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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?

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  • $\begingroup$ Depends on the field you are solving in. Ok for $Z$ not for $Z_n$. $\endgroup$
    – user67773
    May 2, 2013 at 8:16

6 Answers 6

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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 $A$.

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$A^2 - 5A + 6 = 0$ 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 $\begin{pmatrix} 2 & 0\\ 0 & 3 \end{pmatrix}$
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  • $\begingroup$ Don’t you assume that $A$ is diagonalizable? $\endgroup$
    – k.stm
    May 2, 2013 at 8:28
  • $\begingroup$ $A$ is diagonalizable because $P(A) = 0$ and $P = (X-2)(X-3)$ has simple roots. $\endgroup$
    – Arnaud
    May 2, 2013 at 8:38
  • $\begingroup$ Well, yes. Right. $\endgroup$
    – k.stm
    May 2, 2013 at 8:41
  • $\begingroup$ The "e.g." in the first two points are misleading; these are the only possible examples in each case. $\endgroup$ Jul 16, 2013 at 15:34
  • $\begingroup$ Sorry, I wanted to say "i.e.". Thank you. $\endgroup$
    – Arnaud
    Jul 16, 2013 at 15:44
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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 $ad-bc\neq0$.

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    $\begingroup$ This is false : $2I$ is a solution which is not similar to $B$. $\endgroup$
    – Arnaud
    May 2, 2013 at 8:40
  • $\begingroup$ @Arnaud I have corrected the answer to include the more general cases. $\endgroup$
    – Daryl
    May 2, 2013 at 10:09
  • $\begingroup$ But you should keep the two cases with $v_1=v_2$ separate, since it is pointless for those solutions to run over all those values of $a,b,c,d$ only to learn that together they only give a single solution for each of these two cases. $\endgroup$ Jul 16, 2013 at 15:31
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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

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

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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 $A$.

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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 $\begin{pmatrix}2.5\pm\sqrt{0.25-k_1k_2}&k_1\\k_2&2.5\mp\sqrt{0.25-k_1k_2}\end{pmatrix}$ , $k_1,k_2\in\mathbb{C}$

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  • $\begingroup$ The other answers also indicate an infinite number of solutions but state that they must be a similarity transform of a matrix with eigenvalues that are either 2 or 3; your final matrix has eigenvalues that are 2 and 3 and so it is similar to one of the others. I didn't down vote this btw... $\endgroup$ May 19, 2013 at 17:30
  • $\begingroup$ @Graham Hesketh: my answer not only show that it has infinitely many solutions but also show that the general solution has two arbitrary constants. Since I calculated from first principle, the properties come out most realistical. $\endgroup$ May 19, 2013 at 17:41
  • $\begingroup$ @Graham Hesketh: Perhaps the downvoters of my answer are mainly coming from the active users in meta.math.stackexchange.com/questions/9602. $\endgroup$ May 19, 2013 at 23:33

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