1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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 $$ B = U \pmatrix{\sqrt{\lambda_1}\\&\ddots \\ && \sqrt\lambda_n}U^* $$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.