begin{array}{cc} {a_3} & {b_3} \\ {c_3} & {d_3} \\ \end{array} \right) & \left( \begin{array}{cc} {a_4} & {b_4} \\ {c_4} & {d_4} \\ \end{array} \right) \\ \end{array} \right)=\left( \begin{array}{cccc} a & 0 & b & 0 \\ 0 & a & 0 & b \\ c & 0 & d & 0 \\ 0 & c & 0 & d \\ \end{array} \right).\left( \begin{array}{cccc} {a_1} & {b_1} & {a_2} & {b_2} \\ {c_1} & {d_1} & {c_2} & {d_2} \\ {a_3} & {b_3} & {a_4} & {b_4} \\ {c_3} & {d_3} & {c_4} & {d_4} \\ \end{array} \right)$ $=\left( \begin{array}{cc} \left( \begin{array}{cc} a {a_1}+{a_3} b & a {b_1}+b {b_3} \\ a {c_1}+b {c_3} & a {d_1}+b {d_3} \\ \end{array} \right) & \left( \begin{array}{cc} a {a_2}+{a_4} b & a {b_2}+b {b_4} \\ a {c_2}+b {c_4} & a {d_2}+b {d_4} \\ \end{array} \right) \\ \left( \begin{array}{cc} {a_1} c+{a_3} d & {b_1} c+{b_3} d \\ c {c_1}+{c_3} d & c {d_1}+d {d_3} \\ \end{array} \right) & \left( \begin{array}{cc} {a_2} c+{a_4} d & {b_2} c+{b_4} d \\ c {c_2}+{c_4} d & c {d_2}+d {d_4} \\ \end{array} \right) \\ \end{array} \right)$
So, in certain circumstances this is possible. People on Math.Stackexchange commented that this is essentially tensor product, but TensorProduct
in Mathematica gives something different.
So, can such product be somehow realized in Mathematica? Maybe I shoud use TensorProduct in some combination with other transformation?
Also, is there a way to "expand" the matrix increasing its order this way: $\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)\to \left( \begin{array}{cccc} a & 0 & b & 0 \\ 0 & a & 0 & b \\ c & 0 & d & 0 \\ 0 & c & 0 & d \\ \end{array} \right)$
That is to an equivalent matrix of higher order (replace all elements with equivalent matrices)?
KroneckerProduct[{{a, b}, {c, d}}, IdentityMatrix[2]]
. $\endgroup$