## Every Self-Adjoint Transformation has an Eigenvector

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.

[…] off, since is self-adjoint, we know that it has an eigenvector , which we can pick to have unit length (how?). The subspace is then invariant under the action of […]

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