Complex Eigenvalues

Since $\lambda_2$ is complex, its complex conjugate must also be an eigenvalue!

Therefore, $\lambda_3 = \overline{\lambda}_2 \ \ \implies\ \ \boxed{\lambda_3 = 2 \colorFour{+}3i}$

Similarly, since we know the complex eigenvector associated with $\lambda_2$, we simply take its conjugate to find the eigenvector for $\lambda_3$:

$\vec x_3 = \overline{\vec x_2} \ \ \implies \ \ \boxed{ \vec x_3 = \begin{bmatrix} 2 \colorFour{-} i\\ 5 \end{bmatrix} }$

Find the Eigenvalues

Solve the equation:

$\begin{array}{rcl} \detp{A-\lambda I}&=&0\\[1em] \colorFour{ \left|\begin{array}{cc}3-\lambda & -5\\1&-1-\lambda\end{array}\right|} &=&0\\[1em] (3-\lambda)(-1-\lambda)-(-5)(1)&=&0\\[1em] (\lambda^2-2\lambda-3)+5&=&0\\[1em] \lambda^2-2\lambda+2&=&0 \end{array}$

We can solve for $\lambda$ using the quadratic formula:

$\lambda \ \ =\ \ \dfrac{-(-2) \pm \sqrt{(-2)^2-4(1)(2)}}{2(1)} \ \ =\ \ \dfrac{2 \pm \sqrt{-4}}{2} \ \ =\ \ \dfrac{2 \pm 2i}{2} \ \ =\ \ 1\pm i$

Then the eigenvalues are $\boxed{ \lambda_1 = 1+i }$ and its conjugate, $\boxed{ \lambda_2=1-i }$.

Find the Eigenvectors

$\colorThree{\lambda_1 = 1+i}$

Substituting this eigenvalue into the matrix above ($A-\lambda I$), our goal is to solve for $\vec x_1$:

$\colorFour{ \begin{bmatrix} 3-\colorThree{(1+i)} & -5\\ 1 & -1-\colorThree{(1+i)} \end{bmatrix}} \vec x_1 = \begin{bmatrix} 0 \\0 \end{bmatrix}$

Solve by writing the augmented matrix of this system and row-reducing:

$\begin{array}{ll} &\left[\begin{array}{cc|r} 2-i& -5 &0\\ 1&-2-i& 0\\ \end{array}\right] \begin{array}{c} ~\\ R_2 \leftrightarrow R_1\\ \end{array} \\[2em] \longrightarrow &\left[\begin{array}{cc|r} 1&-2-i& 0\\ 2-i& -5 &0\\ \end{array}\right] \begin{array}{c} ~\\ R_2 - (2-i)R_1\\ \end{array} \\[2em] &\text{Note: }\ (-2-i)(2-i) = -4-\cancel{2i}+\cancel{2i}+i^2=-5 \\[1em] \longrightarrow &\left[\begin{array}{cc|r} 1&-2-i& 0\\ 0& 0 &0\\ \end{array}\right] \end{array}$

Including a parameter for the second variable (since there is no leading 1 in the second column), we get the equations:

$\left\{\begin{array}{rcl} 1x_1+(-2-i)x_2&=&0\\ x_2&=&t \end{array}\right. \quad \implies \quad \left\{\begin{array}{rcl} x_1&=&(2+i)t\\ x_2&=&t \end{array}\right. \quad \implies \quad \vec x_1 = t \begin{bmatrix} 2+i\\ 1 \end{bmatrix}$

So an eigenvector associated with $\lambda_1 = 1+i$ is: $\boxed{ \vec x_1 = \left[\begin{array}{c} 2+i\\ 1 \end{array}\right] }$

$\colorThree{\lambda_2 = 1-i}$

Since the second eigenvalue is the conjugate of the first, its eigenvector is simply the conjugate of the first eigenvector!

Therefore, negating all of the imaginary parts, we have: $\boxed{ \vec x_2 = \left[\begin{array}{c} 2-i\\ 1 \end{array}\right] }$

For practice, check your work by applying the definition of eigenvalues/eigenvectors: $A\vec x = \lambda \vec x$.