Okay, this tells us nothing in the complex case, but for real transformations we have no reason to assume that a given transformation has any eigenvalues at all. But if our transformation is self-adjoint it must have one.
When we found this in the complex case we saw that the characteristic polynomial had to have a root, since is algebraically closed. It’s the fact that isn’t algebraically closed that causes our trouble. But since sits inside we can consider any real polynomial as a complex polynomial. That is, the characteristic polynomial of our transformation, considered as a complex polynomial (whose coefficients just happen to all be real) must have a complex root.
This really feels like a dirty trick, so let’s try to put it on a bit firmer ground. We’re looking at a transformation on a vector space over . What we’re going to do is “complexify” our space, so that we can use some things that only work over the complex numbers. To do this, we’ll consider itself as a two-dimensional vector space over and form the tensor product . The transformation immediately induces a transformation by defining . It’s a complex vector space, since given a complex constant we can define the scalar product of by as . Finally, is complex-linear since it commutes with our complex scalar product.
What have we done? Maybe it’ll be clearer if we pick a basis for . That is, any vector in is a linear combination of the in a unique way. Then every (real) vector in is a unique linear combination of and (this latter is the complex number, not the index; try to keep them separate). But as complex vectors, we have , and so every vector is a unique complex linear combination of the . It’s like we’ve kept the same basis, but just decided to allow complex coefficients too.
And what about the matrix of with respect to this (complex) basis of ? Well it’s just the same as the old matrix of with respect to the ! Just write
Then if is self-adjoint its matrix will be symmetric, and so will the matrix of , which must then be self-adjoint as well. And we can calculate the characteristic polynomial of from its matrix, so the characteristic polynomial of will be the same — except it will be a complex polynomial whose coefficients all just happen to be real.
Okay so back to the point. Since is a transformation on a complex vector space it must have an eigenvalue and a corresponding eigenvector . And I say that since is self-adjoint$, the eigenvalue must be real. Indeed, we can calculate
and thus , so is real.
Therefore, we have found a real number so that when we plug it into the characteristic polynomial of , we get zero. But then we also get zero when we plug it into the characteristic polynomial of , and thus it’s also an eigenvalue of .
And so, finally, every self-adjoint transformation on a real vector space has at least one eigenvector.