The Unapologetic Mathematician

Mathematics for the interested outsider

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 \mathbb{C} is algebraically closed. It’s the fact that \mathbb{R} isn’t algebraically closed that causes our trouble. But since \mathbb{R} sits inside \mathbb{C} 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 S:V\rightarrow V on a vector space V over \mathbb{R}. 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 \mathbb{C} itself as a two-dimensional vector space over \mathbb{R} and form the tensor product V^\mathbb{C}=V\otimes_\mathbb{R}\mathbb{C}. The transformation S immediately induces a transformation S^\mathbb{C}:V^\mathbb{C}\rightarrow V^\mathbb{C} by defining S^\mathbb{C}(v\otimes z)=S(v)\otimes z. It’s a complex vector space, since given a complex constant c\in\mathbb{C} we can define the scalar product of v\otimes z by c as v\otimes(cz). Finally, S^\mathbb{C} 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 \left\{e_i\right\}_{i=1}^n for V. That is, any vector in V is a linear combination of the e_i in a unique way. Then every (real) vector in V^\mathbb{C} is a unique linear combination of e_i\otimes1 and e_i\otimes i (this latter i is the complex number, not the index; try to keep them separate). But as complex vectors, we have e_i\otimes i=i(e_i\otimes1), and so every vector is a unique complex linear combination of the e_i\otimes1. It’s like we’ve kept the same basis, but just decided to allow complex coefficients too.

And what about the matrix of S^\mathbb{C} with respect to this (complex) basis of e_i\otimes1? Well it’s just the same as the old matrix of S with respect to the e_i! Just write

\displaystyle S^\mathbb{C}(e_i\otimes1)=S(e_i)\otimes1=(s_i^je_j)\otimes1=s_i^j(e_j\otimes1)

Then if S is self-adjoint its matrix will be symmetric, and so will the matrix of S^\mathbb{C}, which must then be self-adjoint as well. And we can calculate the characteristic polynomial of S from its matrix, so the characteristic polynomial of S^\mathbb{C} 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 S^\mathbb{C} is a transformation on a complex vector space it must have an eigenvalue \lambda and a corresponding eigenvector v. And I say that since S^\mathbb{C} is self-adjoint$, the eigenvalue \lambda must be real. Indeed, we can calculate

\displaystyle\lambda\langle v,v\rangle=\langle v,\lambda v\rangle=\langle v,A(v)\rangle=\langle A(v),v\rangle=\langle\lambda v,v\rangle=\bar{\lambda}\langle v,v\rangle

and thus \lambda=\bar{\lambda}, so \lambda is real.

Therefore, we have found a real number \lambda so that when we plug it into the characteristic polynomial of S^\mathbb{C}, we get zero. But then we also get zero when we plug it into the characteristic polynomial of S, and thus it’s also an eigenvalue of S.

And so, finally, every self-adjoint transformation on a real vector space has at least one eigenvector.

August 12, 2009 Posted by | Algebra, Linear Algebra | 1 Comment



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