Why, in terms of the structure of proofs and proof strategy, is this proof of mine said to be backwards in logic? Last year when I was doing the linear algebra and proof writing course, I was often said by my friends and my professors that the logical flow of my proofs are weird or even backwards.
Recently, I accidentally stumbled upon this site. After analysing it and comparing with the standard approaches in proofs in my course and also the features of my proof strategies identified by my friends and professors, I am suspecting it might be a type of synthetic proof strategy. The result of this analysis is summarized in the figure, (purple box)
http://www.slideshare.net/sultanakhan1/analytico-synthetic-method-of-teaching-mathematics

A more illustrative example of the purple box is given here:
The problem to be proved:
Suppose that $V,W$ are vector spaces over $\mathbb{F}$ and that $T : V \rightarrow W$ is a linear transformation.
Suppose that $T$ is one-to-one, and that $\{\mathbf{v}_1, \dots, \mathbf{v}_n\}$ is linearly independent in $V$. Prove that $\{T(\mathbf{v}_1),\dots,T(\mathbf{v}_n)\}$ is linearly independent in $W$.
Standard approach in the course and also in professional settings:
Suppose that
$$\sum_{i=1}^n \alpha_i T(\mathbf{v}_i)=\mathbf{0}$$
for some $\alpha_1, \dots, \alpha_n \in \mathbb{F}$. Then
$$\mathbf{0}=\sum_{i=1}^n \alpha_i T(\mathbf{v}_i)$$
$$=T\left(\sum_{i=1}^n \alpha_i \mathbf{v}_i\right)\text{  as $T$ is linear.}$$
But $T(\mathbf{0})=\mathbf{0}$ and $T$ is one-to-one. Hence
$$\sum_{i=1}^n \alpha_i \mathbf{v}_i=\mathbf{0}$$
But $\{\mathbf{v}_1, \dots, \mathbf{v}_n\}$ is linearly independent, so
$$\alpha_1=\dots=\alpha_n=0$$
Hence $\{T(\mathbf{v}_1),\dots,T(\mathbf{v}_n)\}$ is linearly independent in $W$.
$\square$
My proof that is said to be weird in the standards of professional mathematicians:
Given $\{\mathbf{v}_1, \dots, \mathbf{v}_n\}$ is linearly independent in $V$, that is if
$$\sum_{i=1}^n \alpha_i\mathbf{v}_i=\mathbf{0}$$
then $\alpha_i=0$ for all $i=1,\dots,n$
Apply $T$ both sides
$$T\left(\sum_{i=1}^n \alpha_i\mathbf{v}_i\right)=T(\mathbf{0})$$
Since $T$ is one-to-one. Then
$$T\left(\sum_{i=1}^n \alpha_i\mathbf{v}_i\right)=\mathbf{0}$$
Since $T$ is linear. Then
$$\sum_{i=1}^n \alpha_iT(\mathbf{v}_i)=\mathbf{0}$$
But $\alpha_i=0$ for all $i=1,\dots,n$. Therefore $\{T(\mathbf{v}_1),\dots,T(\mathbf{v}_n)\}$ is linearly independent in $W$.
$\square$
We then discuss about it, and we found that I should start my proof on the thing I want to prove, and not start from known conditions or general cases and then build my way to the claim that needs to be proved by putting in the givens as I go from one step to the next in my proof. However for some cases, it is still not very clear in terms of formalism and structure, why and how my proof is weird.


*

*Why is the logic flow in my proof seems weird to most professional mathematicians?


*Am I proving the wrong thing because of this weird logic?


*If I am doing it correctly, then what terminology will describe this type of proof style of mine?

 A: Your 'standard' proof of the linear algebra statement can be summarised as follows:  


*

*If a general linear combination of the $T(v_{i})$ is zero, then the same linear combination of the $v_{i}$ is zero - so by linear independence of the $v_{i}$, the coefficients are all zero.  


Whereas your 'weird' proof is like this:  


*

*If a linear combination of the $v_{i}$ is zero, then the same linear combination of the $T(v_{i})$ is zero. And the only such linear combination is the trivial one.


