If the matrix $A = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & -1 & 0 & -1 \\ 0 & 0 & i & i \\ 0 & 0 & 0 & -i \end{bmatrix}$, then find the value of $A^4$ using Cayley-Hamilton theorem.

If the matrix $A = \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & -1 & 0 & -1 \\ 0 & 0 & i & i \\ 0 & 0 & 0 & -i \end{bmatrix}$, then find the value of $A^4$ using Cayley-Hamilton theorem. 

(GATE 1993) 

Answer: $A^4 = I$. 

Explanation: 

Since the given matrix $A$ is an upper triangular matrix, the eigenvalues of $A$ are the main diagonal values. 

Therefore, the eigenvalues of $A$ are $1$, $-1$, $i$, $-i$. 

Hence, the characteristic equation of of $A$ is 

$(\lambda - 1)(\lambda + 1)(\lambda - i)(\lambda + i) = 0$  

$\Rightarrow  (\lambda^2 - 1)(\lambda^2 + 1) = 0$ 

$\Rightarrow  (\lambda^4 - 1) = 0$ 

Therefore, by Calley-Hamilton theorem, we have $A^4 - I = 0$, i.e., $A^4 = I$.