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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?

user424241
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    What tells you that $E_i$ they are the projections on the same subspace for both $A$ and $B$ ? – Lelouch Nov 11 '22 at 21:33

1 Answers1

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$$ Ax = \lambda_A x \implies A^5 x = \lambda_A^5 x $$ $$ Bx = \lambda_B x \implies B^5 x = \lambda_B^5 x $$ 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 $A - B = 0$.

Gregory
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  • What do you mean by "it admits a complete set of eigenvectors and so this relation must hold for arbitrary x"? Can you explain that in more detail please? Thanks! – user424241 Nov 11 '22 at 22:22
  • See these for more information: https://math.stackexchange.com/questions/1317936/why-does-a-symmetric-matrix-have-a-complete-set-of-eigenvectors-and-eigenvalues

    https://math.stackexchange.com/questions/82467/eigenvectors-of-real-symmetric-matrices-are-orthogonal

     – Gregory Nov 11 '22 at 22:23
  • Once you understand what it means to be complete, you should be able to see how an arbitrary vector can be written as a linear combination of the eigenvectors. – Gregory Nov 11 '22 at 22:24
  • So if I undertand, every vector can be written as a linear combination of those eigenvector $x$ but as $(A-B)x=0$ for every eigenvector $x$ then $(A-B)x$ for every $x$? – user424241 Nov 11 '22 at 22:33
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    Correct so the statement $(A-B)x = 0$ for a given eigenvector can be extended to an arbitrary vector, because it can be written as al linear combination of the eigenvectors. – Gregory Nov 11 '22 at 23:56
  • And the use of the symmetry was used in the first place as the eigenvalues of a a symmetric matrix are reals or you used that in another way? – user424241 Nov 12 '22 at 00:20
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    Exactly right. Nice work. – Gregory Nov 12 '22 at 00:22
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    @user424241 please accept answer if you would. – Gregory Nov 12 '22 at 02:12