Let $A,B\in M_n(\mathbb{R})$ be symmetric matrices such that $A^5=B^5$. Prove that $A=B$.
My attempt:
If $A$ and $B$ are symmetric matrices we can use the espectral theorem to write $A=\lambda_1 E_1+\lambda_2 E_2+…+\lambda_k E_k$ where $\lambda_i$ are the eingenvalues, $E_i$ are projections and this writing is unique and $B=\alpha_1 E_1+\alpha_2 E_2+…+\alpha_s E_s$, but as $A^5=B^5$ then, we have that $\lambda^5_1 E_1+\lambda^5_2 E_2+…+\lambda^5_k E_k=\alpha^5_1 E_1+\alpha^5_2 E_2+…+\alpha^5_s E_s$.
As the writing is unique, we have that $k=s$ and $\lambda^5_i=\alpha^5_i$ then $\lambda_i=\alpha_i$ and $A=B$.
Is this correct?
Best Answer
$$ Ax = \lambda_A x \implies A^5 x = \lambda_A^5 x $$ $$ Bx = \lambda_B x \implies B^5 x = \lambda_B^5 x $$ But $A^5 = B^5$ and so $\lambda_A^5 = \lambda_B^5 \implies \lambda_A = \lambda_B = \lambda$. Since the $\lambda$'s were arbitrary this proves that the eigenvalues are identical (can you see where symmetry was used here)?
Utilizing the fact that the eigenvalues are identical, it is clear that we also have $(A-B)x = 0$ for each eigenvector. Since $A-B$ is symmetric, it admits a complete set of eigenvectors and so this relation must hold for arbitrary $x$. The only way this is possible is if $A - B = 0$.