Why is the condition enough for a matrix to be diagonalizable? I've heard that for a matrix $A\in M_n(\mathbb{C})$, if $A^3=A$, then $A$ is diagonalizable. Does there happen to be a proof or reference as to why this is true?
Out of curiosity, is it necessary that the entries be from $\mathbb{C}$? Would any field $F$ work just as well, or is this possibly a special fact related to the properties of $\mathbb{C}$?
 A: This holds because it holds for the Jordan normal form. Every matrix over $\mathbb C$ is similar to a matrix in Jordan normal form. The third power of a Jordan normal block of size at least $2$ is
$$
\pmatrix{\lambda&1\\
&\lambda&\ddots\\
&&\ddots&1\\
&&&\lambda&}^3
=
\pmatrix{\lambda^3&3\lambda^2\\
&\lambda^3&\ddots\\
&&\ddots&3\lambda^2\\
&&&\lambda^3&}
$$
(with further non-zero entries further above the diagonal). If $A^3=A$, then this also holds for any similar matrix, and thus for the Jordan normal form, which implies $\lambda=\lambda^3$ and $3\lambda^2=1$. These equations don't have a common solution, which shows that all Jordan blocks are of size $1$, that is, the Jordan normal form is diagonal. This clearly also works for any other integer power of $A$.
Regarding arbitrary fields, this exposition proves that every matrix over an algebraically closed field is similar to a matrix in Jordan normal form. Thus the result holds in any algebraically closed field $\mathbb F$ in which there is no common solution for $\lambda=\lambda^3$ and $3\lambda^2=1$. For  $\operatorname{char}\mathbb F=3$, the second equation reads $0=1$, which has no solutions. For  $\operatorname{char}\mathbb F\ne3$, we can multiply the first equation by $3$ and substitute for $3\lambda^2$ from the second equation to get $3\lambda=\lambda$ and thus $3=1$ (since $\lambda=0$ doesn't solve the second equation). Subtracting $1$ yields $2=0$, so the statement holds if $\operatorname{char}\mathbb F\ne2$.
For $\operatorname{char}\mathbb F=2$, we have $\pmatrix{1&1\\0&1}^3=\pmatrix{1&1\\0&1}$, and this matrix is not diagonalizable.
A: An $n$ by $n$ matrix $A$ with coefficients in a field $K$ is diagonalizable if and only if its minimal polynomial $f$ has no multiple roots and splits over $K$. 
Clearly, if $A$ is diagonalizable, $f$ has no multiple roots and splits over $K$. 
For the converse, one can use the Chinese Remainder Theorem to reduce the statement to the case $A=0$.
A: Here is an argument/explanation that does not use Jordan normal form or charateristic polynomials. Instead it uses a more linear transformation-based
viewpoint, and properties of projectors.  I find this a useful way to think about these kinds of questions, which is why I'm positing it.

Note first that $A^3 = A$ implies that $A^4 = A^2$.  This means that $A^2$ is a projection.  In general, if $P^2  = P$, for some linear transformation of a vector space $V$ (over any field), then $V$ is the direct sum of the image of $P$ and
the image of $I- P$.  (Here $I$ denotes the identity.)  Furthermore,
on the the image of $P$, the transformation $P$ acts by the identity, 
while on the image of $I - P$, it acts by zero.
So in our case, the $3$-dim'l vector space $V$ on which your matrix $A$ is
acting splits as the direct sum of the image of $A^2$, on which $A^2$ 
acts by the identity, and as the image of $I - A^2$, on which (using the
equation $A - A^3 = 0$) we see that $A$ acts by zero.
So we have partially diagonalized $A$; we have decomposed $V$ into a sum
of two subspaces, each invariant under $A$, with $A^2 = I$ on the first, 
and $A = 0$ on the second.
To complete the diagonalization, we make the same kind of argment, but now
we may assume that $A^2 = I$.
From $A^2 = I$, we find that $\bigl(\dfrac{I-A}{2}\bigr)^2 = \dfrac{I-A}{2}$.
Thus $(I-A)/2$ is again a projector, and so the subspace on which $A^2 = I$
decomposes as a sum of two subspaces, one on which $(I-A)/2 = I,$
which is to say $A = - I$, and one on which $(I-A)/2 = 0$, which is to say $A = I$.
Putting it altogether, we decomposed our original space $V$ as the sum of three
$A$-invariant subspaces, on which $A$ acts by $0$, $-1$, and $1$ respectively.

The argument works over any field where $2$ is invertible.  If we are in char. $2$, then the decomposition into the sum of spaces on which $A = 0$ and $A^2 = I$ is still possible, but (as Joriki points out in his answer) we can't necessarily
diagonalize a matrix satisfying $A^2 = I$.
One way to see this is to note that in char. $2$, $A^2 = I$ is equivalent to
$(A-I)^2 = 0$, and so we can construct matrices $A$ such that $A^2 = I$ by choosing nilpotent matrices $N$ such that $N^2 = 0$, and then setting
$A = I + N$.  If $N \neq 0$ (which is possible for $n\times n$ matrices with $n > 1$), then such an $A$ (identity plus non-zero nilpotent) is not diagonalizable (in any characteristic; but away from char. $2$, matrices of this form can't
satisfy $A^2 = 0$).
A: You might also want to look at the minimal polynomial $\mu_A$ of $A$. If $A^3-A=0$, then $\mu_A \mid X^3-X=X(X^2-1)$. If $\mathrm{char}(K)\neq 2$ (if $K$ is the underlying field) then this polynomial is equal to $X(X-1)(X+1)$ and a matrix with a minimal polynomial which splits into linear factors with multiplicity 1 is diagonalisable.
