Prove this matrix to be unitary This is a homework question so, hints are appreciated. But if someone is generous enough, to show the full calculation, I'd be quite grateful!
Say a matrix B is anti-hermitian:$$\begin{bmatrix}
i & -1\\ 
1 & i
\end{bmatrix}$$
And I want to calculate the matrix exponential $e^{Bt}$, where the exponential is a function of time. Now after I attempted my own calculation for $e^{Bt}$, using Taylor series and Caley Hamilton theorem, I ended up with this:
\begin{bmatrix}
\ \frac{e^{2it}+1}{2} & \frac{e^{2it}-1}{2i}\\ 
 \frac{1- e^{2it}}{2i} & \frac{e^{2it}+1}{2}
\end{bmatrix}
But here's the problem, in theory if the exponent of a unitary operator is anti-hermitian, then the operator is unitary. But now that I have gotten the matrix representation of the matrix exponential, and I want to prove the above matrix to be unitary using the statement:
$$U^{-1}=U^{\dagger}$$ 
I run into some problem. Can some just please check if my matrix exponential is correct or if I am missing something, that prevents me from proving $e^{Bt}$ as unitary using $U^{-1}=U^{\dagger}$. 
Thanks a ton, in advance!
 A: Consider $B$. As user7440 pointed out its eigenvalues are $0$ and $2i$. As the powers of $B$ are linear combinations of $I$ and $B$ (Cayley-Hamilton) so,
taking limits, is $\exp(tB)$, that is
$$\exp(tB)=rI+sB$$
for some $r$ and $s$.
Applying this identity to an eigenvector of $B$ gives $e^{t\lambda}=r+s\lambda$
for each eigenvalue of $B$. So
$$1=r+0s$$
and
$$e^{2it}=r+2is.$$
Then $r=1$ and $s=\frac1{2i}(e^{2it}-1)$. Therefore
$$\exp(tB)=I+\frac{e^{2it}-1}{2i}B$$
which simplifies to user7440's expression.
A: $$
B = \left[ \begin{array}{cc} \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}}
\end{array} \right]
\left[ \begin{array}{cc} 2i & 0 \\ 0 & 0
\end{array} \right]
\left[ \begin{array}{cc} \frac{-i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{-i}{\sqrt{2}}
\end{array} \right]
$$
So
$$
e^{Bt} = \left[ \begin{array}{cc} \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}}
\end{array} \right]
\left[ \begin{array}{cc} e^{2it} & 0 \\ 0 & 1
\end{array} \right]
\left[ \begin{array}{cc} \frac{-i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{-i}{\sqrt{2}}
\end{array} \right]
=
\left[ \begin{array}{cc} \frac{e^{2it}+1}{2} & i\frac{e^{2it}-1}{2} \\ i\frac{1- e^{2it}}{2} & \frac{e^{2it}+1}{2}
\end{array} \right]
$$
For $t = 0$, we recover the identity matrix and, for $t=0$, its derivative is equal to $B$ (while the derivative of your result does not yield $B$).
(EDIT): $e^{Bt}$ is not equal to the matrix whose entries are the exponential of the entries of $B$ (the same remark applies for $B^n$).
