A question from Golan's linear algebra:

Let $A\in M(n,\mathbb R)$ (which denotes the set of all $n\times n$ matrices, for some $n\geq 2$) be symmetric. Does there exist a symmetric matrix $B$ such that $B^2=A$?

It asks again whether it is possible to find such a matrix in case $A$ is symmetric and positive definite? How to do this? Any hints to approach them. I am totally clueless

  • $\begingroup$ I cannot imagine that a book would contain such in incomprehnsible phrase. Can you at least cite it correctly? $\endgroup$ – Marc van Leeuwen Jan 8 '15 at 13:15
  • $\begingroup$ I dont get what you are asking @MarcvanLeeuwen $\endgroup$ – Learnmore Jan 8 '15 at 14:09
  • $\begingroup$ "i.e the set of all $n×n$ matrices $n≥2$ is symmetric" is either false or nonsense. I'm asking you to just copy faithfully what is written in the book. Probably it is said (in the book) that $A$ is a symmetric matrix, but that is not what is written in the question above. $\endgroup$ – Marc van Leeuwen Jan 8 '15 at 14:47
  • $\begingroup$ @MarcvanLeeuwen I assume he meant to write "the set of all $n\times n$ matrices $n\geq 2$ that are symmetric". I understand your frustration, though $\endgroup$ – Omnomnomnom Jan 8 '15 at 18:23
  • $\begingroup$ @Omnomnomnom: actually I think it is slightly different yet. I'll edit, and if it is not that, let OP change it to match the book. $\endgroup$ – Marc van Leeuwen Jan 8 '15 at 18:26

Hint: Every symmetric matrix is unitarily diagonalizable by the spectral theorem. How could you (easily) construct a square root of a diagonal matrix?

If $B^2 = A$, what could the eigenvalues of $B$ be? If $A$ is symmetric, what do we know about its eigenvalues?

The answer will be no in general, but yes if $A$ is also positive definite.

If $A = UDU^*$ where $D$ is diagonal, try $$ B = U \pmatrix{\sqrt{\lambda_1}\\&\ddots \\ && \sqrt\lambda_n}U^* $$

  • $\begingroup$ if $A$ is symmetric all eigen values are real $\endgroup$ – Learnmore Jan 8 '15 at 6:38
  • $\begingroup$ if $A$ is positive definite all eigen values are positive also eigen values of $B$ are square root of that of $B$ $\endgroup$ – Learnmore Jan 8 '15 at 6:39
  • $\begingroup$ is it right? @ Omnomnomnom $\endgroup$ – Learnmore Jan 8 '15 at 6:40
  • $\begingroup$ how to construct $B$ is not clear $\endgroup$ – Learnmore Jan 8 '15 at 6:41
  • $\begingroup$ The correct statement is that if $\lambda$ is an eigenvalue of $A$, then at least one of $\sqrt \lambda$ and $-\sqrt \lambda$ must be an eigenvalue of $B$, so you had the right idea. $\endgroup$ – Omnomnomnom Jan 8 '15 at 18:25

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