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I want to prove an obviously wrong statement: Any set $\{v_1, \ldots , v_n\}$ of non-zero vectors in Rm is linearly independent.

We proceed by induction on the size of the set. Consider the base case in which we have a single vector $v$. Since $v\neq 0$, the set $\{v\}$ is linearly independent. Now assume that for a fixed $n ∈ N$, any set of $n$ vectors is linearly independent. Consider any set of $n + 1$ vectors, say $\{v_1, \ldots , v_{n+1}\}$. By the inductive hypothesis, the sets $\{v_1, \ldots , v_n\}$ and $\{v_2, \ldots , v_{n+1}\}$ are linearly independent. Therefore, the set $\{v_1, \ldots , v_{n+1}\}$ is also linearly independent. By induction, the claim thus holds for a set of non-zero vectors of any size.

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Free Ikham
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    In general, if you're trying to figure out what's going wrong in a proof that proves a result you know is false, you can try running the proof on an example. Then you go through the proof line by line until the first line where you've said a false thing about your example. – Qiaochu Yuan Oct 06 '20 at 20:57
  • For $k=n+1\ge 2$ you can split ${v_1,...., v_k}$ into ${v_1,....,v_{k-1}}$ and ${v_2,....,v_k}$. For $k = n = 1$ you can not You are making assumptions about your case where $n >$ base case, that you are not making about your base case. ... You can't assume anything in you induction step that does not hold in your base case. this leads to all horses are the same color error. – fleablood Oct 06 '20 at 21:50

2 Answers2

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By the inductive hypothesis, the sets {v1, . . . , vn} and {v2, . . . , vn+1} are linearly independent. Therefore, the set {v1, . . . , vn+1} is also linearly independent.

The mistake is here in the "Therefore". There is no reason given why this should be true.

A counterexample for $n=2$ would be $v_1=(1,1)$, $v_2=(1,0)$, $v_3=(0,1)$.

supinf
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The problem is that given $\{v_1\}$ linearly independent and $\{v_2\}$ linearly independent we can't conclude that the set $\{v_1,v_2\}$ is linearly independent, that is we can't use the inductive hypotesis from the base case and therefore induction doesn't work.

Refer also to the related

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