## 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) $u$invariant 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., $\mathrm{ker}(\mathsf{L}) \neq \{ 0 \}$).