%% LyX 2.4.4 created this file.  For more info, see https://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[english]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\setcounter{secnumdepth}{-2}
\setlength{\parindent}{0bp}
\usepackage{amsmath}
\usepackage{amsthm}

\makeatletter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Textclass specific LaTeX commands.
\theoremstyle{definition}
\newtheorem*{problem*}{\protect\problemname}

\@ifundefined{date}{}{\date{}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
\usepackage{graphicx}
\usepackage{geometry}
\newcommand{\limi}{\underset{n\rightarrow\infty}{\lim}}
\newcommand{\limxz}{\underset{x\rightarrow x_{0}}{\lim}}
\newcommand{\limfxz}{\underset{x\rightarrow x_{0}}{\lim}f(x)}
\newcommand{\limfxm}{\underset{x\rightarrow x_{0}^{-}}{\lim}f(x)}
\newcommand{\limfxp}{\underset{x\rightarrow x_{0}^{+}}{\lim}f(x)}
\newcommand{\limxm}{\underset{x\rightarrow x_{0}^{-}}{\lim}}
\newcommand{\limxp}{\underset{x\rightarrow x_{0}^{+}}{\lim}}
\newcommand{\sqrm}{M_{m\times m}(\mathbb{F})}
\newcommand{\seqa}{(a_{n})_{n=1}^{\infty}}
\newcommand{\seqx}{(x_{n})_{n=1}^{\infty}}
\newcommand{\funcdr}{f:D\rightarrow\mathbb{R}}
\newcommand{\der}{\limxz\frac{f(x)-f(x_{0})}{x-x_{0}}}
\usepackage[all]{pdfprivacy}

\makeatother

\usepackage{babel}
\providecommand{\problemname}{Problem}
\begin{document}
\title{{\Large Semester 2026A, Exam A, Q7\vspace{-2em}}}

\maketitle
\begin{problem*}
Let $V$ be a vector space with $\dim V=n$, and
let $T\colon V\to V$ be a linear map such that $\operatorname{Im}(T)\cap\ker(T)\ne\{0_{V}\}$.
Prove that $\operatorname{rk}(T\circ T)\le n-2$.
\end{problem*}

\medskip{}

\rule[0.5ex]{1\columnwidth}{1pt}

\medskip{}

\begin{proof}
\[
\text{rk}(T\circ T)=\dim\left(\text{Im}(T\circ T)\right)=\dim\left(\left\{ T\left(T\left(v\right)\right)\mid v\in V\right\} \right)=\dim\left(\left\{ T(v)\mid v\in\text{Im}T\right\} \right)
\]

Let us look at the function $T':\operatorname{Im}(T)\rightarrow V$
which is defined by $\forall x\in\operatorname{Im}(T),T'(x)=T(x)$
(It's also a linear transformation from inheritance)

\[
x\in\text{ker}(T')\iff x\in\operatorname{Im}(T)\wedge T'(x)=0\iff x\in\operatorname{Im}(T)\wedge T(x)=0\iff x\in\operatorname{Im}(T)\wedge x\in\ker(T)
\]

Then 
\[
\text{ker}(T')=\operatorname{im}(T)\cap\ker(T)\ne\{0_{V}\}
\]
Which means 
\[
\dim\left(\text{ker}(T')\right)\ge1
\]

Furthermore, we know that $\operatorname{Im}(T)\cap\ker(T)\ne\{0_{V}\}$
so in particular $\ker(T)\ne\{0_{V}\}$ and $\dim\left(\text{ker}(T)\right)\ge1$
as well.

According to the second dimensions theorem,

\[
\dim\left(\text{Im}(T)\right)=\dim\left(V\right)-\dim\left(\text{ker}(T)\right)\le n-1
\]

and

\[
\dim\left(\text{Im}(T')\right)=\dim\left(\operatorname{Im}(T)\right)-\dim\left(\text{ker}(T')\right)\le\dim\left(\operatorname{Im}(T)\right)-1\le n-1-1=n-2
\]

Let us notice that 

\[
\operatorname{Im}(T')=\left\{ T'(x)\mid x\in\operatorname{Im}(T)\right\} =\left\{ T(x)\mid x\in\operatorname{Im}(T)\right\} =\left\{ T\left(T\left(x\right)\right)\mid x\in V\right\} =\text{Im}(T\circ T)
\]
Therefore

\[
\text{rk}(T\circ T)=\dim\left(\text{Im}(T\circ T)\right)=\dim\left(\text{Im}(T')\right)\le n-2
\]
\end{proof}
\begin{quote}
This is an exam problem I translated and solved. The original exam
was written by the Linear Algebra 1 course lecturers at HUJI.
\end{quote}

\end{document}
