# Any list of $n+1$ vectors of an $n-$dimensional vector space is not linearly independent

Theorem: Suppose $$V$$ is an $$n-$$dimensional vector space. Then any collection of $$n+1$$ vectors $$\{v_1,v_2,\dots,v_n,v_{n+1}\}$$ is linearly dependent (or not linearly independent).

Proof. Suppose $$V$$ is an $$n-$$dimensional vector space and consider the collection of vectors $$\{v_1,\dots,v_{n+1}\}$$. Since $$V$$ is $$n-$$dimensional, by definition, there exists a basis for $$V$$ with $$n$$ vectors, i.e. there exists a collection of basis vectors $$\{b_1,\dots,b_n\}$$ such that $$\displaystyle\sum_{i=1}^n r_ib_i=v_i$$ for any $$b_i \in V$$ and for some scalar $$r_i \in \mathbb{R}$$ or $$\mathbb{C}$$ (without loss of generality).

Consider the statement

$$a_1v_1+a_2v_2+\dots+a_nv_n+a_{n+1}v_{n+1}=\hat{0}$$

If we can show that some $$v_i$$ is a multiple of another vector(s), we are done. Since $$b_1,\dots,b_n$$ form a basis for $$V$$, each $$v_i$$ can be rewritten as a linear combination of $$b_1,\dots,b_n$$. Thus,

$$a_1(r_1b_1)+a_2(r_2b_2)+\dots+a_n(r_nb_n)+a_{n+1}(r_{n+1}b_{n+1})=\hat{0}$$

$$a_{n+1}(r_{n+1}b_{n+1})=-(a_1r_1)b_1-(a_2r_2)b_2-\dots-(a_nr_n)b_n$$

$$a_{n+1}v_{n+1}=-a_1v_1-\dots-a_nv_n$$

If $$v_{n+1} = \hat{0}$$, then the vectors $$\{v_1,\dots,v_{n+1}\}$$ are linearly dependent since $$\hat{0}$$ is a multiple of every vector. So suppose $$v_{n+1} \neq \hat{0}$$. Since $$v_{n+1} \neq 0, a_{n+1} \neq 0$$, and thus, in either case, the vectors are linearly dependent. $$\hspace{1cm} \square$$

My proof seems like it's in the right ballpark, but my ending paragraph seems to fail to connect to any of my previous work; in other words, my ending conclusion fails to incorporate the notion of a collection of $$n+1$$ vectors. Am I missing anything here?

• Probably your argument would be clearer if you wound up showing a linear dependence relation among the $b_i$'s, which (as a basis) were supposed to be linearly independent. Aug 12, 2021 at 1:25
• $$\sum_{i=1}^{n} r_ib_i=v_i$$ has two conflicting usages of $i.$ What you want is:$$v_i=\sum_{j=1}^n r_{ij}b_j.$$ Aug 12, 2021 at 2:45

1. You wrote "$$\sum_{i = 1}^{n} r_{i}b_{i} = v_{i}$$" which should be $$\sum_{i = 1}^{n} r_{i,j}b_{i} = v_{j}$$
2. It's not clear why $$a_{1}(r_{1}b_{1}) + a_{2}(r_{2}b_{2}) + ... + a_{n}(r_{n}b_{n}) + a_{n + 1}(r_{n + 1}b_{n + 1}) = 0$$ since $$v_{i} \not = r_{i}b_{i}$$
You can start with the assumption that the set $$A = \{v_{1},...,v_{n + 1}\}$$ is linearly independent. Then use the dimension of V to see that the set $$A \setminus \{v_{i}\}$$ is a basis for V, for all $$i = 1,2,..,n + 1$$. From that, conclude that $$v_{i} \in \text{span}(A \setminus \{v_{i}\})$$ for all $$i = 1,...,n + 1$$
• Is the purpose of this proof to arrive at a contradiction? Because, in the end, I seek to show that the set $\{v_1,\dots,v_{n+1}\}$ is linearly dependent. Aug 12, 2021 at 3:08
• Yes, by showing that $A \setminus \{v_{1}\}$ is a basis. You can argue that the set A cannot be linearly independent. This is a contradiction because we assume that A is linearly independent. Aug 13, 2021 at 4:15