>
> Heeey! My multivariable calculus and matrix algebra class comes to use!
> Apparently <1,1,-1> is an eigenvector of the transformation matrix defined
> by the rotation, since Ax = lx (where A is the transformatoin matrix and l
> is some constant). So, the question is, is there such thing as a three by
> three matrix with no eigenvalues?

I'm new to this, but I would say no: To get the eigenvalues of a matrix A you
calculate this:
det(A-zE)=0
(z are the eigenvalues, E is the matrix with ones on main diagonal and zeros
at all other places)
That's a polynom. You can always get values of z (at least one value)
sometimes they are complex, but with a third-degree polynom you get at least
a real one. (just guessing...)
So there's always at least one eigenvalue (that could be zero, but non the
less)

cukk

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