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How many matrices are such that $A^2 =I$, where $A$ is a $2\times2$ matrix and $I$ is a $2\times2$ identity matrix?

I can only think of the identity and it negative are they more? Is it an application of Cayley-Hamilton theorem. I have seen a similarly post by I cannot follow it. Could someone answer in simple and understandable terms.

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Can $A$ be complex matrix? –  Patrick Li Oct 23 '12 at 18:45
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There are more, e.g., $\begin{pmatrix} 0& 1\\1&0\end{pmatrix}$. –  Fabian Oct 23 '12 at 18:46
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4 Answers 4

up vote 2 down vote accepted

Hint:$A=\left(\begin{array}{cc}1&a\\ 0&-1 \end{array}\right)$ check that $A^2=I$

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Still not all, see my comment above. –  Fabian Oct 23 '12 at 18:47
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@Fabian I wanted to say the number is infinite –  clark Oct 23 '12 at 18:48
    
Ok, got your point. –  Fabian Oct 23 '12 at 18:49
    
@clark, I think this is the best answer, thanks very much. –  Vaolter Oct 24 '12 at 13:00
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You can compute this manually if you want:

$$\left[\begin{array}{cc} a & b \\c & d \end{array}\right]^2=\left[\begin{array}{cc} a^2+bc & b(a+d) \\c(a+d) & bc+d^2 \end{array}\right]$$

To get the identity matrix, either $a=-d$, so $a^2+bc=1$ (and these can be picked freely, leaving plenty of options) or $b=0$ and $c=0$, so $a=\pm1, d=\pm 1$.

More conceptually, you're asking this: "What linear transformation, applied twice, brings you back to where you started?" You could swap the $x$ and $y$ axes:

$$\left[\begin{array}{cc} 0 & 1 \\1 & 0 \end{array}\right]$$

flip the space around the $x$ axis:

$$\left[\begin{array}{cc} -1 & 0 \\0 & 1 \end{array}\right]$$

Or a number of other things! Just think of a transformation that is undone by applying it again, and find the matrix that corresponds to it.

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I assumed matrices over $\mathbb{R}$. The connection to other fields is relatively straightforward, though I can elaborate if needed. –  Robert Mastragostino Oct 23 '12 at 18:53
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Let's call $q$ the minimum polynomial such that $q(A)=0$. It's a well known fact that $q$ must divide $t^2-1$. This means that it could be:

  1. $t-1$
  2. $t+1$
  3. $(t-1)(t+1)$

In each one of this cases A is diagonalizable. That's because $q$ is the minimun polynomial which nullify $A$ (So, as example, $A-I$ is enough to nullify the generalized eigenspace for $1$).

So we have just 3 Jordan forms.

$\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$ $\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}$ $\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}$

To answer your question you have those 3 Conjugacy classes.

Notes: This argument can easily be generalized. In $\mathbb{F_2}$ there is just identity.

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so are you saying there are infinite matrices? –  Vaolter Oct 24 '12 at 13:09
    
I'm saying much more. But yes, in infinite field it's as you said. –  Ivan Oct 24 '12 at 17:16
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The possible eigenvalues are $\pm 1$. With the exception of $I$ and $-I$ the matrix will be a reflection. Therefore in a suitable basis it is given by $$\begin{pmatrix} 1 &0 \cr 0 & -1 \end{pmatrix}\ . $$ All other solutions are conjugate to this matrix. e.g. $$\begin{pmatrix}11 & -20 \cr 6 & -11 \end{pmatrix}$$

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