# Thread: Eigenvalues part 2

1. ## Eigenvalues part 2

i need to find the eigenvalues and vectors of this matrix
[0 1]
[-1 0]
lamda - A gives me this:
[Lamda -1]
[1 Lamda]

Solve the det of this to get lamda squared +1

Only solution is imaginary, i and -i
The matrix of ilamda - A:
[i -1]
[1 i]

Not sure what to do here. Never worked with imaginary numbers in a matrix. Can someone show me how to solve this. Thanks

2. Originally Posted by PensFan10
i need to find the eigenvalues and vectors of this matrix
[0 1]
[-1 0]
lamda - A gives me this:
[Lamda -1]
[1 Lamda]

Solve the det of this to get lamda squared +1

Only solution is imaginary, i and -i
The matrix of ilamda - A:
[i -1]
[1 i]

Not sure what to do here. Never worked with imaginary numbers in a matrix. Can someone show me how to solve this. Thanks
Is your matrix over $\mathbb{C}$? Or over $\mathbb{R}$? If the latter, then there are no eigenvectors. If the first, then it is the same process as working with real numbers.

3. its over c. how do you rref with i?

4. You "rref" with complex number exactly like with real numbers! Just remember to multiply and divide correctly!

But I would bother with "rref" here. Just use the definition of "eigenvalue": If $\lambda$ is an eigenvalue of A, then there exist a non-zero vector, v, such that $Av= \lambda v$ and the eigenvectors are the vectors satisfying that.

Since i is an eigenvalue, you must have $\begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= i\begin{bmatrix} x \\ y \end{bmatrix}$

$\begin{bmatrix}y \\-x\end{bmatrix}= \begin{bmatrix}ix \\ iy\end{bmatrix}$.

That gives the two equations y= ix and -x= iy which, since 1/i= -i, are really the same equation. From y= ix, we can write $\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x \\ ix\end{bmatrix}= x\begin{bmatrix}1 \\ i\end{bmatrix}$.

The eigenvectors corresponding to eigenvalue i are multiples of $\begin{bmatrix}1 \\ i\end{bmatrix}$.

Now you find the eigenvectors corresponding to eigenvalue -i.