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## Homework Statement

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.

It's quite long , hope you guys read through it. Thanks!

## Homework Equations

N/A

## The Attempt at a Solution

Let ##\{v_1, \cdots, v_m\}## be linearly independent and ##\{u_1, \cdots, u_m\}## spans V. Note that ##\{v_1, \cdots, v_m\} \subseteq \operatorname{span}(u_1, \cdots, u_n)## which implies ##\forall i=1,\cdots,m## ##v_i=\sum_{j=1}^{n}a_{j}u_{j}##. Suppose that ##n \lt m## and let ##v_1=\sum_{i=1}^{m}a_{i}u_{i}##,##\quad## ##v_2=\sum_{i=1}^{m}b_{i}u_{i}##,##\quad## ##\cdots##, ##\quad## ##v_3=\sum_{i=1}^{m}\eta_{i}u_{i}##. Observe that

\begin{align}

\delta_{1}v_1+\delta_{2}v_2+\cdots +\delta_{n}v_{n}=v_k; \quad k \in \{n+1,\cdots ,m\}\\

\Longleftrightarrow \delta_{1}(\sum_{i=1}^{m}a_{i}u_{i})+\delta_{2}(\sum_{i=1}^{m}b_{i}u_{i})+\cdots +\delta_n(\sum_{i=1}^{m}\eta_{i}u_{i}) &=v_k\\

\Longleftrightarrow \delta_1(a_{1}u_{1}+\cdots +a_{m}u_{m})+\delta_2(b_{1}u_{1}+\cdots +b_{m}u_{m})+\cdots +\delta_n(\eta_{1}u_{1}+\cdots +\eta_{m}u_{m}) &=v_k\\

\Longleftrightarrow (\delta_{1}a_{1}+\cdots +\delta_{n}\eta_{1})u_{1}+\cdots +(\delta_{1}a_{m}+\cdots +\delta_{n}\eta_{m})u_{m} &=v_{k}

\end{align}

Since ##\exists \xi_{i}\in\Bbb{F}## such that ##\xi_{1}u_1+\cdots +\xi_{m}u_m=v_{k}##, then by solving the following system of linear equations

\begin{align}

\delta_{1}a_{1}+\cdots +\delta_{n}\eta_{1} &= \xi_1\\

\vdots\\

\delta_{1}a_{m}+\cdots +\delta_{n}\eta_{m} &= \xi_{m}

\end{align}

we obtain the necessary ##\delta_{i}'s## so that ##\sum_{i=1}^{n}\delta_{i}v_{i}=v_k##. Since ##n \neq m##, there could either be non-unique solutions or no solutions. We consider the two scenarios.

Case 1: infinite solutions; By definition of a basis, ##\{v_1,\cdots ,v_n\}## cannot form a basis due to the required uniqueness of expressing ##v_k## with respect to the set of vectors ##\{v_1,\cdots ,v_n\}##. But ##\operatorname{span}(v_1, \cdots, v_n) \neq V## so it must be that ##\{v_1, \cdots , v_n\}## is not linearly independent. We have a contradiction.

Case 2: No solution; then ##v_k \notin \operatorname{span}(v_1, \cdots, v_n)## which implies ##\delta_{1}v_1 + \delta_{2}v_2 + \cdots + \delta_{n}v_n + \delta_{n+1}v_k = 0## if and only if all ##\delta_{i}'s## equal zero. Equivalently,

\begin{align}

& \Longleftrightarrow (\delta_{1}a_{1}+\cdots +\delta_{n}\eta_{1}+\delta_{n+1}\xi_{1})u_{1}+\cdots +(\delta_{1}a_{m}+\cdots +\delta_{n}\eta_{m}+\delta_{n+1}\xi_{m})u_{m} = 0\\

& \Longleftrightarrow \forall i=1,\cdots, m \quad \delta_{1}a_i+\cdots +\delta_{n}\eta_i+\delta_{n+1}\xi_i =0

\end{align}

But, if we let ##\delta_{n+1}=\frac{-(\delta_{1}a_i+ \cdots +\delta_{n}\eta_i)}{\xi_i}##, ##\forall i=1, \cdots, m##, then ##\exists \delta_i \neq 0## such that ##(\sum_{i=1}^{n}\delta_{i}v_i)+\delta_{n+1}v_k=0##. This implies the set of vectors ##\{v_1, v_2\cdots,v_n, v_k\}## is not linearly independent. We have another contradiction.

##\therefore## ##m \leq n##.