Problem. Are the following matrices similar? \[A=\begin{pmatrix}1 & -1 & 1 & -1 & 1\\ -1 & 1 & -1 & 1 & -1\\ 1 & -1 & 1 & -1 & 1\\ -1 & 1 & -1 & 1 & -1\\ 1 & -1 & 1 & -1 & 1 \end{pmatrix},\qquad B=\begin{pmatrix}1 & 0 & 1 & 0 & 1\\ 0 & 1 & 0 & 1 & 0\\ 1 & 0 & 1 & 0 & 1\\ 0 & 1 & 0 & 1 & 0\\ 1 & 0 & 1 & 0 & 1 \end{pmatrix}\;\in M_{5\times5}(\mathbb{R})\]
Proof. Let us recall the claim that if matrices are similar, they have the same rank; thus, if matrices have different ranks, they are not similar.
\[\text{rk}\text{A =\text{rk} }T_{A}=\text{dim}\left(Ax\mid x\in\mathbb{R}^{5}\right)\]
\[\text{rk}\text{B =\text{rk} }T_{B}=\text{dim}\left(Bx\mid x\in\mathbb{R}^{5}\right)\]
For \(x\in\mathbb{R}^{5}\),
\[Ax=\begin{pmatrix}1 & -1 & 1 & -1 & 1\\ -1 & 1 & -1 & 1 & -1\\ 1 & -1 & 1 & -1 & 1\\ -1 & 1 & -1 & 1 & -1\\ 1 & -1 & 1 & -1 & 1 \end{pmatrix}\begin{pmatrix}x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5} \end{pmatrix}=\begin{pmatrix}x_{1}-x_{2}+x_{3}-x_{4}+x_{5}\\ -x_{1}+x_{2}-x_{3}+x_{4}-x_{5}\\ x_{1}-x_{2}+x_{3}-x_{4}+x_{5}\\ -x_{1}+x_{2}-x_{3}+x_{4}-x_{5}\\ x_{1}-x_{2}+x_{3}-x_{4}+x_{5} \end{pmatrix}\]
so \[\text{Im}\left(T_{A}\right)=\left\{ t\begin{pmatrix}1\\ -1\\ 1\\ -1\\ 1 \end{pmatrix}\mid t=x_{1}-x_{2}+x_{3}-x_{4}+x_{5}\in\mathbb{R}\right\}\]
And since the scalar \(t=x_{1}-x_{2}+x_{3}-x_{4}+x_{5}\in\mathbb{R}\) attains every value in \(\mathbb{R}\),
\[=\text{Span}\left(\begin{pmatrix}1\\ -1\\ 1\\ -1\\ 1 \end{pmatrix}\right)\]
A sequence of one non zero vector is always linearly independent, hence this sequence is a base of \(\text{Im}\left(T_{A}\right)\) , so \(\text{rk}A=1\).
For \(x\in\mathbb{R}^{5}\),
\[Bx=\begin{pmatrix}1 & 0 & 1 & 0 & 1\\ 0 & 1 & 0 & 1 & 0\\ 1 & 0 & 1 & 0 & 1\\ 0 & 1 & 0 & 1 & 0\\ 1 & 0 & 1 & 0 & 1 \end{pmatrix}\begin{pmatrix}x_{1}\\ x_{2}\\ x_{3}\\ x_{4}\\ x_{5} \end{pmatrix}=\begin{pmatrix}x_{1}+x_{3}+x_{5}\\ x_{2}+x_{4}\\ x_{1}+x_{3}+x_{5}\\ x_{2}+x_{4}\\ x_{1}+x_{3}+x_{5} \end{pmatrix}\]
so \[\text{Im}\left(T_{B}\right)=\left\{ t\begin{pmatrix}1\\ 0\\ 1\\ 0\\ 1 \end{pmatrix}+k\begin{pmatrix}0\\ 1\\ 0\\ 1\\ 0 \end{pmatrix}\mid t=x_{1}+x_{3}+x_{5}\in\mathbb{R},k=x_{2}+x_{4}\right\}\]
These scalars reach every value in \(\mathbb{R}\) so
\[=\text{Span }\left\{ \begin{pmatrix}1\\ 0\\ 1\\ 0\\ 1 \end{pmatrix},\begin{pmatrix}0\\ 1\\ 0\\ 1\\ 0 \end{pmatrix}\right\}\]
It’s obvious this sequence is linearly independent (recall that a sequence of two vectors is linearly dependent if and only if one vector is a scalar multiple of the other):
\[\forall x\in\mathbb{R}\quad x\begin{pmatrix}1\\ 0\\ 1\\ 0\\ 1 \end{pmatrix}=\begin{pmatrix}x\\ 0\\ x\\ 0\\ x \end{pmatrix}\neq\begin{pmatrix}0\\ 1\\ 0\\ 1\\ 0 \end{pmatrix},\quad\wedge\quad\forall x\in\mathbb{R}\quad x\begin{pmatrix}0\\ 1\\ 0\\ 1\\ 0 \end{pmatrix}=\begin{pmatrix}0\\ x\\ 0\\ x\\ 0 \end{pmatrix}\neq\begin{pmatrix}1\\ 0\\ 1\\ 0\\ 1 \end{pmatrix}\]
Therefore it’s a basis for \(\text{Im}\left(T_{B}\right)\), and \(\text{rk}B=2\neq\text{rk}A\) which means these matrices are not similar. ∎
This is an exam problem I translated and solved. The original exam was written by the Linear Algebra 1 course lecturers at HUJI.