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Let $U = \{v_1, ..., v_i \} \in \mathbb{R}^n$. Prove if $U$ is not linearly dependent then $i \le n$.

No complete answers, only hints please

So, $U$ is linearly independent. We solve $v_1c_1 + ... + v_ic_i = 0$, getting the system into a matrix I see that $v_j = $

The coefficients are some form of $x_h$ for some $h$'s.

We know the system is consistent because it only has one solution $c_1 = .. = c_i = 0$.

So the system contains $0$ free variables/parameters.

But that doesn't help.

Can I get a hint?

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    Do you know the theorem that if a linear independent set has the same cardinality of the dimension of the vector space, it then follows that the set is a basis? – 2017-01-19

1 Answers 1

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Hint: Try proving the contrapositive. Along the way, use an intermediate result that may help. So the sketch is:

  • Suppose that $i > n$ (that is, suppose that $U$ has more vectors than the dimension of the vector space they live in). Our goal is to show that $U$ is linearly dependent.
  • To this end, let $A$ be the $n \times i$ matrix whose columns are the vectors in $U$. Observe that $\boxed{\text{... fill in the details here}}$. Thus, it follows that $A\vec x = \vec 0$ has infinitely many solutions.
  • But then since $A\vec x = \vec 0$ has infinitely many solutions, we know in particular that it has a nontrivial solution, say $\vec b \neq \vec 0$. Observe that $\boxed{\text{... fill in the details here}}$. Thus, we conclude that $U$ must be linearly dependent, as desired.