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.

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August 12, 2009 - Posted by John Armstrong | Algebra, Linear Algebra

1 Comment »

[...] 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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