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

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.

• It is difficult to answer your question without more context, since you are quoting from the middle of a proof. (What are $v_1, \ldots,v_n$? What is $k$? etc.) Regarding why the coefficients are zero: if $v_1, \ldots, v_n$ are a basis and $\sum_{i=1}^n a_i v_i = 0$, then linear independence implies $a_1 = \cdots \ a_n = 0$. But we can't tell if this is the correct explanation without more context. Jul 3, 2020 at 22:21
• Oh...I just realized that the pics I uploaded didn't transfer...@angryavian, look now Jul 3, 2020 at 22:27

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 $$0$$.