232_7_Thomas_Breydo.tex

\documentclass{amsart}
\usepackage{thomas} % my style file, https://git.io/thomas.sty


\newcommand{\pagenum}{232}
\newcommand{\probnum}{7}

\title{\pagenum.\probnum}
\author{Thomas\ Breydo}

\begin{document}

\maketitle

\begin{problem*}
Suppose $T$ is a positive operator on $V.$ Prove that $T$ 
is invertible if and only if
\begin{align*}
    \iprod{Tv,v}>0
\end{align*}
for all nonzero $v\in V.$
\end{problem*}

\vspace{0.5in}

\begin{claim*}
If $\iprod{Tv,v}>0$ for all nonzero $v\in V,$
then $T$ is invertible.
\end{claim*}
\begin{proof}
Suppose $\iprod{Tv,v}>0$ for all nonzero $v\in V.$
For any $v\in V$ where $Tv=0,$
\begin{align*}
\iprod{Tv,v}=0,
\end{align*}
which implies $v=0,$ since $\iprod{Tv,v}>0$ for all nonzero $v.$ 
Since $Tv=0$ implies $v=0,$ $T$ is injective and
thus invertible.
\end{proof}

\begin{claim*}
If $T$ is invertible, then $\iprod{Tv,v}>0$ for all nonzero
$v\in V.$
\end{claim*}
\begin{proof}
Since $T$ is positive, $\iprod{Tv,v}\ge0$ for all $v\in V.$
Thus, it suffices
to prove that $\iprod{Tv,v}\ne0$ for all nonzero $v\in V.$

Suppose $v\in V$ is nonzero.
Then, $Tv\ne 0$ because $T$ is invertible.
If $\sqrt Tv=0$ then $Tv=0,$ so $\sqrt Tv\ne 0.$
Thus, by the definiteness of the inner product,
\begin{align*}
    \iprod*{\sqrt Tv,\sqrt Tv}>0.
\end{align*}
But
\begin{align*}
    \iprod*{\sqrt Tv,\sqrt Tv} &= \iprod*{\sqrt Tv,\sqrt T^*v} \\
                  &= \iprod*{\sqrt T\sqrt Tv,v} \\
                  &= \iprod{Tv,v}, 
\end{align*}
so $\iprod{Tv,v}>0$ for nonzero $v.$
\end{proof}

\vspace{0.5in}

\begin{note*}
You can view the source code for this solution
\href{https://github.com/thomasbreydo/linalg/blob/main/\pagenum_\probnum_Thomas_Breydo.tex}
{here}.
\end{note*}

\end{document}