Existence of ordered bases $\beta$ and $\gamma$ for $V$ and $W$, such that $[T]_{\beta}^{\gamma}$ is a diagonal - Questions about proof

Question

Let $V$ and $W$ be vector spaces such that $\text{dim}(V) = \text{dim}(W)$, and let $T:V \to W$ be linear. Show that there exist ordered bases $\beta$ and $\gamma$ for $V$ and $W$, respectively, such that $[T]_{\beta}^{\gamma}$ is a diagonal matrix.

My question pertains to two different steps taken in the proof that I do not fully see the jump in logic coming about. Here are the parts where I am having trouble.

My first question has to do with Step 3: What is trying to be accomplished by writing $$\sum_{j = 1}^{k}c_{j}v_{j} - \sum_{i = k+1}^{n}c_{i}v_{i} = 0 \ ?$$

I get that $\sum_{i = k+1}^{n}c_{i}v_{i}$ is in the Null Space of $T$ and as such can be written as a linear combination of the basis of the Null Space. But how does this lead into Step 4? Of which I also ask how is it that $$c_{1} = c_{2} = \dots c_{k} = c_{k+1} = \dots c_{n} = 0 \ ?$$

I get why $c_{1} = c_{2} = \dots c_{k} = 0$. That's because these were the coefficients for the basis vectors from the Null Space. But the other ones?...why are they all $0$'s ?

The other steps of the proof after this make sense to me, it was mainly those two steps. I'm just frustrated because it feels as if I attempt these proofs and then I remember some established results, but never the necessary added results to answer the question...I'm rambling...apologies.

Answer 1

The images contain typoes. In step 3, after the first line, the $u_j$'s are wrongly converted into $v_j$'s. They should still be $u_j$'s, and so you will get a linear combination of elements of base $\beta$ which equate 0 in the end. This justifies all coefficients being deduced to be $0$.