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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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – angryavian
    Jul 3, 2020 at 22:21
  • $\begingroup$ Oh...I just realized that the pics I uploaded didn't transfer...@angryavian, look now $\endgroup$ Jul 3, 2020 at 22:27

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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 $0$.

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  • $\begingroup$ That makes sense to me. And truthfully I should've been more confident in my abilities when I noticed that typo myself, but I second guessed it. Thanks for the clarification. $\endgroup$ Jul 3, 2020 at 22:52
  • $\begingroup$ You are welcome. Glad to have helped. $\endgroup$ Jul 3, 2020 at 23:27

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