Roots of Matrices and Diagonalization

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.
• Your answers are correct. However, your statement "the third condition will only be satisfied when the eigenvalues of $A_i$ have real square roots" is false. As a counterexample, consider $-I$, which has the square root $$\pmatrix{0&-1\\1&0}$$ – Omnomnomnom May 18 '15 at 16:21