\documentclass[12pt]{article}
\input{../macros.tex}
\begin{document}

\thispagestyle{plain}

\setcounter{session}{25}
\begin{center}
\textbf{Exercises on symmetric matrices and positive definiteness}
\end{center}


\begin{problem} (6.4 \#10.  {\em Introduction to Linear Algebra:} Strang) Here is a quick ``proof" that the eigenvalues of all real matrices are real:
$$\text{\textbf{False Proof:} }\; A \vb x = \lambda \vb x \text{ gives }
  \vb x^T A \vb x = \lambda \vb x^T \vb x \text{ so } \lambda =
  \frac{\vb x^T A \vb x}{\vb x^T \vb x} \text{ is real.}$$ There is a
  hidden assumption in this proof which is not justified. Find the
  flaw by testing each step on the $90\,^{\circ}$ rotation matrix:
$$\twobytwo 0 {-1} 1 0 $$
with $\lambda = i$ and $\vb x = (i,1).$
\end{problem}

\solution{
We can esily confirm that $A \vb x = \lambda \vb x = \small \twobyone
{-1}i$. Next, check if $\vb {x^T} A \vb x = \lambda \vb{x^T}
\vb x$ is true for the $90\,^{\circ}$ rotation matrix:
\begin{eqnarray*}
\vb x^T \mb A \vb x &=& \onebytwo i 1 \twobyone {-1} i = 0\\
\lambda \vb x^T \vb x &=& i \onebytwo i 1 \twobyone i 1 = 0\\
\vb x^T \mb A \vb x &=& \lambda \vb x^T \vb x. \checkmark
\end{eqnarray*}
Note that $\vb {x^T} \vb x = 0.$ Since the next and last step involves
dividing by this term, the hidden assumption must be that $\vb {x^T}
\vb x \neq 0.$ If $x = (a,b)$ then
$$\vb {x^T} \vb x = \onebytwo a b \twobyone a b = a^2 + b^2.$$ The ``proof'' assumes that the squares of the components of the
eigenvector cannot sum to zero: $a^2 + b^2 \neq 0$.  This may be false
if the components are complex.
}

\begin{problem}(6.5 \#32.) A \textit{group} of nonsingular matrices includes $AB$ and $A^{-1}$ if it includes $A$ and $B.$ ``Products and inverses stay in the group." Which of these are groups? 

\begin{enumerate}[a)]
\item Positive definite symmetric matrices $A.$

\item Orthogonal matrices $Q.$

\item All exponentials $e^{tA}$ of a fixed matrix $A.$

\item Matrices $D$ with determinant 1.
\end{enumerate}
\end{problem}

\solution{
\begin{enumerate}[a)]

\item The positive definite symmetric matrices $A$ \textbf{do not form a group.}  To show this, we provide a counterexample in the form of two positive definite symmetric matrices $A$ and $B$ whose product is not a positive definite symmetric matrix.

If $A =\small \twobytwo 2111$ and $B = \small \twobytwoc 1 {1/2} {1/2} 1$ then $AB = \small \twobytwoc {2.5} {2} {1.5} {1.5}$ is not symmetric.

\item The orthogonal matrices $Q$ \textbf{form a group}. If $A$ and $B$ are orthogonal matrices, then:
\begin{align*}
&A^T A = I \Rightarrow A^{-1} = A^T \Rightarrow A^{-1 }\text{ is orthogonal, and} \\
&B^T B = I \Rightarrow (AB)^T AB = B^T A^T AB = B^T B = I \Rightarrow AB \text{ is orthogonal.} 
\end{align*}

\item The exponentials $e^{tA}$ of a fixed matrix $A$ \textbf{form a group.} For the elements $e^{p A}$ and  $e^{q A}$: 
\begin{align*}
&(e^{pA})^{-1} = e^{-pA}  \text{ is of the form } e^{tA} \\
&e^{p A} e^{q A} = e^{(p+q) A}  \text{ is of the form } e^{tA}
\end{align*}

\item The matrices $D$ with determinant 1 \textbf{form a group.}  If $\det A = 1$ then $\det A^{-1} = 1$.  If matrices $A$ and $B$ have determinant 1 then their product also has determinant 1:
$$\det(AB) = \det(A)\det(B) = 1.$$
\end{enumerate}
}

\end{document}