Norms of idempotent matrices So if I have a random idempotent matrix (a matrix that when squared equals itself) how do I go about calculating its 2-norm? 
I know that a idempotent matrix has eigenvalues of only 0 or 1 and I know that in most cases the 2-norm is equal to the largest eigenvalue (although this isn't always the case). 
It doesn't make much sense to me. 
 A: 
In most cases the 2-norm is equal to the largest eigenvalue.

I don't know what you mean by "most cases"; it is true for normal matrices if you add the absolute value, but not typically otherwise.  For nonzero idempotent matrices, the norm is $1$ if and only if the matrix is self-adjoint (if and only if it is normal).  In general, the norm of $T$ is the square root of the largest eigenvalue of $T^*T$ (or of $TT^*$).
If $E$ is an $n$-by-$n$ complex idempotent matrix acting on $\mathbb C^n$, and $K\subseteq \mathbb C^n$ is its range, then you can find an orthonormal basis $(e_1,\ldots,e_k)$ for $K$, and complete it to an orthonormal basis $B$ for $\mathbb C^n$; the remaining vectors form an orthonormal basis for $K^\perp$.  With respect to this basis, $E$ has matrix
$$[E]_B =\begin{bmatrix}I & A\\ 0& 0\end{bmatrix},$$
for some $k$-by-$(n-k)$ matrix $A$.  Note that $A=0$ if and only if $E$ is self-adjoint. 
Then the norm of $E$ depends on $A$ according to the equation $\|E\|^2=\|I+AA^*\|=1+\|AA^*\|.$  Since $AA^*$ is positive semidefinite, $\|AA^*\|$ is its largest eigenvalue. 
A: The 2-norm is square root of the largest eigenvalue of $A^TA$, which is the largest absolute value of an eigenvalue of $A$ only if $A$ is normal. Yours is not.
Here, you might be stuck maximizing:
\begin{gather}
\|A\|_2^2 = \sup_{\|x\|\neq 0}{\frac{x^TA^TAx}{x^Tx}}
\end{gather}
