Real Symmetric Matrices

Question

Let $A$ be a real, symmetric, $n$ x $n$ matrix. Suppose $A^m=I$ for some $m$.

Prove $A^2=I$.

I think I want to use the symmetric implies diagonalizability...and take powers from there...correct?

Answer 1

Since $A$ is real symmetric all its eigenvalues are real, and furthermore there exists an orthogonal matrix $O$ such that

$O^TAO = \Lambda$,

where $\Lambda$ is a diagonal matrix formed by placing the eigenvalues of $A$ on its main diagonal, and zeroes everywhere else. If we now take the $m^{\text{th}}$ power of both sides of this equation, we obtain

$(O^TAO)^m = \Lambda^m$,

and since

$(O^TAO)^m = O^TA^mO = O^TIO = O^TO = I$,

it follows that

$\Lambda^m = I$,

so for each eigenvalue $\lambda_i$ on the diagonal of $\Lambda$ we have

$\lambda_i^m = 1$.

Now the fact that the $\lambda_i$ are real forces the conclusion that

$\lambda_i = \pm 1$

for each $i$. Hence

$\lambda_i^2 = 1$

for each $i$, forcing

$\Lambda^2 = I$.

Then

$O^TA^2O = \Lambda^2 = I$,

whence

$A^2 = OIO^T = I$.

Quod Erat Demonstrandum, for you classicists out there!

Answer 2

Hint

$$\forall\lambda\in\mathbb R,\quad\lambda^m=1\iff\lambda=\pm1\Rightarrow\lambda^2=1$$

Answer 3

From $A = PDP^{-1}$ ($D$ is a diagonal matrix) and $A^{m} = I$, we have that

$$P(D^{m} - I)P^{-1} = 0.$$

Thus, $$D^{m} = I.$$

Since each element of $D$ is a real number, diagonal elements of D can be only $1$ or $-1$.

Therefore, $D^{2} = I$ which implies $A^{2} = I.$