Square root of a specific 3x3 matrix From a problem set I'm working on: (Edit 04/11 - I fudged a sign in my matrix...)

Let $A(t) \in M_3(\mathbb{R})$ be defined: $$ A(t) =
 \left( \begin{array}{crc} 1 & 2 & 0 \\
 0 & -1 & 0 \\ t-1 & -2 & t \end{array}
 \right).$$
For which $t$ does there exist a $B \in M_3(\mathbb{R})$
  such that $B^2 = A$?

In a previous part of the problem, I showed that $A(t)$ could be diagonalized into a real diagonal matrix for all $t \in \mathbb{R}$, with eigenvalues $1,-1,t$.
A few things I've thought of:


*

*The matrix is not positive-semidefinite, so the general form of the square root does not work. (Is positive-definiteness a necessary condition for the existence of a square root?)

*Since $A = B^2$, then $\det(B^2) = (\det B)^2 = \det A$. So $\det A \geq 0$ for there to be a real-valued square root, forcing $t \leq 0$ to be necessary.

*My professor suggested that, since $B^2$ fits the characteristic polynomial of $A$, $\mu_A(x) = (x-1)(x+1)(x-a)$, then the minimal polynomial of $B$ must divide $\mu_A(x^2) = (x^2-1)(x^2+1)(x^2-a) = (x-1)(x+1)(x^2+1)(x^2-a)$. Examining the possible minimal polynomials, one can find the rational canonical form, square it, and check whether the eigenvalues match. This probably could get me the right answer, but I am fairly sure that there is an alternative to a "proof by exhaustion".

 A: Assume that there exists a real number $t$ and a real matrix $B$ such that $A(t)=B^2$. 
Note that $-1$ is an eigenvalue of $A(t)$, hence $A(t)+I=(B-\mathrm{i}I)(B+\mathrm{i}I)$ is singular. This implies that $B-\mathrm{i}I$ or $B+\mathrm{i}I$ is singular. Since $B$ is real valued, this means that both $B-\mathrm{i}I$ and $B+\mathrm{i}I$ are singular. Likewise, $1$ is an eigenvalue of $A(t)$, hence $A(t)-I=(B-I)(B+I)$ is singular. This implies that $B-I$ or $B+I$ is singular. Hence the eigenvalues of $B$ are $\{\mathrm{i},-\mathrm{i},1\}$ or $\{\mathrm{i},-\mathrm{i},-1\}$. 
In both cases, $B$ has three distinct eigenvalues hence $B$ is diagonalizable on $\mathbb{C}$. This implies that the eigenvalues of $A(t)$ are $-1$ (twice) and $1$ (once) and that $A(t)$ is diagonalizable as well. Hence $t=-1$. We now look at the matrix $A(-1)$.
One can check that $A(-1)$ is diagonalizable hence $A(-1)$ is similar to a diagonal matrix with diagonal $(1,-1,-1)$. Both $I_1$ (the $1\times1$ matrix with coefficient $1$) and $-I_2$ (the $2\times2$ diagonal matrix with diagonal coefficients $-1$) have square roots: take $I_1$ for $I_1$ and the rotation matrix $\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$ for $-I_2$. Hence $A(-1)$ is a square.
Finally $A(t)$ is a square if and only if $t=-1$.
A: A common exercise is to first diagonalize a matrix with positive eigenvalues, and then find a "square root" for the matrix as you have been asked to do.  This is trivial since if $A=PDP^{-1}$ and $B=P\sqrt{D}P^{-1}$ then $A=B^2$ where $\sqrt{D}$, of course, denotes the diagional matrix D after taking the square root of it's entries (which is why we need positive eigenvalues if you are working with real matricies).  Thus your question follows from what you have already determained about diagionalizing the matrix.
A not so trivial result is to prove that a matrix with positive eigenvalues has a square root regardless of whether or not it can be diagonalized.  
