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Let A be a real $ n \times n$ positive definite symmetric matrix.Show that $A=B^2$ for some positive definite symmetric matrix $B$.

So we can try and diagonalize $A$ such that $A=SJS^{-1}$then let $B=S^{-1}\sqrt{J}S $.

But what if one of the eigenvalues of $A$ is "missing" an eigenvector? Our $J$ and $B$ will no longer be positive definite, right?

This addresses an issue that the following answer: Square root of Positive Definite Matrix, does not address.

David
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  • What do you mean by "one of the eigenvalues missing an eigenvector"? Since $A$ is symmetric there are orthonormal bases of $\mathbb{R}^n$ consisting of eigenvectors of $A$. The situation $\ker :(A - \lambda I) \subsetneqq \ker :(A - \lambda I)^n$ for some $\lambda$, which is the only interpretation of "missing an eigenvector" I can think of that makes sense, cannot occur for symmetric $A$. – Daniel Fischer Nov 19 '19 at 13:52
  • @DanielFischer I mean the situation in which you have to use generalized eigenvectors instead of ordinary eigenvectors, if that ever occurs. – David Nov 20 '19 at 11:25
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    For symmetric (real) matrices, this never occurs. They are (orthogonally, even) diagonalisable, hence the generalised eigenspaces coincide with the ordinary eigenspaces. – Daniel Fischer Nov 20 '19 at 12:31
  • @DanielFischer Thanks for the clarification. – David Nov 20 '19 at 13:13

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