Prove that if $A$ is Hermitian and $A^m=I$, then $A^2=I$ (and $A=I$ if $m$ is odd)

Question

Let $A$ be a Hermitian $n\times n$ matrix over ${\Bbb C}$ such that $A^m = I$ for some natural number $m$.

Prove:

1. $A^2=I$.
2. If $m$ is odd, then $A=I$

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Well, for the first question I did this: Since $A$ is Hermitian then $A$ is diagonalizable. $A$'s eigenvalues are $\{-1,1\}$. The minimal polynomial that reset $A$ and $$M(x) = (x+1)^z (x-1)^k $$ for $k,z$ natural numbers. So the options for the minimal polynomial are:

1. $$(x-1)$$
2. $$(x+1)$$
3. $$(x-1)(x+1)$$
4. $$(x-1)^2 $$

Eventually I got $A$ can be: $$ A=I \quad \mathrm{or}\quad A=-I $$

Is that even right?

For the second part of the question I have no clue.

Answer 1

Your approach is right but in the middle it is not clear what you are doing. First since $A$ is hermitian all the eigenvalues of $A$ are real. Further, $A^m = I$ implies that any eigenvalue $\lambda$ satisfies the relation $\lambda^m - 1 =0$. Since $\lambda \in \mathbb{R}$, if $m$ is even, we get that $\lambda = \pm 1$. If $m$ is odd, we get that $\lambda = 1$.

Also, any hermitian matrix permits a decomposition of the form $$A = U \Lambda U^{\dagger}$$ where $U$ is a unitary matrix and $\Lambda$ is a diagonal matrix with the eigenvalues of $A$ as its diagonal entries. This is so since it is possible to find an orthonormal basis of $\mathbb{C}^n$ consisting of $n$ eigenvectors of $A$.

If $m$ is odd, we showed that all the eigenvalues have to be $1$. Hence, $\Lambda$, the diagonal matrix with the eigenvalues of $A$ as its diagonal entries, is the identity matrix i.e. $\Lambda = I$. This gives us that $$A = U I U^{\dagger} = I$$

If $m$ is even, we showed that all the eigen values have to be $\pm 1$. Hence, $\Lambda^2 = I$. This gives us that $$A^2 = (U \Lambda U^{\dagger})(U \Lambda U^{\dagger}) = U \Lambda (U^{\dagger} U) \Lambda U^{\dagger} = U \Lambda I \Lambda U^{\dagger} = U \Lambda^2 U^{\dagger} = UIU^{\dagger} = U U^{\dagger} = I$$

Answer 2

As $\,A\,$ is diagonalizable the only possible forms for its minimal polynomial are $\,(1)\,$ , if $\,z=0\,$, $\,(2)\,$ , if $\,k=0\,$, or $\,(3)\,$ , if $\, k,z\geq 1\,$ . Case $\,(4)\,$is impossible as a square matrix over some field is diagonalizable iff its minimal pol. is the product of different linear polynomials over that field, and from these possibilities either $$A=I\,\,,\,\,A=-I\,\,,\,\,A^2-I=0$$ from which you get what you wanted.

Answer 3

Since you can compute powers in diagonal form, there is almost nothing to this once you know it is diagonalisable with eigenvalues (diagonal entries) in $\{1,-1\}$: one has $A^2=I$ because $1^2=1=(-1)^2$, and if $m$ is odd then $A^2=1$ and $A^m=I$ easily imply $A=I$. For the latter step you could use the minimal polynomail: by $A^2=I$ it is a divisor of $X^2-1$, and reducing $X^m-1$ modulo $X^2-1$ leaves $X-1$, which annihilates $A$ so $A=I$. Apart from this I don't really see how much the minimal polynomial does for you here. But in any case diagonlisable implies the minimal polynomial has no repeated factors, so its only possible values are $1$, $X-1$, $X+1$ and $X^2-1=(X-1)(X+1)$. And yes, $1$ is a possibility, but only if $n=0$.

But note that it is not true that $A=I$ or $A=-1$; the diagnal form of $A$ could have any mix of entries $1$ and $-1$.