Proof: Max number of Linearly Independent Vectors

[#-Riley-#]

Homework Statement

Prove that a set of linearly independent vectors in Rⁿ can have maximum n elements.

So how would you prove that the maximum number of independent vectors in Rⁿ is n?I can understand why in my head but not sure how to give a mathematical proof. I understand it in terms of the number of independent vectors being equal to the rank of the matrix they create and obviously a matrix of dimension n can only have max n pivots. But I don't think that's really sufficient for a proof.

 

[mfb]

But I don't think that's really sufficient for a proof.

If you write it down properly, that should be fine.
The main point: a n x m matrix (for your m vectors) cannot have rank larger than n.

 

[Ray Vickson]

Homework Statement

Prove that a set of linearly independent vectors in Rⁿ can have maximum n elements.

So how would you prove that the maximum number of independent vectors in Rⁿ is n?I can understand why in my head but not sure how to give a mathematical proof. I understand it in terms of the number of independent vectors being equal to the rank of the matrix they create and obviously a matrix of dimension n can only have max n pivots. But I don't think that's really sufficient for a proof.

Yes, it is (with a bit of tweaking). If vectors $v_1, v_2, \ldots, v_n, v_{n+1} \in R^n$ are linearly independent, we should not be able to find $(c_1, c_2, \ldots, c_{n+1}) \neq(0,0, \ldots, 0)$ giving $c_1 v_2 + c_2 v_2 + \cdots + c_n v_n + c_{n+1} v_{n+1} = \vec{0}$. Assuming, instead, that you CAN find such $c_i$, you should be able to get a contradiction.