MATH 3210 Class Notes 02-03-2025

notevector-spaces

Recall Span and Linear Combination

Discussed Polynomials as a Vector Space

Proved Linear Dependence Lemma

Theorem

In a finite-dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.

Proof

Suppose $u_1,\dots ,u_m\in V$ is linearly independent and $w_1,\dots ,w_m\in V$ spans $V$. We need to show that $m\le n$.

Let $B$ be the list $w_1,\dots ,w_n$, which spans $V$. Then the list $u_1,w_1,\dots ,w_n$ is linearly dependent since $u_1\in \mathrm{span}(w_1,\dots ,w_n)$ by assumption. By the linear dependence lemma, one of $u_1,w_1,\dots ,w_n$ is a linear combination of the previous vectors in the list. $u_1\ne 0$ since $u_1,\dots ,u_m$ is linearly independent. So $k\ne 1$ in the lemma. Therefore, we can remove some $w_i$ from the list, and preserve the span. So $V=\mathrm{span}(u_1,(n-1)\text{ remaining w’s})=B$.

Step $k$, for $k=2,\dots ,m$: After step $k-1$, the list $B$ looks like $u_1,\dots ,u_{k-1},((n-k+1)\text{ w’s})$. Also $B$ spans $V$. Thus, $u_k\in \mathrm{span}(B)$, so $u_1,\dots ,u_{k-1},u_k,(n-k+1\text{ w’s})$ is linearly dependent.

Since $u_1,\dots ,u_k$ is linearly independent, our index from the linear dependent lemma must “point to” some $w_i$. Thus, we can remove this $w$ and preserve the span.

Since we can do this for all the $u$‘s, $m\le n$.

Theorem

Every subspace of a finite-dimensional vector space is finite dimensional.

$$f(x)=\sum _{n=1}^{\infty }\frac{x^2}{4}+\pi +4!$$