0
$\begingroup$

Let $ A $ be an $n\times n$ matrix with real entries. Then is it always possible to find a real symmetric $n \times n$ matrix $B$ such that $B^ 2$ = $A \times A.$?

It would be great if someone could give me some hints.I am currently reading linear algebra

$\endgroup$
5
  • $\begingroup$ Do you assume that $A$ is also symmetric $\endgroup$
    – orangeskid
    Jan 20, 2015 at 6:24
  • $\begingroup$ No ,No sir I dont assume it @orangeskid $\endgroup$
    – Bobthe builder
    Jan 20, 2015 at 6:26
  • $\begingroup$ Is $A\times A$ just $A \cdot A$ ? $\endgroup$
    – orangeskid
    Jan 20, 2015 at 6:48
  • $\begingroup$ yes you are right $\endgroup$
    – Bobthe builder
    Jan 20, 2015 at 6:51
  • $\begingroup$ Why not write $B^2=A^2$ then? $\endgroup$
    – Marc van Leeuwen
    Jan 20, 2015 at 7:32

3 Answers 3

Reset to default
2
$\begingroup$

Note that if $B$ is symmetric, then $B^2$ is also symmetric. Therefore, if $A^2$ is not symmetric, there exists no such $B$.

The condition really is :such $B$ exists if and only if $A^2$ symmetric and the eigenvalues of $A$ are real. That will make $A^2$ positive semi-definite, and for these there exists a square root $B$ symmetric (even positive semi-definite).

Therefore: two checks: $A^2$ symmetric and eigenvalues of $A$ real.

$\endgroup$
4
  • $\begingroup$ is the condition that $A$ has either real or purely imaginary eigenvalues sufficient for $A^2$ to be symmetric? $\endgroup$
    – abel
    Jan 20, 2015 at 9:20
  • $\begingroup$ @abel No; for instance $$ A = \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right). $$ $\endgroup$
    – mollyerin
    Jan 20, 2015 at 9:31
  • $\begingroup$ @mollyerin, thanks. do you know of a sufficient condition for $A^2$ to be symmetric but $A$ is not. $\endgroup$
    – abel
    Jan 20, 2015 at 9:34
  • $\begingroup$ @abel Not offhand; maybe post it as a Math:SE question? :D $\endgroup$
    – mollyerin
    Jan 20, 2015 at 9:43
0
$\begingroup$

Hint

If this was true then $B^{T}=B$. Which means $A^{2}=[A^{T}]^{2}$. Now try to see if you can find a matrix $A$ which might not follow this last condition. Try with $2 \times 2$ examples.

$\endgroup$
3
  • $\begingroup$ yes will this do ? @Anurag A $\endgroup$
    – Bobthe builder
    Jan 20, 2015 at 6:24
  • $\begingroup$ a_11=0,a_12=1,a_21=2,a_22=1 $\endgroup$
    – Bobthe builder
    Jan 20, 2015 at 6:25
  • $\begingroup$ are u there ?@ Anurag $\endgroup$
    – Bobthe builder
    Jan 20, 2015 at 6:26
0
$\begingroup$

If $A$ be a symmetric matrix then the problem will be solved.Let $A=\left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$ then $A^2=\left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$ but if $B=\left( \begin{array}{cc} a & b \\ b & c \end{array} \right)$ be a symmetric matrix then $B^2=\left( \begin{array}{cc} a^2+b^2 & ab+bc \\ ab+bc & b^2+c^2 \end{array} \right)$ imply $b=c=0$ and $a=1$. this is an counterexample for above problem.

$\endgroup$
2
  • $\begingroup$ so u have got a symmetric matrix .what' s your point $\endgroup$
    – Learnmore
    Jan 20, 2015 at 6:38
  • $\begingroup$ For $A$ anti-symmetric there will in general not be a symmetric solution $B$ to $B^2=A^2$. $\endgroup$
    – Marc van Leeuwen
    Jan 20, 2015 at 7:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .