Eigenvalues part 2

PensFan10 · #1 $Old$ November 6th, 2009, 09:25 AM

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

Defunkt · #2 $Old$ November 6th, 2009, 10:02 AM

Quote:

Originally Posted by PensFan10 $View Post$

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.

PensFan10 · #3 $Old$ November 6th, 2009, 10:22 AM

its over c. how do you rref with i?

HallsofIvy · #4 $Old$ November 6th, 2009, 09:13 PM

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.