# Number of indie vectors ##\leq ## cardinality of spanning set

• Terrell
So you could start by saying "Let ##u_1, \dots, u_n## be a minimal spanning set for the space. Then for every spanning set ##v_1, \dots, v_m##, with ##m > n##, we have the following contradiction..."In summary, the given statement is proven by assuming a minimal spanning set and any other spanning list, with a greater length, to show a contradiction. This implies that the length of every linearly independent list of vectors is less than or equal to the length of every spanning list.

## 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! 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##. It might be easier to assume that you have a set of m lin. independent vectors and a spanning set with n vectors, with m > n. If you get a contradiction (which should be easier to do than what you did), you're done.

Also, it looks like you have a typo in the first line of your proof. The two sets -- the linearly independent vectors and the spanning set, shouldn't have the same final index of m.

• Terrell
Mark44 said:
it looks like you have a typo in the first line of your proof.
Yes. It should be ##\{u_1,\cdots, u_n\}##
Mark44 said:
It might be easier to assume that you have a set of m lin. independent vectors and a spanning set with n vectors, with m > n.
I believe this is what I did. At line 2, my third sentence starts with suppose ##m \lt n##. Then I got a contradiction in the end. But as what you've probably already notice, my proof seems more complicated than it should be. So I have a feeling that I might have an incorrect use of contradiction. Terrell said:
Yes. It should be ##\{u_1,\cdots, u_n\}##

I believe this is what I did. At line 2, my third sentence starts with suppose ##m \lt n##.
Actually what you wrote is "Suppose that n < m..." which is correct for a proof by contradiction if your spanning set is ##\{u_1, u_2, \dots, u_n \}##
Terrell said:
Then I got a contradiction in the end. But as what you've probably already notice, my proof seems more complicated than it should be. So I have a feeling that I might have an incorrect use of contradiction. Yes, your proof seems more complicated than it needs to be, in part because of the use of lots more letters than are needed.

For example, instead of having coefficients of ##a_i, b_i,## and ##\eta_i##, just use two subscripts. So ##v_i = \sum_{j = 1}^n a_{i~j}u_j##. If you write the equations for ##v_1, v_2, \dots, v_n## in the form of a matrix product, you have ##\vec v = A \vec u##, where A is a matrix with m rows and n columns.

I think you can assume without loss of generality that the u vectors are a minimal spanning set; i.e., a basis for the space. Since you need to prove the given statement for every spanning list, it doesn't matter if you prove it for the smallest spanning set.

Last edited:
• Terrell