Proving set of vectors is linearly independent

Question

The question asks:

If $\{v_{1}, v_{2}, \cdots, v_{n}, v_{n+1}\}$ is linearly independent, prove $ \{v_{1}, v_{2},\cdots, v_{n}\}$ is linearly independent.

My attempt at a proof:

If $\{v_{1}, v_{2},\cdots, v_{n}\}$ is linearly dependent. Then $ t_{1}v_{1} + t_{2}v_{2} + \cdots + t_{n}v_{n} = 0$ has some solutions where $t_{i} \neq 0 \forall i$.

But $\{v_{1}, v_{2},\cdots, v_{n}, v_{n+1}\}$ being linearly independent implies $ t_{1}v_{1} + t_{2}v_{2} + \cdots + t_{n}v_{n} + t_{n+1}v_{n+1} = 0$ where $t_{i} = 0$ $\forall i$. So we have derived a contradiction.

So $ \{v_{1}, v_{2},\cdots, v_{n}\}$ is linearly independent.

$\square$

Not sure if I've performed this proof by contradiction correctly so any insight would be nice. Or alternative proofs even !

Answer 1

Linear independence of $\{v_1,\dots,v_n,v_{n+1}\}$ is not expressed by the fact that $t_1v_1+\dots+t_nv_n+t_{n+1}v_{n+1}=0$ when $t_i=0$ for all $i$.

The statement

$t_1v_1+\dots+t_nv_n+t_{n+1}v_{n+1}=0$ when $t_i=0$ for all $i$

is true for every (finite) set of vectors.

The correct condition is

if $t_1v_1+\dots+t_nv_n+t_{n+1}v_{n+1}=0$, then $t_i=0$, for all $i$.

Maybe you have expressed badly your knowledge, but that misunderstanding of linear independence is very common.

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Avoid contradiction, if you can. In order to show that $\{v_1,\dots,v_n\}$ is linearly independent, suppose $t_1v_1+\dots+t_nv_n=0$. If you set $t_{n+1}=0$, then also $$ t_1v_1+\dots+t_nv_n+t_{n+1}v_{n+1}=0 $$ so, by assumption, $t_i=0$ for $i=1,2,\dots,n,n+1$, in particular for $i=1,2,\dots,n$.

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If you want to go by contradiction, suppose $t_1v_1+\dots+t_nv_n=0$, with the $t_i$ not all zero. Then, with $t_{n+1}=0$, we have $$ t_1v_1+\dots+t_nv_n+t_{n+1}v_{n+1}=0, $$ a contradiction.

Actually this is not really a proof by contradiction: we have proved that if $\{v_1,\dots,v_n\}$ is linearly dependent, then also $\{v_1,\dots,v_n,v_{n+1}\}$ is linearly dependent.

Answer 2

$t_1v_1+t_2v_2+\cdots+t_nv_n+0\cdot v_{n+1}\Rightarrow t_1,\ldots,t_n=0$

This is a quick version of your proof. Also, the idea behind your proof is right, but strictly speaking it is wrong. Just note that "$t_1v_1+t_2v_2+\cdots+t_nv_n+t_{n+1}v_{n+1}=0$ where $t_i=0 \forall i.$" always hold. I suppose this is just an expression error.