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Let $V, W$ be finite dimensional vector spaces over a field $F$. Let $\mathcal{B}_{V} = \{\mathbf{v_1, \cdots, v_n} \}$ and $\mathcal{B}_{W} = \{\mathbf{w_1, \cdots, w_m} \}$ be corresponding bases. Let $\psi: V \to W$ a linear map and let $A$ be the matrix representing $\psi$ with respect to the above bases.

(a) Let $\chi: F^n \to V$ be given by $\chi(\mathbf{x}) = \displaystyle \sum_{i=1}^{n} x_i\mathbf{v}_i$. In one line, explain why $\chi$ is an isomorphism.

(b) Prove that:

$$\mathbf{x} \in \ker(A) \iff \left (\displaystyle \sum_{i=1}^{n} x_i\mathbf{v}_i \right )\in \ker(\psi)$$

(c) Conclude that $$\dim(\ker(A)) = \dim(\ker(\psi))$$

For (a) it suffices to notice that $\chi$ is linear. Since $\dim(F^n) = \dim(V) = n$, $\chi$ is forced to be an isomorphism. For (b) I really need help. It really feels like it should not be that hard, but apparently I am missing some small step that ruins everything. Any help will be much appreciated!

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    $\begingroup$ The key here is that $A$ is a matrix with respect to the basis $\{\mathbf{v}_1,\ldots,\mathbf{v}_n\}$. In other words if you take a vector $y$ in $V$ with respect to the basis above, then $\mathbf{y}=\sum_{i=1}^ny_i\mathbf{v}_i$ and the $i$'th column of $A$ represents what $A$ takes $\mathbf{v}_i$ to. $\endgroup$
    – Alex R.
    Mar 11 '13 at 0:22
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    $\begingroup$ For (a), it's not quite enough. For example, the $0$ map from $F^n$ to $V$ is linear and the dimensions match. You also need to argue that either $\chi$ is injective or surjective. (Since the dimensions match, $\chi$ is injective iff it is surjective.) $\endgroup$
    – Jason DeVito
    Mar 13 '13 at 0:09
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I will give a few hints, because these exercises are good to do yourself. Don't take part a lightly, if you get that one, the rest will follow almost immediately.

For a) notice that the elements of $F^n$ are the coordinate vectors for a n-dimensional vector space over $F$. Now try to look carefully what $\chi$ does.

For b) notice that the matrix is itself a linear map: $F^n -> F^n$. Now look at $\chi_w$ which is defined the same way as $\chi$ but with the the basis $\mathcal{B}_{W}$. Now try looking at what the map $ \chi_w \circ A \circ \chi^{-1}: V -> W$ amounts to. Notice that $\chi$ is an isomorphism by a) and analogously $\chi_w$ as well. Therefore their kernel is trivial.

For c) try to apply $\chi$ on the whole kernel of $A$. What do you get?

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