Finding eigenvalues

Question

Find all eigenvalues and eigenvectors:

a.) $\pmatrix{i&1\\0&-1+i}$

b.) $\pmatrix{\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta}$

For a I got: $$\operatorname{det} \pmatrix{i-\lambda&1\\0&-1+i-\lambda}= \lambda^{2} - 2\lambda i + \lambda - i - 1 $$

For b I got: $$\operatorname{det} \pmatrix{\cos\theta - \lambda & -\sin\theta \\ \sin\theta & \cos\theta - \lambda}= \cos^2\theta + \sin^2\theta + \lambda^2 -2\lambda \cos\theta = \lambda^2 -2\lambda \cos\theta +1$$

But how can I find the corresponding eigenvalues for a and b?

Answer 1

For $a$ you can note that the matrix in case is upper triangular, or use the fact the the quadratic formula is also valid over $\mathbb{C}$.

For $b$ the last equality you have is not true, how did the $\cos(\theta)$ coefficient of $\lambda$ disappeared ? you should apply the quadratic formula in this case too and use a simple trigonometric identity.

Answer 2

You can find eigen values by putting $\det(A-\lambda E)=0$.

If you want to find corresponding eigenvectors, too, try solving this equation:

$Av=\lambda v$, where v is an eigenvector for $\lambda$ in this equation. In other termss: $Av_i=\lambda_i v_i$