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So I'm studying linear algebra and one of the self-study exercises has a set of true or false questions. One of the questions is this:

If $A^2 = I$ (Identity Matrix), then $A = \pm I$ ?

I'm pretty sure it is true but the answer says it's false. How can this be false (maybe it's a typography error in the book)?

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    $\begingroup$ Try $$ A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. $$ $\endgroup$ Feb 5, 2012 at 20:13
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    $\begingroup$ I'd point out that it is true if you're working with $1$-by-$1$ matrices (over $\mathbb C$, or any other integral domain). But for $n \geq 2$ the ring of $n$-by-$n$ matrices over any non-trivial ring is not an integral domain: this means that $(A+I)(A-I) = 0$ doesn't necessarily imply that $A + I = 0 $ or $A - I = 0$. $\endgroup$
    – Matt
    Feb 5, 2012 at 20:30
  • $\begingroup$ possible duplicate of Finding number of matrices whose square is the identity matrix $\endgroup$ Feb 5, 2012 at 20:56
  • $\begingroup$ There's an entire family of so-called involutory matrices. Look up Householder reflectors, for instance. $\endgroup$ Feb 6, 2012 at 5:11
  • $\begingroup$ What book is that exercise from? $\endgroup$
    – Akash Gaur
    Jan 22, 2017 at 17:43

5 Answers 5

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A simple counterexample is $$A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $$ We have $A \neq \pm I$, but $A^{2} = I$.

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In dimension $\geq 2$ take the matrix that exchanges two basis vectors ("a transposition")

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  • $\begingroup$ If you want to exchange the (standard) basis vectors $e_{i}$ and $e_{j}$ ($1 \leq i,j \leq n$), then use the matrix $A = [m_{ij}]$ with $m_{kk} = 1, k\neq i,j$, $m_{ij} = m_{ji} = 1$ and $m_{kl} = 0$ for all other values of $k$ and $l$. For example, if you want $e_2$ and $e_3$ exhanged in $\mathbb{R}^{3}$, take $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$$ It is clear that such a matrix always satisfies $A^2 = I$, since applying it twice always gets you back to where you started. $\endgroup$ Feb 5, 2012 at 21:01
  • $\begingroup$ Thank you @Martin Wanvik, pretty clear explanation. $\endgroup$ Feb 5, 2012 at 21:52
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I know $2·\mathbb C^2$ many counterexamples, namely

$$A=c_1\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}+c_2\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}\pm\sqrt{c_1^2+c_2^2\pm1}\begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix},$$

see Pauli Matrices $\sigma_i$.

These are all such matrices and can be written as $A=\vec e· \vec \sigma$, where $\vec e^2=\pm1$.

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The following matrix is a conterexample $ A = \left( {\begin{array}{cc} -1 & 0 \\ 0 & 1 \\ \end{array} } \right) $

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"Most" (read: diagonalizable) matrices can be viewed simply as a list of numbers -- its eigenvalues -- in the right basis. When doing arithmetic with just this matrix (or with other matrices that diagonalize in the same basis), you just do arithmetic on the eigenvalues.

So, to find diagonalizable solutions to $A^2 = I$, we just need to write down a matrix whose eigenvalues satisfy $\lambda^2 = 1$ -- and any such matrix will do.

When thinking about matrices in this way -- as a list of independent numbers -- it makes it easy to think your way through problems like this.

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    $\begingroup$ Every matrix satisfying $A^2=I$ is diagonalizable, because either it is $\pm I$ or its minimal polynomial is $(x-1)(x+1)$. The general solution is obtained by taking all diagonal matrices with entries $\pm 1$ on the diagonal and conjugating by invertible matrices. $\endgroup$ Feb 6, 2012 at 5:03
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    $\begingroup$ Jonas Meyer, this is only true if $char F \ne 2$. Otherwise, there are such matrices which are not diagonalizable, $\endgroup$
    – the L
    Feb 6, 2012 at 8:18
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    $\begingroup$ @Jonas: That's a good point to mention as an appendix, but dealing properly with non-diagonalizable matrices in this fashion is somewhat more sophisticated. The only reason I mentioned the word was so that I didn't mislead Randolf into thinking this method works (unmodified) for all matrices; e.g. that the argument I gave isn't sufficient to tell us that this (or any!) equation has only diagonalizable solutions. $\endgroup$
    – user14972
    Feb 6, 2012 at 9:53
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    $\begingroup$ @anonymous: Good point, e.g. $\begin{bmatrix}1&1\\ 0&1\end{bmatrix}$. @@Hurkyl: I agree, it is best as an appendix. I appreciate your caution, but wanted to point out that your method does lead to the general solution (in the characteristic $0$ case that the OP is probably working in). $\endgroup$ Feb 6, 2012 at 15:49

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