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?
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 $A$.
$A^2 - 5A + 6 = 0$ is equivalent to $(A-2)(A-3) = 0$, which is equivalent to $Sp(A) \subset \{2, 3\}$.
Three cases are possible :
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 $ad-bc\neq0$.
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
$$ \pmatrix{0 & 1 \\ 1 & 0} \pmatrix{3 & 0 \\ 0 & 2} \pmatrix{0 & 1 \\ 1 & 0} = \pmatrix{ 2 & 0 \\ 0 & 3}. \tag 4 $$
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 $A$.
In fact there are infinitely many solutions.
Similar to the question $1$ in http://hk.knowledge.yahoo.com/question/question?qid=7010022700078:
Let $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ ,
Then $\begin{pmatrix}a&b\\c&d\end{pmatrix}^2-5\begin{pmatrix}a&b\\c&d\end{pmatrix}+6\begin{pmatrix}1&0\\0&1\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$
$\begin{pmatrix}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{pmatrix}-\begin{pmatrix}5a&5b\\5c&5d\end{pmatrix}+\begin{pmatrix}6&0\\0&6\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$
$\begin{pmatrix}a^2+bc-5a+6&ab+bd-5b\\ac+cd-5c&bc+d^2-5d+6\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$
$\begin{pmatrix}a^2-5a+bc+6&b(a+d-5)\\c(a+d-5)&d^2-5d+bc+6\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$
$\therefore\begin{cases}a^2-5a+bc+6=0~......(1)\\b(a+d-5)=0~.........(2)\\c(a+d-5)=0~.........(3)\\d^2-5d+bc+6=0~......(4)\end{cases}$
From $(2)$ and $(3)$ ,
$\begin{cases}b=0\\c=0\\a+d-5=k_1\end{cases}$ or $\begin{cases}b=k_1\\c=k_2\\a+d-5=0\end{cases}$ , $k_1,k_2\in\mathbb{C}$
Case $1$: $\begin{cases}b=0\\c=0\\a+d-5=k_1\end{cases}$
Put it into $(1)$ and $(4)$ :
$a^2-5a+6=0$
$(a-2)(a-3)=0$
$a=2$ or $3$
$d^2-5d+6=0$
$(d-2)(d-3)=0$
$d=2$ or $3$
They are automatically satisflied $a+d-5=k_1$
$\therefore A=\begin{pmatrix}2&0\\0&2\end{pmatrix}$ or $\begin{pmatrix}2&0\\0&3\end{pmatrix}$ or $\begin{pmatrix}3&0\\0&2\end{pmatrix}$ or $\begin{pmatrix}3&0\\0&3\end{pmatrix}$
Case $2$: $\begin{cases}b=k_1\\c=k_2\\a+d-5=0\end{cases}$
$(1)-(4)$ :
$a^2-d^2-5a+5d=0$
$(a+d)(a-d)-5(a-d)=0$
$(a+d-5)(a-d)=0$
$\therefore\begin{cases}a+d-5=0~......(5)\\a-d=2k_3~........(6)\end{cases}$ , $k_3\in\mathbb{C}$
$(5)+(6)$ :
$2a-5=2k_3$
$a=2.5+k_3~......(7)$
$(5)-(6)$ :
$2d-5=-2k_3$
$d=2.5-k_3~......(8)$
Put $(7)$ into $(1)$ and $(8)$ into $(4)$ :
$\begin{cases}(2.5+k_3)^2-5(2.5+k_3)+k_1k_2+6=0\\(2.5-k_3)^2-5(2.5-k_3)+k_1k_2+6=0\end{cases}$
$\begin{cases}6.25+5k_3+k_3^2-12.5-5k_3+k_1k_2+6=0\\6.25-5k_3+k_3^2-12.5+5k_3+k_1k_2+6=0\end{cases}$
$\begin{cases}k_3^2+k_1k_2-0.25=0\\k_3^2+k_1k_2-0.25=0\end{cases}$
$k_3^2+k_1k_2-0.25=0$
$k_3^2=0.25-k_1k_2$
$k_3=\pm\sqrt{0.25-k_1k_2}$
$\therefore A=\begin{pmatrix}2.5\pm\sqrt{0.25-k_1k_2}&k_1\\k_2&2.5\mp\sqrt{0.25-k_1k_2}\end{pmatrix}$ , $k_1,k_2\in\mathbb{C}$
Hence $A=\begin{pmatrix}2&0\\0&2\end{pmatrix}$ or $\begin{pmatrix}2&0\\0&3\end{pmatrix}$ or $\begin{pmatrix}3&0\\0&2\end{pmatrix}$ or $\begin{pmatrix}3&0\\0&3\end{pmatrix}$ or $\begin{pmatrix}2.5\pm\sqrt{0.25-k_1k_2}&k_1\\k_2&2.5\mp\sqrt{0.25-k_1k_2}\end{pmatrix}$ , $k_1,k_2\in\mathbb{C}$
