If $A^4=I$ then $A$ must be diagonalizable? Suppose we have a real matrix $A$ which satisfies $A^4=I$, can we determine if $A$ is diagonalizable?
I believe the answer is that we can't because all we know about the matrix $A$ is that it is invertible (otherwise $A^4$ couldn't be an invertible matrix)..
How can I prove it? How can I find such a matrix $A$ which isn't diagonalizable but $A^4 = I$?
The only matrix $A$ I was able to find which satisfies $A^4=I$ is the identity matrix itself but the identity matrix is diagonalizable.
 A: I suppose we are considering $\mathbb{R}$ or $\mathbb{C}$.
The minimal polynomial of $A$ divides $$x^4 -1 = (x-1) (x+1) (x^2 +1) $$
A matrix is diagonalizable over the field $F$ if and only if its minimal polynomial is a product of distinct linear factors over $F$, moreover the minimal and characteristic polynomial have the same roots. 
Call $p(x)$ the minimal polynomial of $A$ in the field we are considering.
Thus if $F = \mathbb{C}$ the matrix is always diagonalizable, because the roots of $x^4 -1$ are all differents and so $p(x)$ is a product of distinct linear factors over $\mathbb{C}$ because $p(x) \mid x^4 -1$.
If $F = \mathbb{R}$ we need further information to decide if $A$ is diagonalizable or not.  
A: $\text{Hint (for the complex matrices):}$
actually your claim is true. If $p(A)=0$ where $p$ is a polynom with simple roots, then $A$ is diagonalizable. It follows from Jordan form of a matrix.
Let
$$A = Q^{-1}J_A Q=Q^{-1}
\begin{pmatrix}
J_A^1 & 0 &\dots& 0\\
0& J_A^2&\dots& 0\\
&\dots&\dots\\
0&0&\dots&J_A^n
 \end{pmatrix}
Q$$
where $J_A^m, m=1,\dots,n$ are the Jordan cells of $J_A$:
$$J_A^m=
\begin{pmatrix}
\lambda_m & 1 & 0 & \dots&  0\\
0&\lambda_m & 1 & \dots&  0\\
&\dots& &\dots\\
0 & 0 & 0 & \dots&  \lambda_m\\
\end{pmatrix}
$$
We claim that if $p(A)=0$ then all cells $J_A^m$ have the sizes $1\times 1$, otherwise
$$p(A) = Q^{-1}p(J_A) Q=Q^{-1}
\begin{pmatrix}
p(J_A^1) & 0 &\dots& 0\\
0& p(J_A^2)&\dots& 0\\
&\dots&\dots\\
0&0&\dots&p(J_A^n)
 \end{pmatrix}
Q$$
$$p(J_A^m)=
\begin{pmatrix}
p(\lambda_m) & \frac{p'(\lambda_m)}{1!} & \frac{p''(\lambda_m)}{2!} & \dots\\
0&p(\lambda_m) & \frac{p'(\lambda_m)}{1!} & \dots\\
&\dots& &\dots\\
0 & 0 & 0 & \dots&  p(\lambda_m)\\
\end{pmatrix}\ne 0
$$
since $p$ doesn't have multiple roots: $p(\lambda)=p'(\lambda)=0$ has no solution.
In your case $p(x)=x^4-1$ has 4 distict roots.
A: The polynomial $P(x)=x^4-1$  annihilates $A$ so the minimal polynomial $\mu_A$ of $A$ divides $P$ so $\mu_A$ has simple roots so $A$ is diagonalizable.
A: No, consider the rotation by $\pi/2$ in $\mathbb R^2$:
$$A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$
Which must satisfy $A^4=I$, but has no (real!) eigenvectors, so it's not diagonalizable.
A: The term diagonalisable needs to be qualified by the field$~K$ over which one is working; for this reasons it is best to talk of diagonalisable linear operators (where $K$ is implicit in the vector space$~V$ that the operator acts upon) rather than matrices; the latter can be interpreted over various fields, and may be diagonalisable over some but not over others. The question stresses that that matrix has real entries, but that does not prevent it from being the matrix of a complex linear operator; I will however take this as an indication that being diagonalisable as a real linear operator is meant.
Given an annihilating polynomial for the linear operator, here $X^4-1$, one can always (in principle) factor it into pairwise relatively prime factors in $K[X]$, and obtain a direct sum decomposition of$~V$ into the kernels of those factors evaluated at the linear operator (or matrix). Here the finest such decomposition for $K=\Bbb R$ is $X^4-1=(X-1)(X+1)(X^2+1)$, so we get a decomposition $V=\ker(A-I)\oplus\ker(A+I)\oplus\ker(A^2+I)$. Each of these spaces might be $\{0\}$. When they are not, the first two spaces are the eigenspaces for $\lambda=1$ and $\lambda=-1$, respectively. However the third subspace does not contain any eigenvectors of$~A$, since the corresponding eigenvalue would have to satisfy $\lambda^2+1=0$ which is impossible for real values$~\lambda$. Having a nonzero dimensional $A$-stable subspace without eigenvectors ensures that $A$ is not diagonalisable (the restriction to a stable subspace of a diagonalisable operator is always diagonalisable), so one gets that $A$ satisfying $A^4=I$ is diagonalisable over$~\Bbb R$ if and only if $\ker(A^2+I)=\{0\}$. Equivalently, since this condition means that one has $V=\ker(A-I)\oplus\ker(A+I)$, it is diagonalisable if and only if $A^2=I$.
Just for completeness, such $A$ is always diagonalisable over $\Bbb C$, since the complex vector space $W$ acted upon decomposes $W=\ker(A-I)\oplus\ker(A+I)\oplus\ker(A-\mathbf iI)\oplus\ker(A+\mathbf iI)$, in which each of the summands is an eigenspace if it differs from$~\{0\}$.
