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

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.

## 3 Answers

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.

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

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.

• Andrew What does "any real matrix square root must have four eigenvalues" means?
– xldd
Jul 28 at 14:42
• 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 Jul 28 at 14:51

$$\begin{pmatrix}a & b \\ c & d \end{pmatrix}\begin{pmatrix}a & b \\ c & d \end{pmatrix}= \begin{pmatrix}a^2+ bc & ab+ bd \\ ac+ cd & bc+ d^2 \end{pmatrix}$$ 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 $$A^2= \begin{pmatrix}-1 & 0 \\ 0 & -2\end{pmatrix}$$.