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


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,




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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – hardmath
    Aug 12, 2021 at 1:25
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    $\begingroup$ $$\sum_{i=1}^{n} r_ib_i=v_i$$ has two conflicting usages of $i.$ What you want is:$$v_i=\sum_{j=1}^n r_{ij}b_j.$$ $\endgroup$ Aug 12, 2021 at 2:45

1 Answer 1

  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 $i = 1,...,n + 1$

  • $\begingroup$ 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. $\endgroup$
    – SunRoad2
    Aug 12, 2021 at 3:08
  • $\begingroup$ 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. $\endgroup$ Aug 13, 2021 at 4:15

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