Prove this matrix to be unitary

Question

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!

Answer 1

$$ B = \left[ \begin{array}{cc} \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{i}{\sqrt{2}} \end{array} \right] \left[ \begin{array}{cc} 2i & 0 \\ 0 & 0 \end{array} \right] \left[ \begin{array}{cc} \frac{-i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{-i}{\sqrt{2}} \end{array} \right] $$ 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 $B^n$).

Answer 2

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.