Question

find the eigenvalues of the following matrix

\\a = \

$$\begin{bmatrix} 3 & 2 - i & -3i \\\\ 2 + i & 0 & 1 - i \\\\ 3i & 1 + i & 0 \\end{bmatrix}$$

\\

-6
-2
-1
1
2
6

Explanation:

Verify matrix properties

Using the Hermitian Matrix and Eigenvalues of Hermitian Matrix knowledge points

$$ A =
LATEXBLOCK0
$$

$$ A^H = \overline{A^T} =
LATEXBLOCK1
= A $$

All eigenvalues \(\lambda_1, \lambda_2, \lambda_3\) must be real.

Calculate trace and determinant

Using the Matrix Trace knowledge point

$$ \text{tr}(A) = 3 + 0 + 0 = 3 \implies \lambda_1 + \lambda_2 + \lambda_3 = 3 $$

$$
LATEXBLOCK2
$$

$$ \lambda_1 \lambda_2 \lambda_3 = 12 $$

Find sum of principal minors

We compute the sum of the principal \(2 \times 2\) minors \(M_{11} + M_{22} + M_{33}\):

$$ M_{11} = \det
LATEXBLOCK3
= -(1-i)(1+i) = -2 $$

$$ M_{22} = \det
LATEXBLOCK4
= -(-3i)(3i) = -9 $$

$$ M_{33} = \det
LATEXBLOCK5
= -(2-i)(2+i) = -5 $$

$$ M_{11} + M_{22} + M_{33} = -2 - 9 - 5 = -16 $$

Solve characteristic equation

The characteristic equation is:

$$ \lambda^3 - \text{tr}(A)\lambda^2 + (M_{11} + M_{22} + M_{33})\lambda - \det(A) = 0 $$

$$ \lambda^3 - 3\lambda^2 - 16\lambda - 12 = 0 $$

Testing integer roots among divisors of \(-12\):
For \(\lambda = -1\):

$$ (-1)^3 - 3(-1)^2 - 16(-1) - 12 = -1 - 3 + 16 - 12 = 0 $$

Factoring out \((\lambda + 1)\):

$$ (\lambda + 1)(\lambda^2 - 4\lambda - 12) = 0 $$

$$ (\lambda + 1)(\lambda - 6)(\lambda + 2) = 0 $$

The eigenvalues are \(\lambda = -2, -1, 6\).

Match with given options

The set of eigenvalues is \(\{-2, -1, 6\}\).
Looking at the options provided:

• \(-6\) (not an eigenvalue)
• \(-2\) (is an eigenvalue)
• \(-1\) (is an eigenvalue)
• \(1\) (not an eigenvalue)
• \(2\) (not an eigenvalue)
• \(6\) (is an eigenvalue)

Answer:

• O -6
• O -2 (Correct answer)
• O -1 (Correct answer)
• O 1
• O 2
• O 6 (Correct answer)