# Is this a correct way to do this proof?

Prove that if $\{\vec{v}_1,\dots,\vec{v}_k\}$ is linearly independent, then $\{\vec{v}_1,\dots{},\vec{v}_{k-1}\}$ is also linearly independent.

My proof:

Since $\{\vec{v}_1,\dots,\vec{v}_k\}$ is linearly independent, then the equation $$c_1\vec{v}_1+\cdots+c_k\vec{v}_k=\vec{0}$$ has only the trivial solution. This then implies that $$c_1\vec{v}_1+\cdots+c_{k-1}\vec{v}_{k-1}=-c_k\vec{v}_k$$ also has only the trivial solution by basic algebra. $-c_k\vec{v}_k=0$ so $\{\vec v_1, \ldots,\vec{v}_{k-1}\}$ is also linearly independent.

Is this proof correct?

-
Needs to be tidied up - there is a curly right brace hanging around without a friend - but your main idea is right. – in_wolframAlpha_we_trust Apr 12 '13 at 13:37
@in_wolfram_we_trust: I think I just fixed that. Eric, please edit again if I misinterpreted. – Tomas Lycken Apr 12 '13 at 13:38
@in_wolfram_we_trust: Hm... well, my edit will be visible when it's peer-reviewed... but I changed it to "so $\{\vec v_1,\dots,\vec v_{k-1}\}$ is..." – Tomas Lycken Apr 12 '13 at 13:39
@EricDong Personally I think that your proof to be acceptable you must show that $c_1v_1+\ldots+c_{k-1}v_{k-1}=0$ has only the trivial solution. – user63181 Apr 12 '13 at 13:50

If we suppose that $\{\vec{v}_1,\ldots,\vec{v}_{k-1}\}$ is linearly dependent then there's $c_1,\ldots,c_{k-1}$ not all zero such that $$c_1\vec{v}_1+\cdots+c_{k-1}\vec{v}_{k-1}=0$$ then take $c_k=0$ and we find $$c_1\vec{v}_1+\cdots+c_k\vec{v}_k=\vec{0}$$ and this a contradiction with the fact that $\{\vec{v}_1,\ldots,\vec{v}_{k}\}$ is linearly independent.
This is the way I'd do it, although this really shouldn't be written as a proof by contradiction - it's a proof by contrapositive. You assume the smaller set is linearly dependent and show that the larger set must also be, which is logically equivalent to "$\{v_1,\ldots, v_k\}$ linearly independent implies $\{v_1,\ldots,v_{k-1}\}$ linearly independent." – Stahl Apr 12 '13 at 13:41
Suppose $c_1\mathbf{v}_1 + \ldots + c_{k-1}\mathbf{v}_{k-1} = \mathbf{0}$. Then $c_1\mathbf{v}_1 + \ldots + c_{k-1}\mathbf{v}_{k-1} + 0\mathbf{v}_k = \mathbf{0}$. Since $\{\mathbf{v}_1, \ldots,\mathbf{v}_k\}$ are linearly independent, $c_1 = c_2 = \cdots = c_{k-1} = 0$. QED