MATH 3210 Class Notes 02-24-2025

notevector-spaces

Invertibility and Isomorphisms

Theorem

Suppose $V,W$ are finite-dimensional vector spaces and $\dim V=\dim W$. If $T:\in \mathcal{L}(V,W)$, then $T$ is invertible if and only if $T$ is injective if and only if $T$ is surjective.

Proof

Invertible $⟹$ injective and surjective is trivial because we have already proved an earlier theorem (invertible $⟺$ injective and surjective).

Also, ETS injective $⟺$ surjective because of the theorem.

(Injective $⟹$ Surjective) Recall: Fundamental Theorem of Linear Maps

Injectivity $⟹$ $\dim \mathrm{null}(T)=0$, so then the FTLM implies that $\dim V=0+\dim \mathrm{range}(T)=\dim W$ by assumption. Therefore, $\dim W=\dim \mathrm{range}(T)$, so $T$ is surjective.

(Surjective $⟹$ Injective) FTLM gives us that $\dim V=\dim \mathrm{null}(T)=\dim \mathrm{range}(T)$, so $\dim W=\dim \mathrm{null}(T)+\dim W$, so $0=\dim \mathrm{null}(T)⟹\mathrm{null}(T)=\left\{0\right\}$, so $T$ is injective.

Theorem

Suppose $\dim V=\dim W<\infty$, $S\in \mathcal{L}(W,V)$, $T\in \mathcal{L}(V,W)$. Then $ST=I_V⟺TS=I_W$.

Defined Isomorphic Vector Spaces

Theorem

Two finite dimensional vector spaces over some field $\mathbb{F}$, are isomorphic if and only if they have the same dimension.

Proof

("$⟹$") Suppose $V,W$ are isomorphic. Then there exists some map $T:V\to W$ that is an isomorphism (i.e. invertible). By the Fundamental Theorem of Linear Maps, we have $\dim V=\dim \mathrm{null}(T)+\dim \mathrm{range}(T)=0+\dim W$.

("$⟸$") Assume $\dim V=\dim W=n$. Let $v_1,\dots ,v_n$ and $w_1,\dots ,w_n$ be bases of $V,W$, respectively. Define $T\in \mathcal{L}(V,W)$ by $T(c_1{v}_1+\cdots +c_n{v}_n)=c_1{w}_1+\cdots +c_n{w}_n$. $T$ is well-defined, since $v_i$‘s form a basis. Also, $T$ is surjective, since every element in $W$ can be written as a linear combination of $w_1,\dots ,w_n$, which corresponds to some $T(v)$ by the definition of $T$. Furthermore, $T$ is injective since the nullspace of $T$ is $\left\{0\right\}$. Therefore, $T$ is invertible, so $T$ is an isomorphism.

Corollary

If $\dim V=n$, then $V\simeq {\mathbb{F}}^n$.

Example

1. $P_m(\mathbb{R})\simeq {\mathbb{R}}^{m+1}$. An example isomorphism is the one below,

a_{0} \\ a_{1} \\ \vdots \\ a_{m} \end{bmatrix} $$ ##### Notation $\mathbb{F}^{m,n} :=$ vector space of $m$-by-$n$ matrices with coefficients in $\mathbb{F}$.

\dim \mathbb{F}^{m,n} = m \cdot n

If $v_{1},\ldots, v_{n}$ is a basis of $V$ and $w_{1}, \ldots, w_{m}$ is a basis of $W$, then each $T \in \mathscr{L}(V,W)$ has a matrix representation

\begin{align*} M(T) = \begin{matrix} w_{1} \ \vdots \ w_{m} \end{matrix}&\ \begin{pmatrix} \ \qquad \ \ \ \vdots\ \ \ \qquad \ \end{pmatrix} \in \mathbb{F}^{m,n} \ & \begin{matrix} \quad v_{1} \cdots v_{n} \end{matrix} \end{align*}

So $M: \mathscr{L}(V,W) \to \mathbb{F}^{m,n}$. #### Theorem If $\dim V = n$, $\dim W = m$, then $\mathscr{L}(V,W) \simeq \mathbb{F}^{m,n}$. ##### Corollary $\dim \mathscr{L}(V,W) = (\dim V) ( \dim W)$ if both are finite. ### Products and Quotients #### Defined [[Product (Vector Spaces)]] ##### Notation - $V_{1} + \cdots + V_{n} = \sum_{i=1}^{n}V_{i}$ . - $V_{1} \oplus \cdots \oplus V_{n} = \bigoplus_{i = 1}^n V_{i}$. - $V_{1} \times \cdots \times V_{n} = \prod_{i=1}^{n}V_{i}$. ##### Proposition $\prod_{i=1}^{n}V_{i}$ is a vector space.