$\sin(A)$ , where $A$ is a matrix

Firstly, thank you for this interesting follow up problem.

I used the fact that any square matrix can be factorised according to this link .

So, the matrix $A$ can be written as:

$A = V\Lambda V^{-1}$

Here, $\Lambda$ is a diagonal matrix comprising of the eigenvalues of the matrix $A$ while $V$ is a square matrix comprising of the eigen vectors of $A$ in each column.

$\Lambda = \begin{bmatrix} \lambda_1 & 0\\ 0 & \lambda_2\\ \end{bmatrix}$

Having decomposed the above matrix, we now look at the sine series expansion for matrices:

$\sin(A) = A -\frac{A^3}{3!} + \frac{A^5}{5!} + ...$

$\sin(A) = V\Lambda V^{-1} - \frac{(V\Lambda V^{-1})^3}{3!} + \frac{(V\Lambda V^{-1})^5}{5!} + \dots$

Remember that

$VV^{-1} = I$

$V^{-1}V = I$

The above matrix exponential can be rewritten as:

$\sin(A) = V\Lambda V^{-1} - \frac{V\Lambda V^{-1}V\Lambda V^{-1}V\Lambda V^{-1}}{3!} + \dots$

All the terms having $V^{-1}V$ combined cancel to be the identity matrix. The expression simplifies to:

$\sin(A) = V\Lambda V^{-1} - \frac{V\Lambda^3V^{-1}}{3!} + \dots$

or:

$\sin(A) = V(\Lambda - \frac{\Lambda^3}{3!} + \dots)V^{-1}$

This gives us the result:

$\sin(A) = V\sin(\Lambda)V^{-1}$

Using a similar analysis, for a diagonal matrix it can also be proved that:

$\sin(\Lambda) = \begin{bmatrix} \sin(\lambda_1) & 0\\ 0 & \sin(\lambda_2)\\ \end{bmatrix}$

Finally, the required Final result is:

$\sin(A) = V\begin{bmatrix} \sin(\lambda_1) & 0\\ 0 & \sin(\lambda_2)\\ \end{bmatrix}V^{-1}$

From here, it is just a matter of computation.

• First, Compute the eigen values ( $\lambda_1$ and $\lambda_2$ ) of the matrix $A$ and its corresponding eigen vectors.
• Then combine the eigen vectors as column entries to obtain the square matrix $V$ .
• Use the final result to get the answer