Spectra of linear operators #1

Today, I’m going to talk about linear operators on vector spaces. To start, we’ll let V be a finite-dimensional (say dimension n) vector space over a field K, and consider a linear operator u : V \to V. We will also write u \in \mathcal{L}(V),  since \mathcal{L}(V) is the K-algebra of linear operators on V.

The first thing I’ll talk about is invariant subspaces. A subspace W \subseteq V is called invariant or (more specifically) uinvariant if u(W) \subseteq W. Note that certainly V and \ker(u) are both invariant subspaces (easy to verify).

We know that if we fix two ordered bases \alpha, \beta for V then our operator u has an associated matrix with respect to this choice. What is interesting is that if W is u-invariant of dimension d, we can choose a basis \beta whose first d elements make up a basis for W. In this manner, using the u-invariance, we actually obtain a matrix for u of the form

\begin{bmatrix} A & B \\ O & C \end{bmatrix}

where, as you might have guessed, A is a d \times d square matrix, and O is the n-d \times d zero matrix. In other words, the last n-d rows of the first d columns are zero, precisely because u is transforming those basis vectors in \beta to vectors that lie within W to vectors that again lie within W.

We say that \lambda \in K is an eigenvalue of u if u(v) = \lambda v for some vector v \in V. In this case, \lambda is called the eigenvalue associated to the eigenvector v. So in other words, an eigenvector is a vector which is simply rescaled by the transformation u! This is a very nice property, and so it is in our interest to analyze this phenomenon.

Firstly, we note that if v is an eigenvector with eigenvalue \lambda then any vector in \mathrm{span} \{v\} is also an eigenvector with that same eigenvalue: u(\mu v) = \mu \cdot u(v) = \mu \cdot \lambda v = \lambda(\mu v). So if two eigenvectors have different associated eigenvalues, they are linearly independent. As a corollary, we see already that u can only have n distinct eigenvalues, for otherwise V contains a linearly independent set of more than n vectors, which is absurd.

Next I’ll introduce something called the characteristic polynomial \mathsf{c}(\lambda) of u. This is the polynomial given by \det(\tilde{u} - \lambda I_n) where I_n is the n \times n identity matrix, and \tilde{u} is the matrix of the operator u under some basis for V. It turns out that the characteristic polynomial of the operator u does not depend on which basis we choose. So instead of talking about matrices, we can just talk about the determinant of an operator. Hence we can just define \mathsf{c}(\lambda) = \det(u - \lambda I).

Theorem. The roots of \mathsf{c}(\lambda) are the eigenvalues of u.
Proof. If \lambda is a root of \mathsf{c}(\lambda), then \det(u - \lambda I) = 0, hence u - \lambda I is not an invertible linear operator. Invoking the rank-nullity theorem, we see that u has a nontrivial kernel, and hence there is nonzero v \in V so that (u - \lambda I)(v) = u(v) - \lambda v = \mathbf{0}, and hence u(v) = \lambda v. The existence of such v means that by definition, \lambda is an eigenvalue. The proof that every eigenvalue is a root of \mathsf{c}(\lambda) is just a reversal of this argument.

We should be careful to note that while it seems we have given an “intrinsic” (basis-independent) definition of the characteristic polynomial, we are still relying on the fact that our linear operator u does have a matrix representation. Thus, we are still relying on the finite-dimensionality of V. The reason why the characteristic polynomial is important is because its roots are precisely the finite set of scalars \lambda such that u - \lambda I is not an invertible linear operator. We will denote this set \mathrm{Spec}(u), the spectrum of the linear operator u.

In the infinite-dimensional case, we can no longer talk about determinants or characteristic polynomials, so the spectrum takes its place. Consider the vector space \mathfrak{C} of \mathcal{C}^\infty real-valued functions, defined on the real line. Note that if D : \mathfrak{C} \to \mathfrak{C} is the differentiation operator, then for each \lambda \in \mathbb{R}, the function \exp(\lambda x) is an eigenfunction with eigenvalue \lambda. Since there are infinitely many choices for \lambda, the differentiation operator has infinitely many distinct eigenvalues. It follows that our vector space \mathfrak{C} must surely be infinite-dimensional.

Interesting remark: If a linear operator \mathsf{L} on a vector space \mathsf{U} has a non-trivial kernel, then it is certainly not invertible. However, if a linear operator is non-invertible, its kernel might still in general be trivial (!!!), but luckily when \mathsf{U} is finite-dimensional, the rank-nullity theorem tells us that this cannot happen (i.e., \mathrm{ker}(\mathsf{L}) \neq \{ 0 \}).


About mlbaker

just another guy trying to make the diagrams commute.
This entry was posted in linear algebra and tagged , , , , , . Bookmark the permalink.

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