MATH 3210 Class Notes 03-03-2025

notevector-spaces

Duality

Dual Vector Space

Defined the dual vector space, dual basis, and dual transformation.

Example 1

Let $D\in \mathcal{L}(P(\mathbb{R}),P(\mathbb{R}))$ be differentiation. How does $D^{\prime }$ behave?

1. Pick $\phi (p)=p(3)\in W^{\prime }=(P(\mathbb{R}){)}^{\prime }$. $D^{\prime }(\phi )=\phi \circ D⟹D^{\prime }(\phi )(p)=p^{\prime }(3)$.
2. $\phi (p)={\int }_0^{1}p\in (P(\mathbb{R}){)}^{\prime }$. $D^{\prime }(\phi )(p)=(\phi \circ D)(p)={\int }_0^1{p}^{\prime }=p(1)-p(0)$.

Theorem 1

Suppose $T\in \mathcal{L}(V,W)$. Then,

1. $(S+T{)}^{\prime }=S^{\prime }+T^{\prime }$ for all $S\in \mathcal{L}(V,W)$.
2. $(\lambda T{)}^{\prime }=\lambda T^{\prime }$.
3. $(ST{)}^{\prime }=T^{\prime }{S}^{\prime }$ for all $S\in \mathcal{L}(W,U)$.

Null Space and Range of Dual Maps

Defined the Annihilator

Example 2

Let $U\subseteq P(\mathbb{R})$ be all the multiples of $x^2$. What’s an example of $\phi \in U^0$?

$\phi (p)=p^{\prime }(0)\in U^0$.

Recall Fundamental Theorem of Linear Maps

Example 3

Let $e_1,e_2,e_3,e_4,e_5$ be the standard basis of ${\mathbb{R}}^5$; let ${\phi }_1,\dots ,{\phi }_5$ be the dual basis in $({\mathbb{R}}^5{)}^{\prime }$ (note ${\phi }_j(x_1,\dots ,x_5)=x_j$).

Suppose $U=\mathrm{span}(e_1,e_2)=\left\{(x_1,x_2,0,0,0)\subseteq {\mathbb{R}}^5\right\}$.

What is $U^0$? Claim: $U^0=\mathrm{span}({\phi }_3,{\phi }_4,{\phi }_5)$.

Theorem 2

If $U\subseteq V$, then $U^0$ is a subspace of $V^{\prime }$.

Proof

1. $0$ is in $V^{\prime }$ and $U^0$.
2. ${\phi }_1,{\phi }_2\in U^0$, then $({\phi }_1+{\phi }_2)(u)={\phi }_1(u)+{\phi }_2(u)=0$, so ${\phi }_1+{\phi }_2\in U^0$.
3. $\lambda {\phi }_1\in U^0$.

Thus, $U^0$ is a subspace of $V^{\prime }$.

Theorem 3

Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$, then

$$\dim U^0=\dim V-\dim U.$$

Proof

1. Take a basis of $U$: $u_1,\dots ,u_m$, extend to a basis of $V$: $u_1,\dots ,u_m,\dots ,u_n$. Then show ${\phi }_{m+1},\dots ,{\phi }_n$ is a basis of $U^0$.
2. Let $i\in \mathcal{L}(U,V)$ be the inclusion map ($i(u)=u\in V$). Thus, $i^{\prime }\in \mathcal{L}(V^{\prime },U^{\prime })$. The FTLM implies that $\dim \mathrm{range}(i^{\prime })+\dim \mathrm{null}(i^{\prime })=\dim V^{\prime }$. However, $\mathrm{null}(i^{\prime })=U^0=\left\{\phi \in V^{\prime }:\phi \circ i=0\text{ on }U\right\}$, and $\dim V^{\prime }=\dim V$, so $\dim \mathrm{range}{i}^{\prime }+\dim U^0=\dim V$. If $\phi \in U^0$, then $\phi$ can be extended to a linear functional $\psi$ on $V$. Then the definition of $i^{\prime }$ shows $i^{\prime }(\psi )=\phi =\psi \circ i$. Thus, $\phi \in \mathrm{range}{i}^{\prime }$, so $\mathrm{range}{i}^{\prime }=U^{\prime }$. So, $\dim \mathrm{range}{i}^{\prime }=\dim U^{\prime }=\dim U$.

With 1) and 2) together, we have $\dim U^0=\dim V-\dim U$.

Theorem 4

Suppose $V,W$ are finite-dimensional and $T\in \mathcal{L}(V,W)$, then

1. $\mathrm{null}(T^{\prime })=(\mathrm{range}(T){)}^0$.
2. $\dim \mathrm{null}(T^{\prime })=\dim \mathrm{null}(T)+\dim W-\dim V$.

Proof

In the book.

Corollary

Suppose $V,W$ are finite-dimensional and $T\in \mathcal{L}(V,W)$, then $T$ is surjective if and only if $T^{\prime }$ is injective.

Theorem 5

Suppose $V,W$ are finite=dimensional and $T\in \mathcal{L}(V,W)$. Then,

1. $\dim \mathrm{range}(T^{\prime })=\dim \mathrm{range}(T)$.
2. $\mathrm{range}(T^{\prime })=(\mathrm{null}T{)}^0$.

Corollary

Suppose $V,W$ are finite-dimensional and $T\in \mathcal{L}(V,W)$, then $T$ is injective if and only if $T^{\prime }$ is surjective.

Theorem 6

Suppose $V,W$ are finite-dimensional and $T\in \mathcal{L}(V,W)$, then the matrix $M(T^{\prime })=M(T{)}^t$ (transpose).