Determinant of a matrix that is its own inverse

Question

I need help in showing that when computing the determinant the inverse of an $n \times n$ matrix with the property

$$M=M^{-1}$$

that is

$$M^2 = I$$

the determinant is either $1$ or $-1$.

I've tried showing it in a couple ways and the way I'm trying to show it has me stuck

$$K^2 = I$$

$$K^2 - I = 0$$

$$\det(K^2 - I) = 0$$

$$\det(I - I) = 0$$

I get here and I am hopelessly stuck. Could I go on to prove it this way? Is there any elementary way to prove this?

Answer 1

We have that $$1 = \det(I) = \det(AA^{-1}) = \det(AA) = \det(A)\det(A) = \det(A)^2.$$ So, $\det(A)^2 - 1 = 0$. This is a polynomial in $\det(A)$, what are its solutions?

Answer 2

If

$M = M^{-1}, \tag 1$

then

$M^2 = I, \tag 2$

so by the multiplicative property of determinants,

$(\det M)^2 = \det (M^2) = \det I = 1, \tag 3$

which implies that

$\det M = \pm 1. \tag 4$

Now in fact, we can go a little further with only a little more work and show that every eigenvalue or $M$ is in the set $S = \{-1, 1\}$. For if

$Mv = \mu v \tag 5$

for some non-zero vector $v$, then

$\mu^2 v = \mu(\mu v) = \mu Mv = M(\mu v) = M(Mv)= M^2 v = Iv = v; \tag 6$

thus

$\mu^2 = 1, \tag 7$

or

$\mu = \pm 1. \tag 8$

Since the eigenvalues of $M$ lie in the set $S$, and $\det M$ is the product of its eigenvalues, we again see that we must have (4).

Finally, we can also write

$(M + I)(M - I) = M^2 - I = 0, \tag 9$

whence

$\det(M + I) \det(M - I) = 0; \tag {10}$

thus

$\det(M + I) = 0 \tag{11}$

or

$\det(M - I) = 0; \tag{12}$

in the former case, there exists a vector $v$ with

$Mv = -v; \tag{13}$

in the latter

$Mv = v, \tag{14}$

which gives a quick and easy proof of the existence of eigenvectors corresponding the eigenvalues $\mu = \pm 1$.