Question

The matrix

\[ A= \begin{bmatrix} 1 & 1+i & 2i \\ 1-i & 3 & 4 \\ -2i & 4 & 5 \end{bmatrix} \]

has

• A. only real eigen values
• B. eigen values of the form \(a+ib,\; a\neq0,\; b\neq0\)
• C. eigen values of the form \(x+iy,\; x=0\)
• D. no inverse

Hint:

For complex matrices, \[ \boxed{ A=A^{H} \Longrightarrow \text{All eigen values are real.} } \] This is one of the most important properties of Hermitian matrices.

Answer 1

The Correct Option is A

Step 1: Check whether the matrix is Hermitian. A matrix is Hermitian if \[ A=A^{H}, \] where \(A^{H}\) denotes the conjugate transpose. For the given matrix, \[ \overline{(1+i)}=1-i,\qquad \overline{(2i)}=-2i, \] and the corresponding symmetric entries satisfy \[ a_{12}=\overline{a_{21}},\qquad a_{13}=\overline{a_{31}},\qquad a_{23}=\overline{a_{32}}. \] Hence, \[ \boxed{A=A^{H}.} \]

Step 2: Use the property of Hermitian matrices. A Hermitian matrix always has \[ \boxed{\text{all eigen values real}.} \] Therefore, \[ \boxed{\text{only real eigen values}} \] is the correct answer. Thus, \[ \boxed{(A)} \] is the correct answer.