Find all matrices $A$ of order $2 \times 2$ that satisfy the equation $A^2-5A+6I = O$ 
Find all matrices $A$ of order $2 \times 2$ that satisfy the equation
$$
A^2-5A+6I = O
$$

My Attempt:
We can separate the $A$ term of the given equality:
$$
\begin{align}
A^2-5A+6I &= O\\
A^2-3A-2A+6I^2 &= O
\end{align}
$$
This implies that $A\in\{3I,2I\} = \left\{\begin{pmatrix}
3 & 0\\ 
0 & 3
\end{pmatrix},
\begin{pmatrix}
2 & 0\\ 
0 & 2
\end{pmatrix}\right\}$.
Are these the only two possible values for $A$, or are there other solutions?If there are other solutions, how can I find them?
 A: The Cayley-Hamilton theorem states that every matrix $A$ satisfies its own characteristic polynomial; that is the polynomial for which the roots are the eigenvalues of the matrix:
$p(\lambda)=\det[A-\lambda\mathbb{I}]$.
If you view the polynomial:
$a^2-5a+6=0$,
as a characteristic polynomial with roots $a=2,3$, then any matrix with eigenvalues that are any combination of 2 or 3 will satisfy the matrix polynomial:
$A^2-5A+6\mathbb{I}=0$,
that is any matrix similar to:
$\begin{pmatrix}3 & 0\\ 0 & 3\end{pmatrix}$,$\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix}$,$\begin{pmatrix}2 & 0\\ 0 & 3\end{pmatrix}$.    Note:$\begin{pmatrix}3 & 0\\ 0 & 2\end{pmatrix}$ is similar to $\begin{pmatrix}2 & 0\\ 0 & 3\end{pmatrix}$.
To see why this is true, imagine $A$ is diagonalized by some matrix $S$ to give a diagonal matrix $D$ containing the eigenvalues $D_{i,i}=e_i$, $i=1..n$, that is:
$A=SDS^{-1}$, $SS^{-1}=\mathbb{I}$.
This implies:
$A^2-5A+6\mathbb{I}=0$,
$SDS^{-1}SDS^{-1}-5SDS^{-1}+6\mathbb{I}=0$,
$S^{-1}\left(SD^2S^{-1}-5SDS^{-1}+6\mathbb{I}\right)S=0$,
$D^2-5D+6\mathbb{I}=0$,
and because $D$ is diagonal, for this to hold each diagonal entry of $D$ must satisfy this polynomial:
$D_{i,i}^2-5D_{i,i}+6=0$,
but the diagonal entries are the eigenvalues of $A$ and thus it follows that the polynomial is satisfied by $A$ iff the polynomial is satisfied by the eigenvalues of $A$.
A: $A^2 - 5A + 6 = 0$ is equivalent to $(A-2)(A-3) = 0$, which is equivalent to $Sp(A) \subset \{2, 3\}$.
Three cases are possible :


*

*$Sp(A) = \{2\}$, i.e. $A = 2I$

*$Sp(A) = \{3\}$, i.e. $A = 3I$

*$Sp(A) = \{2, 3\}$, i.e. $A$ is similar to $\begin{pmatrix}
2 & 0\\ 
0 & 3
\end{pmatrix}$

A: Two matrices $A$ and $B$ are similar if there exists a matrix $P$ such that $A=PBP^{-1}$.
The solutions to your equation are $x=2,3$. Thus, all matrices which satisfy your equation must be similar to $B=\begin{bmatrix}v_1&0\\0&v_2\end{bmatrix}$, where $v_1$ and $v_2$ are either $2$ or $3$.
Choosing $P=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, all solutions to your equation are
$$
A=PBP^{-1}=\frac{1}{ad-bc}\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}v_1&0\\0&v_2\end{bmatrix}\begin{bmatrix}d&-b\\-c&a\end{bmatrix},
$$
for any choice of $a,b,c,d$ where $ad-bc\neq0$.
A: As in general we have

$$
\mathbf{A}^2 - \big( \lambda_1 + \lambda_2 \big) \mathbf{A}
+ \lambda_1 \lambda_2 \mathbf{I} = 0, \tag 1
$$

we see that in this case

$$
\lambda_1 = 2, \quad \lambda_2 = 3. \tag 2
$$

So the general solution is given by

$$
\bbox[16px,border:2px solid #800000]
 { \mathbf{B} \pmatrix{ 3 & 0 \\ 0 & 2} \mathbf{B}^{-1}, } \tag 3
$$

for any matrix such that $\det(\mathbf{B}) \ne 0$.
Note that

$$
\pmatrix{0 & 1 \\ 1 & 0} \pmatrix{3 & 0 \\ 0 & 2} \pmatrix{0 & 1 \\ 1 & 0}
= \pmatrix{ 2 & 0 \\ 0 & 3}. \tag 4
$$

A: Let $p(A)$ be the minimal polynomial of $A$. Then $p(A) \mid (x-2)(x-3)$, so $p(A) = (x-2), (x-3)$, OR $(x-2)(x-3)$. 
If you don't have access to eigenvalues for this homework, perhaps just try plugging in a,b,c,d for values of $A$. 
