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L07E03.tex
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L07E03.tex
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\documentclass[solutions.tex]{subfiles}
\xtitle
\begin{document}
\maketitle
\begin{exercise} a) Rewrite Eq. $7.10$ in component form,
replacing the symbols $A$, $B$, $a$, and $b$ with the matrices
and column vectors from Eqs. $7.7$ and $7.8$. \\
b) Perform the matrix multiplications $Aa$ and $Bb$ on the
right-hand side. Verify that each result is a $4\times1$ matrix. \\
c) Expand all three Kronecker products. \\
d) Verify the row and column sizes of each Kronecker product:
\begin{itemize}
\item $A\otimes B$ : $4\times4$
\item $a\otimes b$ : $4\times1$
\item $Aa\otimes Bb$ : $4\times 1$
\end{itemize}
e) Perform the matrix multiplication on the left-hand side,
resulting in a $4\times1$ column vector. Each row should be the
sum of four separate terms. \\
f) Finally, verify that the resulting column vectors on the left
and right sides are identical.
\end{exercise}
Recall Eq. $7.10$
\[
(A\otimes B)(a\otimes b) = (Aa\otimes Bb)
\]
And Eq. $7.7$ and $7.8$:
\[
A\otimes B = \begin{pmatrix}
A_{11}B_{11} & A_{11}B_{12} & A_{12}B_{11} & A_{12}B_{12} \\
A_{11}B_{21} & A_{11}B_{22} & A_{12}B_{21} & A_{12}B_{22} \\
A_{21}B_{11} & A_{21}B_{12} & A_{22}B_{11} & A_{22}B_{12} \\
A_{21}B_{21} & A_{21}B_{22} & A_{22}B_{21} & A_{22}B_{22} \\
\end{pmatrix};\qquad
\begin{pmatrix}
a_{11} \\
a_{21} \\
\end{pmatrix}\otimes\begin{pmatrix}
b_{11} \\
b_{21} \\
\end{pmatrix} = \begin{pmatrix}
a_{11}b_{11} \\
a_{11}b_{21} \\
a_{21}b_{11} \\
a_{21}b_{21} \\
\end{pmatrix}
\]
Our goal is to prove Eq. $7.10$ by following all the recommended steps. It's
a bit tedious, but otherwise presents no major difficulties. \\
\hrr
a) Let's rewrite the equation (that's still to be proved)
in component form:
\[
(A\otimes B)(a\otimes b) = (Aa\otimes Bb)
\]
\[
\Leftrightarrow
\left(\begin{pmatrix}
A_{11} & A_{12} \\
A_{21} & A_{22} \\
\end{pmatrix}\otimes\begin{pmatrix}
B_{11} & B_{12} \\
B_{21} & B_{22} \\
\end{pmatrix}\right)\left(
\begin{pmatrix}
a_{11} \\
a_{21} \\
\end{pmatrix}\otimes\begin{pmatrix}
b_{11} \\
b_{21} \\
\end{pmatrix}\right) = \left(
\left(\begin{pmatrix}
A_{11} & A_{12} \\
A_{21} & A_{22} \\
\end{pmatrix}\begin{pmatrix}
a_{11} \\
a_{21} \\
\end{pmatrix}\right)
\otimes
\left(\begin{pmatrix}
B_{11} & B_{12} \\
B_{21} & B_{22} \\
\end{pmatrix}\begin{pmatrix}
b_{11} \\
b_{21} \\
\end{pmatrix}\right)
\right)
\]
\hrr
b) Let's expand $Aa$ and $Bb$:
\[
Aa = \begin{pmatrix}
A_{11} & A_{12} \\
A_{21} & A_{22} \\
\end{pmatrix}\begin{pmatrix}
a_{11} \\
a_{21} \\
\end{pmatrix} = \begin{pmatrix}
A_{11}a_{11}+A_{12}a_{21} \\
A_{21}a_{11}+A_{22}a_{21} \\
\end{pmatrix};\quad
Bb = \begin{pmatrix}
B_{11} & B_{12} \\
B_{21} & B_{22} \\
\end{pmatrix}\begin{pmatrix}
b_{11} \\
b_{21} \\
\end{pmatrix} = \begin{pmatrix}
B_{11}b_{11}+B_{12}b_{21} \\
B_{21}b_{11}+B_{22}b_{21} \\
\end{pmatrix};
\]
From Eqs. $7.7$ and $7.8$, we can see that all Kronecker
products indeed expand to $4\times1$ matrices.
