Proof that $A\mapsto A^3$ is one to one where $A$ is a symmetric matrix

Question

Let $ S_n$ denote the set of real symmetric matrices and $f: S_n\to S_n$, $A\mapsto A^3$. I want to prove that $f$ is a bijective function. Using spectral theorem I manage to prove that $f$ is onto. Now for the one to one stuff I found some difficulties and I doubt that it's not true but I can't find a counterexample. This is my try: let $A,B\in S_n$ such that $S=A^3=B^3$. Since $S$ is symmetric then by spectral theorem I prove that $A$ and $B$ have the same spectrum (the set of eigenvalues) and the same eigenspaces but I can't from this point prove that $A=B$. Any suggestion or idea?

Answer 1

Since $A$ is symmetric it has a full set of eigenvalues $v_k$.

Suppose $A v = \lambda v$, then $A^3 v = B^3 v = \lambda^3 v$.

If $\lambda = 0$ then since $A,B$ are symmetric we have $Av=Bv = 0$.

Suppose $\lambda \neq 0$. Then $v \in \ker (B^3 -\lambda^3 I)$. Note that $B^3 -\lambda^3 I = (B^2 +\lambda B +\lambda^2) (B-\lambda I)$ and $B^2 +\lambda B +\lambda^2$ is invertible (since $\lambda$ is real and non zero). Hence $v \in \ker(B-\lambda I)$ and so $Bv= \lambda v$.

In particular, $Ax=Bx$ for all $x$ and so $A=B$.

Answer 2

If $A=\sum_i \lambda_i P_i$ (where $\lambda_i$ are the eigenvalues and $P_i$ are the projections) then $A^3=\sum_i \lambda_i^3 P_i$.

If $B=\sum_i \mu_i Q_i$ then $B^3=\sum_i \mu_i^3 Q_i$; if this equals $A^3$ then there is a choice of ordering in which $P_i=Q_i$ for each $i$ (since spectral decomposition is unique). Then $\lambda_i^3=\mu_i^3$, but since symmetric matrices have real eigenvalues this implies $\lambda_i=\mu_i$ which means $A=B$.