GO Classes DPP | GATE CS, DA | Linear Algebra | Eigen Value

Answer 1

Let, $A=\begin{bmatrix}B & C\\ 0 & D\end{bmatrix}$
 

where $B,C,D$ are $3\times 3$ matrices. The eigenvalues of $A$ are the roots of its characteristic polynomial

$p_A(\lambda)=\det(\lambda I_6-A)$.
 

Now,

$\lambda I_6-A=
\begin{bmatrix}
\lambda I_3-B & -C\\
0 & \lambda I_3-D
\end{bmatrix}$

This is a block upper triangular matrix, so its determinant is the product of the determinants of the diagonal blocks:

$p_A(\lambda)=\det(\lambda I_3-B)\,\det(\lambda I_3-D)$
 

Therefore the roots of $p_A(\lambda)$ are exactly the roots of

$\det(\lambda I_3-B)=0$ $\quad$ and $\quad$ $\det(\lambda I_3-D)=0$.

So the eigenvalues of $A$ are the eigenvalues of $B$ together with the eigenvalues of $D$.
 

Given eigenvalues of $B$ are $1,2,3$ and eigenvalues of $D$ are $7,8,9$, the eigenvalues of $A$ are $\boxed{1,2,3,7,8,9}$.

Answer 2

Answer: C

Answer 3

• As eigen values of Block Upper Triangular matrix are eigen values of diagonal submatrices. In this case those sub matrices are B & D.
• So, the eigen values of A would be 1,2,3,7,8,9 hence option 'C' would be the right answer.