# Theorem 2.20 in Friedberg's Linear Algebra

Theorem 2.20 Let $V$ and $W$ be finite-dimensional vector spaces over $F$ of dimensions $n$ and $m$, respectively, and let $\beta$ and $\gamma$ be ordered bases for $V$ and $W$, respectively. Then the function $\phi : \mathcal{L}(V,W) \to M_{m \times n} (F)$, defined by $\phi(T) = [T]_\beta^\gamma$, is an isomorphism.

In proving the theorem, the author merely shows that $\phi$ is linear and surjective. However, this seems to be fallacious. By just showing surjectivity, it seems that the author is implicitly relying on the finite dimensionality of $\mathcal{L}(V,W)$ to conclude that it must also be injective. This is initially what I thought of doing when I tried proving theorem myself; I was going to use the rank-nullity theorem but then realized I do not have finite dimensionality. Instead, I chose to prove the two following two lemmas:

First I proved that $v \in V \setminus \{0\}$ implies $[v]_\beta \neq 0$. Here is my proof:

Let $\beta = \{v_1,...,v_n\}$. Since $v$ is not zero, there exist $a_1,...,a_n$ not all equal to zero such that $v = \sum a_i v_i$. By definition, $[v]_\beta = (a_1,...,a_n)^T$, which cannot be the zero vector in $F^n$, since not every entry is zero.

Then I proved that $T \neq 0$ implies $[T]_\beta^\gamma \neq 0$. Here is my proof:

Let $\gamma = \{w_1,...,w_m\}$ and suppose that $T \neq 0$. Then there exists a $v \in V \setminus \{0\}$ such that $T(v)\neq 0$. Hence $0 \neq [T(v)]_\gamma = [T]_\beta^\gamma [v]_\beta$. If $[T]_\beta^\gamma$ were zero, then we would have a contradiction. Hence, $[T]_\beta^\gamma \neq 0$

From this it follows that if $[T]_\beta^\gamma = 0$, then $T=0$, proving that the kernel of $\phi$ is trivial and therefore injective.

Now, as far as I can tell, $\mathcal{L}(V,W)$ being finite dimensional has not been demonstrated yet; in fact, the corollary of this theorem is that $dim (\mathcal{L}(V,W))$, which of course entails finite dimensionality. Thus, it seems that the proof is lacking since injectivity does not follow from what has been proven heretofore. To put it straightforwardly, how do the authors prove the theorem without explicitly showing injectivity? Perhaps those with a copy of Friedberg might be able to answer the question.

EDIT: As Li Chun Min points out, a crucial part of the proof is that there exists a unique linear operator $T : V \to W$ such that

$$T(v_j) = \sum_{I=1}^m A_{ij} w_j,$$

which gives one surjectivity and injectivity.

• Why the downvote? Please justify the downvote. – user193319 Jun 29 '17 at 14:02
• Yes. You are right that finiteness of $\dim L(V,W)$ haven't been demonstrated yet. Yet Friedberg indeed proven $\phi$ is bijective by showing that for each $A \in M_{m \times n}$, there is (surjective) a unique (injective) $T$ such that $\phi(T)=A$. A bijective linear mapping is an isomorphism. In fact that is a direct proof without using nullity rank! (but it is broken into lemmas 2.8 and 2.6 – Li Chun Min Jun 29 '17 at 14:33
• @LiChunMin Ah! Yes! You are right. I completely overlooked that very important word 'unique.' Thanks for the help! – user193319 Jun 29 '17 at 23:23