Equation $7.10$ is then equivalent to:
\[
\left(\begin{pmatrix}
A_{11} & A_{12} \\
A_{21} & A_{22} \\
\end{pmatrix}\otimes\begin{pmatrix}
B_{11} & B_{12} \\
B_{21} & B_{22} \\
\end{pmatrix}\right)\left(
\begin{pmatrix}
a_{11} \\
a_{21} \\
\end{pmatrix}\otimes\begin{pmatrix}
b_{11} \\
b_{21} \\
\end{pmatrix}\right) = \left(
\begin{pmatrix}
A_{11}a_{11}+A_{12}a_{21} \\
A_{21}a_{11}+A_{22}a_{21} \\
\end{pmatrix}\otimes\begin{pmatrix}
B_{11}b_{11}+B_{12}b_{21} \\
B_{21}b_{11}+B_{22}b_{21} \\
\end{pmatrix}
\right)
\]
\hrr
c), d), e), f) I'll be mixing all those steps together, because
this is fairly trivial. First, $A\otimes B$ and $a\otimes b$ are
respectively Eqs. $7.7$ and $7.8$. This gives us already:
\[
\begin{pmatrix}
A_{11}B_{11} & A_{11}B_{12} & A_{12}B_{11} & A_{12}B_{12} \\
A_{11}B_{21} & A_{11}B_{22} & A_{12}B_{21} & A_{12}B_{22} \\
A_{21}B_{11} & A_{21}B_{12} & A_{22}B_{11} & A_{22}B_{12} \\
A_{21}B_{21} & A_{21}B_{22} & A_{22}B_{21} & A_{22}B_{22} \\
\end{pmatrix}\begin{pmatrix}
a_{11}b_{11} \\
a_{11}b_{21} \\
a_{21}b_{11} \\
a_{21}b_{21} \\
\end{pmatrix} = \left(
\begin{pmatrix}
A_{11}a_{11}+A_{12}a_{21} \\
A_{21}a_{11}+A_{22}a_{21} \\
\end{pmatrix}\otimes\begin{pmatrix}
B_{11}b_{11}+B_{12}b_{21} \\
B_{21}b_{11}+B_{22}b_{21} \\
\end{pmatrix}
\right)
\]
It remains to expand the last Kronecker product, for which we
can use $7.8$:
\[
\begin{pmatrix}
A_{11}B_{11} & A_{11}B_{12} & A_{12}B_{11} & A_{12}B_{12} \\
A_{11}B_{21} & A_{11}B_{22} & A_{12}B_{21} & A_{12}B_{22} \\
A_{21}B_{11} & A_{21}B_{12} & A_{22}B_{11} & A_{22}B_{12} \\
A_{21}B_{21} & A_{21}B_{22} & A_{22}B_{21} & A_{22}B_{22} \\
\end{pmatrix}\begin{pmatrix}
a_{11}b_{11} \\
a_{11}b_{21} \\
a_{21}b_{11} \\
a_{21}b_{21} \\
\end{pmatrix} = \begin{pmatrix}
(A_{11}a_{11}+A_{12}a_{21})(B_{11}b_{11}+B_{12}b_{21}) \\
(A_{11}a_{11}+A_{12}a_{21})(B_{21}b_{11}+B_{22}b_{21}) \\
(A_{21}a_{11}+A_{22}a_{21})(B_{11}b_{11}+B_{12}b_{21}) \\
(A_{21}a_{11}+A_{22}a_{21})(B_{21}b_{11}+B_{22}b_{21}) \\
\end{pmatrix} \\
\]
\[
= \begin{pmatrix}
A_{11}B_{11}a_{11}b_{11}
+ A_{11}B_{12}a_{11}b_{21}
+ A_{12}B_{11}a_{21}b_{11}
+ A_{12}B_{12}a_{21}b_{21} \\
%
A_{11}B_{21}a_{11}b_{11}
+A_{11}B_{22}a_{11}b_{21}
+A_{12}B_{21}a_{21}b_{11}
+A_{12}B_{22}a_{21}b_{21} \\
%
A_{21}B_{11}a_{11}b_{11}
+A_{21}B_{12}a_{11}b_{21}
+A_{22}B_{11}a_{21}b_{11}
+A_{22}B_{12}a_{21}b_{21} \\
%
A_{21}B_{21}a_{11}b_{11}
+A_{21}B_{22}a_{11}b_{21}
+A_{22}B_{21}a_{21}b_{11}
+A_{22}B_{22}a_{21}b_{21} \\
\end{pmatrix}
\]
And it's now trivial to verify that this holds, as expected. $\qed$
\end{document}