Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So my suggestion was:

Assume: Let $S = \{u_1, u_2, ....u_r\}$ be a set of vectors in $\mathbb{R}^n$ . If $r > n$ then the vectors $u_1, u_2 \ldots u_r$ must be linearly dependent.

If one writes the linear system corresponding to $c_1u_1 + c_2u_2 + ... + c_ru_r$ , one will have a homogeneous system of $n$ equations in $r$ unknowns. We know that such a system has infinitely many solutions. Thus in $\mathbb{R}^n$ , a set which is linearly independent cannot contain more than $n$ vectors.

share|cite|improve this question
Look at In this case, one of the vectors, say $v_n$ is the zero vector. Set $a_1=a_2=...=a_{n-1}=0$ and let $a_n$ be free, then you have satisfied the condition for linear dependence. – mathmath8128 Sep 23 '11 at 5:53

First of all, the definition of linear dependence: A set of $n$ vectors in a vector space $U$ is said to be linearly dependent if there exist constants $a_1 \ldots a_n$ not all zero such $a_1u_1 + \ldots a_nu_n = 0$.

So if we have a set of vectors $\{u_1, u_2 \ldots u_n\}$, if one of them is the zero vector, then we when we multiply the zero vector by a constant it is still the zero vector. What can you deduce from here?

Assume without loss of generality that $u_1$ is the zero vector in the list $\{u_1 \ldots u_n\}$. Then to make the linear combination zero, what constants can I attach to $u_1 \ldots u_n$ that will make this zero? Can I choose constants $a_1, \ldots a_n$ such that at least one of them is non-zero?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.