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Let $\phi: V \to W$ a linear transformation between vector spaces and $v_1, \cdots, v_k \in V$. Suppose that the following condition is fulfilled:

$$\phi \left (\sum_{n=1}^{k}c_n\mathbf{v_n} \right) = \mathbf{0} \iff c_n = 0 \ \forall n$$ where $c_n$ are elements of the underlying field. Show that $v_1, \cdots, v_n$ form a basis for $V$.

Now, I know I am missing something small. What is it?

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up vote 4 down vote accepted

You can easily deduce that $v_1,\ldots,v_k$ are linearly independent. Take a linear combination of them. Apply $\phi$. Conclude.

But you can't conclude that they span $V$, so they don't necessarily form a basis.

Assume $k\geq 2$.

Note that if your condition holds, it holds for any choice of $v_j$'s.

In particular, it holds for $(v_1)$ alone and for $(v_1,v_2)$ alone.

If your statement was true, we would end up $(v_1)$ and $(v_1,v_2)$ bases for $V$, hence $\dim V=1=2$. A contradiction.

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I am just proving that if your claim was true, we could prove that $1=2$, a contradiction. If $(v_1)$ is a basis $\dim V=1$ and if $(v_1,v_2)$ is a basis, $\dim V=2$. We can't have these simultaneously, right? – 1015 Feb 17 '13 at 1:39

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