MATH 3210 Class Notes 03-10-2025

notevector-spaces

Eigenvectors and Values

Recall

For $\lambda \in \mathbb{F}$, an eigenvalue of $T\in \mathcal{L}(V)$, $v\ne 0$ an eigenvector associated to $\lambda ⟹Tv=\lambda v$ if and only if $T-\lambda I$ is not injective, surjective, or invertible.

Theorem 1

Eigenvectors corresponding to distinct eigenvalues are linearly independent.

Proof

Suppose the above is not true for the sake of contradiction. Then there exists some smallest $m\in {\mathbb{Z}}_{\ge 0}$ such that the list of eigenvectors $v_1,\dots ,v_m$ corresponding to distinct eigenvalues ${\lambda }_1,\dots ,{\lambda }_n$ is linearly dependent. Note that $m\ge 2$ since eigenvectors are non-zero. Thus, $\exists a_i\in \mathbb{F}$ (none zero) such that

$$\begin{array}{rl}{a}_1{v}_1+\cdots +a_m{v}_m& =0,\\ (T-{\lambda }_{m}I)(a_1{v}_1+\cdots +a_m{v}_m)& =(T-{\lambda }_{m}I)0,\\ a_1(T-{\lambda }_{1}I)v_1+\cdots +a_m(T-{\lambda }_{m}I)v_m& =0,\\ a_1({\lambda }_1-{\lambda }_m)v_1+\cdots +a_m({\lambda }_m-{\lambda }_m)v_m& =0,\\ a_1({\lambda }_1-{\lambda }_m)v_1+\cdots +a_{m-1}({\lambda }_{m-1}-{\lambda }_m)v_{m-1}& =0.\end{array}$$

Thus, we have found a smallest linearly independent list, a contradiction to the minimality of $m$. $\square$

Corollary

If $V$ is finite-dimensional, then any operator on $V$ has at most $\dim V$ distinct eigenvalues.

Operators

Operators are interesting because we can raise them to powers. If $T\in \mathcal{L}(V)$, then $TT$ makes sense, and is also an operator on $V$.

Notation: Suppose $T\in \mathcal{L}(V)$, and $m\in {\mathbb{Z}}_{\ge 0}$,then,

• $T^m=\underset{m\text{ times}}{T\cdots T}$ .
• $T^0=I$.
• If $T$ is invertible, then $T^{-m}=(T^{-1}{)}^m$.

Properties:

• $T^m{T}^n=T^{m+n}$.
• $(T^m{)}^n=T^{mn}$

If $T\in \mathcal{L}(V)$ and $p\in P(\mathbb{F})$ given by $p(z)=a_0+a_{1}z+a_2{z}^2+\cdots +a_m{z}^m$, then $p(T)=a_{0}I+a_{1}T+a_2{T}^2+\cdots +a_m{T}^m\in \mathcal{L}(V)$.

Example 1

Let $D\in \mathcal{L}(P(\mathbb{R}))$ be the differentiation operator $p(x)=7-3x+5x^2$. Then $P(D)=7I-3D+5D^2$. If $q\in P(\mathbb{R})$, then $(p(D))(q)=7q-3q^{\prime }-5q^{\prime \prime }$.

Definition 1

If $p,q\in P(\mathbb{F})$, then $pq\in P(\mathbb{F})$ and is defined by $(pq)(z)=p(z)q(z)$.

Proposition 1

$(pq)(T)=p(T)q(T)=q(T)p(T)=(qp)(T)$, so this is commutative.

We saw earlier, $\mathrm{null}(T)$, $\mathrm{range}(T)$ are invariant subspaces under $T\in \mathcal{L}(V)$.

Proposition 2

If $T\in \mathcal{L}(V)$ and $p\in P(\mathbb{F})$, then $\mathrm{null}(T)$ and $\mathrm{range}(T)$ are invariant under $T$.

Example 2

Let $T\in \mathcal{L}({\mathbb{F}}^{\infty })$ be the backshift operator. What are the eigenvectors of $T$? $(a_1,a_2,\dots )$ an eigenvector if $(a_1,a_2,\dots )\ne (0,0,\dots )$ and $T(a_1,a_2,a_3,\dots )=\lambda (a_2,a_3,\dots )$.

Note

$T(c,c,\dots )=1\cdot (c,c,\dots )$. So anything of the form $(c,c,\dots )$ is an eigenvector associated to $1=\lambda$. Also, $\lambda =0$ is an eigenvalue for everything else.

Example 3

Let $D\in \mathcal{L}(P(\mathbb{R}))$ be the differentiation operator. What are the eigenvalues? If $\lambda$ is an eigenvalue, then,

$$Dp=\lambda p\text{ for some }p\ne 0.$$

Thus, $p^{\prime }=\lambda p$ for some $p\ne 0$. No nonzero $p$ satisfies this, so $\lambda =0$ is the only eigenvalue.

Example 4

Let $T\in \mathcal{L}(P(\mathbb{R}))$ be $(Tp)(x)=x{p}^{\prime }(x)$. What are the eigenvalues and vectors?

$$x{p}^{\prime }(x)=\lambda p(x)?$$

$x^{\lambda }$ (vectors) works for $\lambda =1,2,3,\dots$ (values).