Matrix exponential of a skew-symmetric Toeplitz matrix

Question

From today's exam:

Given the following matrix, $$B = \begin{pmatrix} 0 & -2/3 & 1/3\\ 2/3 & 0 & -2/3\\ -1/3 & 2/3 & 0\end{pmatrix}$$ Prove that $$\exp(aB) = I + \sin(a) B + (1 -\cos(a)) B^2$$

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I have tried expanding the exponential $$\exp(aB)=I+aB+\frac{a^2}{2}B^2+\frac{a^3}{3!}B^3 + \cdots$$

But could not manage to find the $B^k$ expression. Can anyone help with this?

Answer 1

The characteristic polynomial of the given matrix is $x^3 +x$. Hence by Cayley-Hamilton theorem, \begin{align*} B^3+B&=0\\ B^3 &= -B\\ B^4 &= -B^2\\ B^5 &= B\\ \vdots\\ \exp(aB)&= I + aB+\frac{a^2B^2}{2!}+\frac{a^3B^3}{3!}+\cdots\\ \exp(aB) &= I + \left(a-\frac{a^3}{3!}+\cdots \right) B + \left( \frac{a^2}{2!}-\frac{a^4}{4!}+\cdots \right)B^2 &&\square\\ \end{align*}