Similar Matrices

Say we have two

n\times n

square matrices

\(A\)

and

\(B\)

\(A\)

and

\(B\)

are similar matrices if there exists some

\(M\)

such that,

\begin{matrix} B=M^{-1}AM \end{matrix}

$\Lambda$ and $$A$$ are similar matrices.

Explanation:

Say we have a $n\times n$ matrix $$A$$ , $\vec{x}_1,\vec{x}_2,\cdots,\vec{x}_n$ are eigenvectors of $$A$$ , and $\lambda_1, \lambda_2, \cdots, \lambda_n$ are the eigenvalues of $$A$$ .
$S=\begin{bmatrix} \vdots & \vdots & \cdots & \vdots \\ \vec{x}_{1} & \vec{x}_{2} & \cdots & \vec{x}_{n} \\ \vdots & \vdots & \cdots & \vdots \\ \end{bmatrix},$ $\Lambda=\begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \\ \end{bmatrix}$
We have seen HERE that, $\Lambda =S^{-1}AS$ So, $\Lambda$ and $$A$$ are similar matrices.
$$A$$ has a family of similar matrices and $\Lambda$ is the best(simplest) similar matrix among them.

Let's take an example,
Say

A= \begin{bmatrix} 2&1\\ 1&2 \end{bmatrix}

Eigenvalues of

\(A\)

are

\(3,1\)

\Lambda= \begin{bmatrix} 3&0\\ 0&1 \end{bmatrix}

\begin{bmatrix} 2&1\\ 1&2 \end{bmatrix}

and

\begin{bmatrix} 3&0\\ 0&1 \end{bmatrix}

are similar matrices.
Let's get another similar matrix using

B=M^{-1}AM

.
Let's choose a random full rank(invertible) matrix

\(M\)

Say

M= \begin{bmatrix} 1&4\\ 0&1 \end{bmatrix}

\underbrace{\begin{bmatrix} 1&-4\\ 0&1 \end{bmatrix}}_{M^{-1}} \underbrace{\begin{bmatrix} 2&1\\ 1&2 \end{bmatrix}}_{A} \underbrace{\begin{bmatrix} 1&4\\ 0&1 \end{bmatrix}}_{M} = \underbrace{\begin{bmatrix} -2&-15\\ 1&6 \end{bmatrix}}_{B}

\begin{bmatrix} 2&1\\ 1&2 \end{bmatrix}

and

\begin{bmatrix} -2&-15\\ 1&6 \end{bmatrix}

are similar matrices.
Notice that here also eigenvalues are

\(3,1\)

Here we can see that

\underbrace{\begin{bmatrix} 2&1\\ 1&2 \end{bmatrix}}_{A},\quad

\underbrace{\begin{bmatrix} 3&0\\ 0&1 \end{bmatrix}}_{\Lambda},\quad

and

\underbrace{\begin{bmatrix} -2&-15\\ 1&6 \end{bmatrix}}_{B}

have same eigenvalues

\(3,1\)

we can verify it by

\text{trace}(\cdot) = 4

and

\text{det}(\cdot) = 3

Similar matrices have same eigenvalues

Explanation:

Say we have a $n\times n$ matrices $$A$$ , $\vec{x}_1,\vec{x}_2,\cdots,\vec{x}_n$ are eigenvectors of $$A$$ , and $\lambda_1, \lambda_2, \cdots, \lambda_n$ are the eigenvalues of $$A$$ .
So $A\vec{x}_i=\lambda_i \vec{x}$
( $MM^{-1}=\mathcal{I}_n$ )
$A\mathcal{I}_n\vec{x}_i=\lambda_i \vec{x}$
$AMM^{-1}\vec{x}_i=\lambda_i \vec{x}$
$\underbrace{(M^{-1}AM)}_{B} M^{-1}\vec{x}_i=\lambda_i M^{-1}\vec{x}$
$B \underbrace{M^{-1}\vec{x}_i}_{\text{(say) }\vec{v}}=\lambda_i \underbrace{M^{-1}\vec{x}_i}_{\text{(say) }\vec{v}}$
$M^{-1}$ is just a transformation of $\vec{x}$ from $\mathbb{R}^n$ to $\mathbb{R}^n$
So $M^{-1}\vec{x}$ is a vector in $\mathbb{R}^n$
$B \vec{v}_i=\lambda_i \vec{v}$
So now we can see that eigenvalues of $B=M^{-1}AM$ is same as eigenvalues of $$A$$
And $$($$ eigenvectors of $B)=M^{-1}$ (eigenvectors of $$A$$ )

What if we have same eigenvalues ?

If we don't have all eigenvalues to be different then, there might not be $$n$$ independent eigenvectors, so the matrix might not be diagonalizable.

Example
Say $A=\begin{bmatrix} 4&0\\ 0&4 \end{bmatrix}$ here eigenvalues are $$4,4$$ and every vector is an eigenvector because all vectors are only scaled by $$4$$ $$A$$ doesn't change the direction
Is there any matrices similar to $$A$$ ?
Say that matrix $$B$$ is similar to $$A$$ .
$B=M^{-1}AM$
$B=M^{-1}\begin{bmatrix} 4&0\\ 0&4 \end{bmatrix}M$
$B=4M^{-1}\begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}M$
$B=4M^{-1}\mathcal{I}M$
$B=4M^{-1}M$
$B=4\mathcal{I}$
$B=\begin{bmatrix} 4&0\\ 0&4 \end{bmatrix}$
So similar matrix for $$A$$ is only $$A$$ itself.

It is not necessary that if our eigenvalues repeat then we can't have similar matrices.

Example
Say $A=\begin{bmatrix} 4&1\\ 0&4 \end{bmatrix}$ here eigenvalues are $$4,4$$
It has a bunch of similar matrices, like
$\begin{bmatrix} 4&1\\ 0&4 \end{bmatrix},\quad$ $\begin{bmatrix} 4&0\\ 7&4 \end{bmatrix},\quad$ $\begin{bmatrix} 7&-2\\ 1&2 \end{bmatrix},\cdots$ these all are similar matrices.