GATE Breaker

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., $A^4 = I$. 

Suppose that the eigenvalues of matrix $A$ are 1, 2, 4. Find the determinant of $(A^{-1})^T$.  Answer: 0.125  Explanation:  The eigenvalues of matrix $A$ are 1, 2, 4.  Therefore, determinant of $A$, $|A|$ = product of the  eigenvalues of $A$ = 8.  Now, $|A^{-1}| = \frac{1}{|A|} = \frac{1}{8}$.  Hence, $|(A^{-1})^T| = |A^{-1}| = \frac{1}{8} = 0.125$. 

Two eigenvalues of a $3 \times 3$ real matrix $P$ are $(2 + \sqrt{-1})$ and $3$. Find the determinant of $P$.  (GATE 2016)  Answer: 15  Explanation:  Here, $P$ is a $3 \times 3$ real matrix.  The given eigenvalues are $2 + i$ and $3$.  Since $P$ is a real matrix, the third eigenvalue must be $2 - i$.  We know that the determinant of a real matrix is equal to the product of its eigenvalues.  Therefore, the determinant of $P$ = product of eigenvalues = $(2 + i) \times 3 \times (2 - i)$ = $15$.