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,

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

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

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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 $3,1$ so $\Lambda= \begin{bmatrix} 3&0\\ 0&1 \end{bmatrix}$
So $\begin{bmatrix} 2&1\\ 1&2 \end{bmatrix}$ 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 $M= \begin{bmatrix} 1&4\\ 0&1 \end{bmatrix}$
So

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

$\quad$
So $\begin{bmatrix} 2&1\\ 1&2 \end{bmatrix}$ and $\begin{bmatrix} -2&-15\\ 1&6 \end{bmatrix}$ are similar matrices.
Notice that here also eigenvalues are $3,1$
$\quad$
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 $\text{det}(\cdot) = 3$

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Similar matrices have same eigenvalues

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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 ?

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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.

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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.

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

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Example
Say $A=\begin{bmatrix} 4&1\\ 0&4 \end{bmatrix}$ here eigenvalues are $4,4$
It has a bunch of similar matrices, like

$$\begin{bmatrix} 4&1\\ 0&4 \end{bmatrix}, \ \begin{bmatrix} 4&0\\ 7&4 \end{bmatrix}, \ \begin{bmatrix} 7&-2\\ 1&2 \end{bmatrix},\cdots$$

these all are similar matrices.