Predicting if a $3 \times 3$ matrix has a real square root

Question

Here are a few $3 \times 3$ matrices

A)$$ \begin{pmatrix} 1& 0& 0\\ 0 & 1 & 0\\ 0& 0 & 1\end{pmatrix}$$B)$$ \begin{pmatrix} 1& 0& 0\\ 0 & 1 & 0\\ 0& 0 & -1\end{pmatrix}$$C)$$ \begin{pmatrix} 1& 0& 0\\ 0 & -1 & 0\\ 0& 0 & -1\end{pmatrix}$$D)$$ \begin{pmatrix} -1& 0& 0\\ 0 & -1 & 0\\ 0& 0 & -1\end{pmatrix}$$

How can say if these matrices are the square of a $3 \times 3$ matrix with real entries or not?

Partial duplicate of https://math.stackexchange.com/questions/2290545/prove-that-matrix-can-be-square-of-matrix-with-real-entries?noredirect=1&lq=1 — Arnaud D., Jun 01 '17 at 15:48

C. Falcon · Accepted Answer · 2017-06-01T15:51:29.210

4

A necessary condition for a real matrix to have a real square root is to have a non-negative determinant.

Indeed, if $A=B^2$, then $\det(A)=\det(B)^2\geqslant 0$.

Method. If $A$ is diagonalizable and has positive eigenvalue, a square root can be found taking the square root of its eigenvalues, namely if $A=P\Lambda P^{-1}$, then $B=P\Lambda^{1/2}P^{-1}$ is a square root of $A$.

edited Jun 01 '17 at 15:51

answered Jun 01 '17 at 15:46

C. Falcon

19,042

This only helps in $B)$. – Jun 01 '17 at 15:47
Thanks.. Is there any way by which I can find the square root also?? – Fullatron Jun 01 '17 at 15:48
@OpenBall Sure, I just dropped a useful observation for more general matrices. When the matrices are diagonalizable or even diagonal things are a lot easier. – C. Falcon Jun 01 '17 at 15:49
(Non-negative determinant) – Jun 01 '17 at 15:49
@OpenBall True, but $A)$ is trivial and $C)=D)$ are covered in the question I linked to in the comments. – Arnaud D. Jun 01 '17 at 15:49
@ZacharySelk As a non-native English speaker, I always forget which term is used to say $\geqslant 0$. Thank you for the reminder! – C. Falcon Jun 01 '17 at 15:52
1

No worries, +1 for the good answer! – Jun 01 '17 at 16:02
(Foreign English speaker) @ZacharySelk – Jun 01 '17 at 16:06

Predicting if a $3 \times 3$ matrix has a real square root

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