Question:
For which of the following matrices $A_i$ is there
- A complex matrix $B$ such that $B^2 = A_i$;
- A self-adjoint complex matrix $B$ such that $B^2 = A_i$;
- A real matrix $B$ such that $B^2 = A_i$?
$A_1 = \begin{pmatrix} 2 & 1\\1 & 2\end{pmatrix}$, $A_2 = \begin{pmatrix} 1 & 2\\2 & 1\end{pmatrix}$, $A_3 = \begin{pmatrix} 1 & 4\\1 & 1\end{pmatrix}$.
Working:
$A_1$ and $A_2$ are both real symmetric matrices, so by the Real Spectral Theorem, there exists orthogonal matrices $P_1$ and $P_2$ such that $P_1^TA_1P_1$ and $P_2^TA_2P_2$ are both diagonal.
A self-adjoint matrix must have real eigenvalues.
The spectra of $A_1$, $A_2$ and $A_3$ are $\{1,3\}$, $\{-1,3\}$ and $\{-1,3\}$ respectively.
If we can find an invertible matrix $M_i$ such that $M_i^{-1}A_iM=D_i$, for some diagonal matrix $D_i$, then $B = \pm M_i \sqrt{D_i}M_i^{-1}$.
The Real Spectral Theorem says that $A_i$ is a real symmetric matrix precisely when there exists an orthogonal matrix $P_i$ such that $P_i^TA_iP_i$ is diagonal.
$A_1$ is a real symmetric matrix with a spectrum of $\{1,3\}$. So, there exists a matrix $B$ satisfying all three conditions.
$A_2$ is a real symmetric matrix with a spectrum of $\{-1,3\}$. So, there exists a matrix $B$ satisfying only the first condition.
$A_3$ is not a real symmetric matrix, and has a spectrum of $\{-1,3\}$. So, there exists a matrix $B$ satisfying only the first condition.