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Let $A^2$ be a symmetric matrice such that the spectral theorem allows us to write $$A^2 = P\operatorname{Diag}(\lambda_1 \dots \lambda_n)P^T$$ Suppose $\forall i \in [\![ 1, n]\!], \lambda_i \geq 0$.

I want to know if the following proposition is true: $$A^2 = P\operatorname{Diag}(\lambda_1 \dots \lambda_n)P^T \implies A = P\operatorname{Diag}(\sqrt{\lambda_1} \dots \sqrt{\lambda_n})P^T$$

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  • $\begingroup$ yes, it is indeed true $\endgroup$ – pointguard0 Jan 30 at 22:30
  • $\begingroup$ Please notice that my answer was initially only half-true; I presented the correct one now. $\endgroup$ – Blazej Jan 31 at 9:16
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This is partially wrong; see discussion below This is false. If you know that $A^2= \mathrm{diag}(x_1,...,x_n)$ in some basis, then in this basis $A=\mathrm{diag}( \pm \sqrt{x_1},....,\pm \sqrt{x_n})$. Thus in general knowing $A^2$ there are $2^n$ choices for (symmetric) $A$. Of course taking square roots becomes unique if you restrict attention to positive matrices.

Correction It was pointed out to me in the comments that answer above is wrong without additional assumptions. Let's take a closer look. We know that $A$ commutes with $A^2$, so (since both are diagonalizable), there exists a common eigenbasis. If $A^2$ has nondegenerate spectrum then this already implies that $A$ is diagonal in the same basis as $A^2$. Otherwise we can group eigenvalues into blocks, so that (perhaps after reshuffling rows and columns) $A^2= \mathrm{diag}(y_1,...,y_1,y_2,...,y_2,...,y_n,...y_k)$, where $y_i \neq y_j$ for $i \neq j$ and $y_i \geq 0$ is repeated $g_i$ times. Now $A$ has to leave each eigenspace of $A^2$ invariant, so it is a direct sum of $g_i \times g_i$ matrices of the form $\sqrt{y_i} S_i$, where $S_i^2 = 1$, $S_i^T = S_i$. For $g_i=1$ the only matrices of this form are $\pm 1$ (reproducing the previous answer), but in general this is not the only possibility. For example for $g_1 = 2$ there is the Pauli $x$ matrix $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.

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  • $\begingroup$ Your result works only if the $(x_i)$ are distinct; otherwise, even if $A$ is assumed to be symmetric, there are non-diagonal solutions in the form of orthogonal symmetries (up to a homothety). $\endgroup$ – loup blanc Jan 30 at 23:53
  • $\begingroup$ @loupblanc thank you, I corrected mu answer. $\endgroup$ – Blazej Jan 31 at 9:16

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