Trace of a matrix

Question

What is the trace of $e^{A}$ where A is a $4 \times4$ matrix

$$\begin{bmatrix}0 & 0 & 0 & t\\ 0 & 0 &-t & 0\\ 0 & t & 0 & 0\\ -t & 0 & 0 & 0 \end{bmatrix}$$

where $t=\dfrac{\pi}{4}$.

Answer 1

Looking at the basis $(e_1,e_4,-e_2,e_3)$, $A$ is similar to a block-diagonal matrix made of two identical blocks $$ B=\pmatrix{0&t\\-t&0}. $$ Note that $A^n$ is similar to the block-diagonal matrix made of two copies of $B^n$, so $\mbox{tr} \,A^n=2\,\mbox{tr}\,B^n$ for every $n\in \mathbb{N}$. Then an easy induction shows that $$ B^{2n}=\pmatrix{(-1)^nt^{2n}&0\\ 0&(-1)^nt^{2n}}\qquad B^{2n+1}=\pmatrix{0&(-1)^nt^{2n+1}\\ (-1)^{n+1}t^{2n+1}&0}. $$ Therefore $$ \mbox{tr}\,e^A=\mbox{tr}\sum_{n=0}^{+\infty} \frac{1}{n!}A^n=\sum_{n=0}^{+\infty} \frac{1}{n!}\mbox{tr}A^n=\sum_{n=0}^{+\infty} \frac{2}{n!}\mbox{tr}B^n=4\sum_{n=0}^{+\infty} \frac{(-1)^nt^{2n}}{(2n)!}=4\cos t=2\sqrt{2}.$$

Answer 2

By definition:

$$ Tr(e^A)=\sum_{k=1}^4 e^{\lambda_k},$$

where, $\lambda_k$ are eigenvalues of $A$, which solve

$$ \det(A-\lambda I)=\lambda^4+2t^2\lambda^2+t^4=(\lambda^2+t^2)^2=0. $$

So,

$$ Tr(e^A)=2(e^{it}+e^{-it})=4\cos t|_{\pi/4}=2\sqrt{2}.$$

Btw,

$$ \det(e^A)=\prod_{k=1}^4 e^{\lambda_k}=e^{\sum_{k=1}^4\lambda_k}=e^{Tr(A)}=e^0=1.$$