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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