|
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? |
|||
|
|
|
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. |
||||
|
