Calculate trace of matrix with complex eigenvalues

Question

I'm trying to calculate $\text{tr}(\exp(A)) $ for matrix $A$, and I found several topics, but not sure if I've get it all right. So I wonder if this is a correct way of doing this.

This is matrix $A$, $$ \begin{bmatrix} 0 & -1 & 1 \\ -1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{bmatrix} $$ Following the answer in here, I calculated the eigenvalues, two of which are complex numbers,

$$\ \lambda_1 = 1.52 , \lambda_2= -0.76+0.85 i , \lambda_3= -0.76-0.85 i \\ \ tr(e^A) = e^{1.52} + e^{-0.76+0.85 i} + e^{-0.76-0.85 i} \\ \ $$

As suggested here, I've written the following for the complex parts

$$\ \ e^{-0.76+0.85 i} + e^{-0.76-0.85 i}= e^{-0.76}.e^{0.85i}+e^{-0.76}.e^{-0.85i}\\ =e^{-0.76}(\cos0.85 + i \sin 0.85+\cos-0.85+i\sin-0.85) = e^{-0.76}.2\cos(0.85) \ $$

So finally we have, $\text{tr}(\exp(A)) = e^{1.52}+e^{-0.76}.2\cos(0.85) $

Thanks in advance!

Answer 1

You made a computational error somewhere. Since $A$ is a real symmetric matrix, its eigenvalues must be real. The eigenvalues of $A$ are all $-2$ and $1$ (twice). So, the eigenvalues of $e^A$ are $e^{-2}$ and $e$ (again, twice), and therefore$$\operatorname{tr}(e^A)=e^{-2}+2e.$$

Answer 2

$A$ is a diagonalizable matrix. Hence, $$ A=Q\Lambda Q^{\mathrm T}, $$ where $Q$ is orthogonal and $\Lambda$ is diagonal with the eigenvalues of $A$ on the diagonal. We have that \begin{align*} \operatorname{tr}(\exp(A)) &=\sum_{k=0}^\infty\frac1{k!}\operatorname{tr}(A^k)\\ &=\sum_{k=0}^\infty\frac1{k!}\operatorname{tr}(Q\Lambda^k Q^{\mathrm T})\\ &=\sum_{k=0}^\infty\frac1{k!}\operatorname{tr}(\Lambda^k Q^{\mathrm T}Q)\\ &=\sum_{k=0}^\infty\frac1{k!}\operatorname{tr}(\Lambda^k)\\ &=\sum_{k=0}^\infty\frac{2+(-2)^k}{k!}\\ &=2e+\frac1{e^{2}} \end{align*} using the fact that $\operatorname{tr}(ABC)=\operatorname{tr}(BCA)$ and the fact that the eigenvalues of $A$ are $1$, $1$ and $-2$.

I hope this is helpful.