This is my homework's problem:
Let $T$ be a linear transformation that maps $V$ to $W$. Prove that this transformation maps every basis of $V$ to the basis of $W$.
I think I knew the answer. I want to go through prove by contradiction. Assume $v$ be a basis of $V$. We transform this using $T$. If we can write this as a linear combination of other transformation of subspace $V$ in this way $$T(v) = c_{1}T(v_{1}) + c_{2}T(v_{2}) + … + c_{n}T(v_{n})$$ so this is contradiction. But what if we can write this $T(v)$ as linear combination of other vectors of $W$ that is can't be achieved via this transformation.
Thanks from your help.
You can't prove that statement since it is false. If, for instance, you define$$\begin{array}{rccc}T\colon&\mathbb{R}^2&\longrightarrow&\mathbb{R}^2\\&(x,y)&\mapsto&(x-y,0),\end{array}$$then the basis $\bigl((1,0),(0,1)\bigr)$ is not maped into a basis of $\mathbb{R}^2$. In fact, $T$ maps no basis of $\mathbb{R}^2$ into another basis of $\mathbb{R}^2$.