They give me the matrix \begin{bmatrix}3 \ -2 \\ \ 3 \ \ \ \ \ \ 1\end{bmatrix}
Until now I have done
Det(λI – A) = Det of\begin{bmatrix}λ -3 \ \ \ \ \ \ 2 \\ \ -3 \ \ \ \ \ \ λ-1\end{bmatrix}
I get that the eigen values are:
λ = 2+$i\sqrt{5}$
λ = 2-$i\sqrt{5}$
And now I am stuck when I try to find the basis for each of the eigenspaces. I know I have to do
:
I$λ_1 – A$ and find the basis for the resulting matrix but the $i$ in the eigenvalues are giving me a hard time.
Best Answer
As you have found, the eigenvalues of $A = \begin{bmatrix} 3 & -2 \\ 3 & 1 \end{bmatrix}$ are $\lambda_1 = 2 - \sqrt{5} i$ and $\lambda_2 = 2 + \sqrt{5} i$. And so, we can find the corresponding eigenvectors by solving $(A - \lambda I) = \mathbf{0}$. For example, for $\lambda_1$,
$$\begin{bmatrix} 3 - (2 - \sqrt{5}i) & -2 \\ 3 & 1 - (2 - \sqrt{5}i) \end{bmatrix} \mathbf{x} = \mathbf{0}$$
As you have probably found out, doing row reduction on the above is difficult. However, there's a trick: $(A - \lambda I) = \mathbf{0}$ must have non-trivial solution, which implies that the matrix $(A - \lambda I)$ is not invertible. That means, $(A - \lambda I)$'s rows are linearly dependent. What does that mean? We can pick one of the equivalent linear equations below, because they both describe the same relationship between $x_1$ and $x_2$:
$$3 - (2 - \sqrt{5}i) x_1 - 2 x_2 = 0$$
$$3 x_1 + 1 - (2 - \sqrt{5}i) x_2 = 0$$
Suppose we pick the first one:
$$3 - (2 - \sqrt{5}i) x_1 - 2 x_2 = 0$$
$$1 + \sqrt{5}i x_1 - 2 x_2 = 0$$
$$x_1 = \frac{2 x_2}{1 + \sqrt{5}i}$$
therefore, the eigenvectors corresponding to $\lambda_1$ have the following form:
$$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \frac{2 x_2}{1 + \sqrt{5}i} \\ x_2 \end{bmatrix} = x_2 \begin{bmatrix} \frac{2}{1 + \sqrt{5}i} \\ 1 \end{bmatrix}$$
And so, the basis for eigenspace corresponding to $\lambda_1$ is $\Bigg\{\begin{bmatrix} \frac{2}{1 + \sqrt{5}i} \\ 1 \end{bmatrix}\Bigg\}$. You can use the same technique to find the basis for eigenspace corresponding to $\lambda_2$.