-2
$\begingroup$

Let $A$ be a real matrix, $\det A>0$, is there a real matrix $B$, such that $A=B^2$?

Related problems can be located in How to find a matrix square root with all real entries (if it exists). Here, we are in real, the det is $>0$. I guess it is wrong, say $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$? Is this a counterexample? I could not figure out.

$\endgroup$
1
2
$\begingroup$

It isn't a counterexample. We have $$\left(\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1\end{pmatrix}\right)^2 = \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}.$$ Here's how I got this counterexample: a complex number $a + ib$ can be represented by the matrix $$\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$$ in the sense that there is a field isomorphism between the set of these matrices and the complex numbers. Your matrix corresponds to the complex number $-i$, so it was a matter of computing a square root of $-i$, in this case, $\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$, then turning it into its matrix form.

$\endgroup$
2
  • $\begingroup$ Then the statement is right or wrong? $\endgroup$
    – xldd
    Jul 28 at 14:32
  • $\begingroup$ Wrong, apparently. See Thomas's answer. $\endgroup$ Jul 28 at 14:37
1
$\begingroup$

There is no real matrix square root for $$\begin{pmatrix}-1&0\\0&-2\end{pmatrix}$$

This is because any real matrix square root must have four eigenvalues.

$\endgroup$
2
  • $\begingroup$ Andrew What does "any real matrix square root must have four eigenvalues" means? $\endgroup$
    – xldd
    Jul 28 at 14:42
  • $\begingroup$ If a real matrix has $a+bi$ as a complex eigenvalue, then it must have the conjugate $a-bi$ as an eigenvalue. One of the eigenvalues must be a square root of $-1,$ and another must be a square root of $-2.$ But if the square root is a real matrix, that means the conjugates must also be eigenvalues for the square root, and hence, four eigenvalues. @xldd $\endgroup$ Jul 28 at 14:51
0
$\begingroup$

$\begin{pmatrix}a & b \\ c & d \end{pmatrix}\begin{pmatrix}a & b \\ c & d \end{pmatrix}= \begin{pmatrix}a^2+ bc & ab+ bd \\ ac+ cd & bc+ d^2 \end{pmatrix}$ so you are looking for a, b, c, and d satisfying the four equations $a^2+ bc= -1$, ab+ bd= 0, ac+ cd= 0, and $bc+ d^2= -2$.

ac+ cd= c(a+ d)= 0 so either c=0 or a+ d= 0. ab+ bd= b(a+ d)= 0 so either b=0 or a+ d= 0.

If either b and c is 0 then the $a^2= -1$ or $d^2= -2$ so there are no real values.

If a+ d= 0 then d= -a so $bc+ d^2= bc+ a^2= -2$ which is impossible since $a^2+ bc= -1$.

Yes, there is no matrix, $A$, such that $A^2= \begin{pmatrix}-1 & 0 \\ 0 & -2\end{pmatrix}$.

$\endgroup$

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