1. ## No real eigenvalues

Show that the following matrix has no real eignvalues, and thus no eigenvectors. Interpret your result geometrically.
[0 -1]
[-1 0]

2. a^2-1=0 where a=eigen value.

3. I am still not sure what this means.

4. i get 1 and -1 as the eigenvalues for your matrix and these are real eigenvalues.

5. Are you sure you have written the matrix correctly?
Charikar is right. The matrix has real eigenvalues and eigenvectors. The matrix is a reflection - the eigenvectors are the bisectors of the axes.
[0 -1]
[1 0]
as that would have been a rotation, which does not have real eigenvectors.

6. Yes I figured it out and got the same thing as Charikar. I did have the wrong matrix. I get x^2+1 as the determinant. I don't know where to go from there at all.

7. Originally Posted by lm6485
Yes I figured it out and got the same thing as Charikar. I did have the wrong matrix. I get x^2+1 as the determinant. I don't know where to go from there at all.
If $det(A-\lambda I) = f(\lambda)$ for some function f, then the eigenvalues of A are the roots of $f(\lambda)$. If it only has solutions in $F = \mathbb{C} - \mathbb{R}$ then its eigenvalues are not real, therefore it has no eigenvectors as your question described.

In this case, $f(\lambda) = \lambda^2 + 1$.

Can you find the roots of f?

8. sqrt of -1?

9. Yes, And therefore the eigenvalues are ?