For arbitary dimensions, showing that the square root of the identity matrix is diagonalizable For arbitrary size, let $A$ be a square matrix such that $A^2=I$ where $I$ is the identity matrix. Then how do I show that $A$ is diagonalizable? Also, is $A$ diagonalizable when $A^2=B$ for some arbitrary diagonalizable matrix $B$?
 A: The polynomial $x^2-1$ annihilates $A$, so its minimal polynomial must be one of $x-1$, $x+1$ or $x^2-1=(x-1)(x+1)$. All of these are products of distinct linear factors, therefore $A$ is diagonalizable.  
One the other hand, consider $B=-I_2$. For $A=\tiny{\begin{bmatrix}0&-1\\1&0\end{bmatrix}}$ we have $A^2=B$, but $A$ is not diagonalizable over the reals.
A: Every vector $x \in \mathbb{R}^n$ can be written as $x = \frac{1}{2}(x + Ax) + \frac{1}{2}(x - Ax)$ where $\frac{1}{2}(x + Ax)$ is in the eigenspace for $\lambda = 1$ and $\frac{1}{2}(x - Ax)$ is in the eigenspace for $\lambda = -1$.  Therefore, the eigenvectors for $\lambda = \pm 1$ span $\mathbb{R}^n$, implying that $A$ is diagonalizable.

More generally, we can apply a useful decomposition theorem: suppose $T \in L(V)$ is a linear operator, and we have pairwise relatively prime polynomials $p_1, \ldots, p_r \in F[t]$, where $F$ is the scalar field, and let $p(t) := p_1(t) \cdots p_r(t)$.  Then $\ker p(T) = \ker p_1(T) \oplus \cdots \oplus \ker p_r(T)$.
In this particular case, we can use $p_1(t) = t-1$, $p_2(t) = t+1$; so $p(t) = t^2-1$, $\ker p(A) = \ker 0 = \mathbb{R}^n$, and thus $\mathbb{R}^n = \ker(T-I) \oplus \ker(T+I)$.

(Do note, however, that these proofs only work if the scalar field does not have characteristic 2.  In fact, the statement is false over characteristic 2 scalar fields; for example, $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^2 = I$ in this case but $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ is not diagonalizable.)
A: For every matrix $A$, there exists an invertible matrix $T$ such that $J:=TAT^{-1}$ is in Jordan form. It is then easily seen that
$$J^2 = TAT^{-1}TAT^{-1} = TA^2T^{-1} = TIT^{-1} = TT^{-1} = I$$
The diagonal of $J^2$ contains just the diagonal values of $J$ squared, therefore on the diagonal, so to get $I$, there can only be $1$ or $-1$ on the diagonal. It is also easily checked that on squaring, for each Jordan block, the subdiagonal is replaced by $2\lambda$ where $\lambda$ is the eigenvalue ($=$ value on the diagonal). But to be equal to the identity matrix, the subdiagonal of the square must be zero;but we've seen before that $\lambda=\pm1\ne 0$. Therefore the only way the condition can be fulfilled is if all Jordan blocks have no subdiagonal, that is, are of size $1$, which means that $J$ is actually a diagonal matrix. But that means that $A$ is diagonalizable.
A: Let $K$ be an algebraically closed field with characteristic not $2$ and $A\in M_n(K)$ be  a diagonalizable invertible matrix. We consider the equation $(*)$ $X^2=A$ where $X\in M_n(K)$. 
$\textbf{Proposition}$. Then $(*)$ has at least one  solution $X$ and each solution is diagonalizable.
$\textbf{Proof}$. We may assume that $A=diag(\lambda_1I_{i_1},\cdots,\lambda_kI_{i_k})$ where the $(\lambda_i)$ are non-zero distinct. 
i) Thus, $2^k$ obvious solutions are $X=diag(\pm \sqrt{\lambda_1}I_{i_1},\cdots,\pm \sqrt{\lambda_k}I_{i_k})$. Obviously, the existence stands even if $A$ is singular. 
ii) Conversely, $X$ is in the form $X=diag(X_1,\cdots,X_k)$ where ${X_j}^2=\lambda_jI_{i_j}$. If $X$ is not diagonalizable, then some $X_j$ is not diagonalizable; since we may assume that $\lambda_j=1$, that implies that there is $m\geq 2$ s.t. $(\pm I_m+J_m)^2=I_m$, where $J_m$ is the nilpotent Jordan matrix of dimension $m$. Note that $(\pm I_m+J_m)^2=I_m\pm 2J_m+{J_m}^2\not =I_m$ because $charac(K)\not= 2$. That is contradictory.  $\square$
