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I was attempting to prove a statement in which I want to include that if some condition doesn't hold, then there is a a setting in which we would get a contradiction. I realise this is not a good way to phrase the question, so maybe a specific example might help.

I have two vectors $$v=\begin{bmatrix}v_1 \\ \vdots \\ v_{n-1}\\0 \end{bmatrix} ,u=\begin{bmatrix}u_1 \\ \vdots \\ u_{n-1}\\u_n \end{bmatrix}$$

where $v_{n-1}$ is non-zero. I am confused as to how to formally write that if $v_n=0$, then there is a possibility that we could write $u$ as a scalar multiple of $v$. But I am not sure if saying "there is a possibility" is rigorous enough. How could I incorporate this into my proof?

Seeker
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    Is it a situation where you want to consider two cases? "Case 1: $u$ is a scalar multiple of $v$. Then [...]. Case 2: $u$ is not a scalar multiple of $v$. Then [...]" – Misha Lavrov Oct 24 '22 at 02:49
  • @MishaLavrov The situation is that I have an $n\times n$ upper triangular matrix $A$ whose columns are linearly independent. I want to show that the entries on the diagonal are non-zero. In doing so, I want to show that if $A_{n,n}=0$ and $A_{n-1,n-1}$ is non-zero, then there is a possibility that we could write $A_{.,n}$ as a scalar multiple of $A_{.,n-1}$. Here $A_{.,k}$ denotes the $k^{th}$ column of $A$. – Seeker Oct 24 '22 at 02:53
  • @MishaLavrov I am just not exactly sure how to word this properly. – Seeker Oct 24 '22 at 02:54
  • In your situation, your problem is not the wording. Your problem is that your approach is not actually enough to carry out the proof you want! The proof is supposed to work even if the $n^{\text{th}}$ column is not a scalar multiple of the $(n-1)^{\text{th}}$ column; it doesn't really help to consider the possibility that it is. – Misha Lavrov Oct 24 '22 at 03:01
  • @MishaLavrov I will try a different approach then. But what I had in mind was that if $A_{n,n}=0$, and there is a possibility that we could write $A_{.,n}$ as a scalar multiple of $A_{.,n-1}$ because $A_{n-1,n-1}\ne 0$(because of my induction hypothesis), then that would contradict the assumption that the columns are linearly independent. But I will try to come up with a different approach. – Seeker Oct 24 '22 at 03:06
  • @Seeker are you looking for straight forward proof of “If $A$ is an upper-triangular $n\times n$ matrix whose columns are linearly independent, then its diagonal entries are nonzero”? – user264745 Oct 24 '22 at 06:39
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    @user264745 That would depend on what you mean by straight forward. I mean I can go the classic way and just say that a square matrix has linearly independent columns if an only if it is invertible. Which happens if and only if the determinant is non-zero. Which happens if and only if the diagonals of an upper triangular matrix is non-zero as the determinant of an upper triangular matrix is the product of its diagonal entries. – Seeker Oct 24 '22 at 06:52
  • @user264745 But if you had a simpler proof in your mind that doesn't use determinants, I would like to know what it is. – Seeker Oct 24 '22 at 06:53

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In general, you should not be trying to include such a situation in your proof.

It would make sense to discuss the possibility that something happens, if your proof should be conducted differently when that thing happens. In your example, we might imagine a proof that goes as follows:

Theorem. If $A$ is an upper-triangular $n\times n$ matrix whose columns are linearly independent, then its diagonal entries are nonzero.

Proof. Let $A$ be such a matrix, and suppose for the sake of contradiction that some of $A$'s diagonal entries are zero.

[Some argument that reduces this to the case where $A_{n,n} = 0$.]

Now, we consider two cases. First, suppose that the $n^{\text{th}}$ column of $A$ is a scalar multiple of the $(n-1)^{\text{th}}$ column. In this case, the columns of $A$ are linearly dependent, contrary to our initial choice of $A$. Second, suppose that the $n^{\text{th}}$ column of $A$ is not a scalar multiple of the $(n-1)^{\text{th}}$ column. In this case, [argument we have yet to come up with].

This is a reasonable way to structure a proof. The problem is that in this particular example, it will turn out that [argument we have yet to come up with] does not benefit from knowing that the $n^{\text{th}}$ column of $A$ is not a scalar multiple of the $(n-1)^{\text{th}}$ column. Once we come up with an argument to handle that situation, that argument applies equally well when the $n^{\text{th}}$ column of $A$ is a scalar multiple of the $(n-1)^{\text{th}}$ column, and we could simplify our proof to avoid considering those cases separately.

When looking for a proof, you should not be trying to find special cases that make your task unusually easy, and dealing with them separately. However, if you found a general argument, and it turned out that the general argument does not work when the $n^{\text{th}}$ column of $A$ is a scalar multiple of the $(n-1)^{\text{th}}$ column, then you might treat that situation as a separate case.

Misha Lavrov
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