But this hasn't proved the statement! More specifically, how do you know that there isn't a set of coefficients $\alpha_{i}$ such that $$\sum_{i=1}^{n}\alpha_{i}T(v_{i})=0 \quad \text{and}\quad \sum_{i=1}^{n}\alpha_{i}v_{i} \ne 0$$
Slightly more formally, define the following logical statements, given a fixed linear combination: 


*

*$P_{v}$="The linear combination of the $v_{i}$ is $0$"

*$P_{T}$="The linear combination of the $T(v_{i})$ is $0$"

*$Q$="All of the coefficients are zero"  


You want to prove that $P_{T} \implies Q$, and you know that $P_{v} \implies Q$ by linear independence.  
Your standard proof shows that $P_{T} \implies P_{v}$, from which we deduce $P_{T} \implies Q$.  
But your 'proof' shows that $P_{v} \implies P_{T}$, from which we cannot deduce anything. This is the sense in which your proof is backwards.  
In more detail your 'proof' is (line by line):
$P_{v}$
$P_{v} \implies Q$
$Q$
$Q \implies P_{T}$
$P_{T}$  
But what you want to deduce is $P_{T} \implies Q$
A: This is me trying to read your proof :

My proof that is said to be weird in the standards of professional mathematicians:
Given $\{\mathbf{v}_1, \dots, \mathbf{v}_n\}$ is linearly independent in $V$, that is if
$$\sum_{i=1}^n \alpha_i\mathbf{v}_i=\mathbf{0}$$
then $\alpha_i=0$ for all $i=1,\dots,n$

so far so good

Apply $T$ both sides
$$T\left(\sum_{i=1}^n \alpha_i\mathbf{v}_i\right)=T(\mathbf{0})$$

I have no idea what you are claiming. You aren't given that $\sum \alpha_iv_i = 0$, so if you are claiming that $T(\sum \alpha_i v_i) = T(0)$, this is false.
Maybe you're claiming that if $A = B \implies P$ holds then $T(A) = T(B) \implies P$ holds but this is not logically sound. But I have no idea what you're trying to do here anyway.
For now I'll suppose that somehow you meant to write "suppose $\sum \alpha_i v_i = 0$" and you didn't write it, even though this still makes no sense as a proof strategy since that's not what you are asked to prove.

Since $T$ is one-to-one. Then 
  $$T\left(\sum_{i=1}^n \alpha_i\mathbf{v}_i\right)=\mathbf{0}$$

bad grammar. I wouldn't know if the "since" applies to the previous step or the next step. I flipped a coin and I will decide it's for the next step, but the reason you gave for it is simply wrong. $T(0) = 0$ because $T$ is linear.  
Injectivity of $T$ means that if $T(x)=0$ then $x=0$ and I don't see any place where you used it between those two statements.
Also you shouldn't put a full stop, neither use "then". "since [blabla], we have / we can deduce / we get [...]" are more readable.

Since $T$ is linear. Then
  $$\sum_{i=1}^n \alpha_iT(\mathbf{v}_i)=\mathbf{0}$$

okay I guess ?

But $\alpha_i=0$ for all $i=1,\dots,n$. 

Huh ? What ? Why ? Since when ?
Possibly this is a consequence of the hypothesis you wrote at the top and your unwritten assumption that $\sum \alpha_i v_i = 0$. At this point I realise that you've been writing $0=0$ in all kinds of obscure and complicated ways.

Therefore $\{T(\mathbf{v}_1),\dots,T(\mathbf{v}_n)\}$ is linearly independent in $W$.

No, so far you only have shown something like "if $x=0$ then $T(x)=0$".
To prove that $\{T(v_i)\}$ is linearly independant you have to suppose that you have some scalars $\alpha_i$ such that $\sum \alpha_i T(v_i) = 0$, and show that all $\alpha_i$ are $0$. I didn't see anything even remotely looking like a proof of that.

Now a correct proof that somehow follow your steps but with all the explanation that you didn't care to put in there, possibly that whenever you wrote a sentence $S$, you didn't mean to claim that $S$ was true (how does this sound like a good idea, I will never know), but only that $S$ implied that the $\alpha_i$ were zero.

We know that $\sum \alpha_i v_i = 0 \implies \alpha_i = 0 \forall i$
Because T is injective, $T(\sum \alpha_i v_i) = T(0) \implies \sum \alpha_i v_i = 0$, and thus $T(\sum \alpha_i v_i) = T(0) \implies \alpha_i = 0 \forall i$
Because T is linear, $T(\sum \alpha_i v_i) = \sum \alpha_iT(v_i)$ and $T(0) = 0$, and thus $\sum \alpha_i T(v_i) = 0 \implies \alpha_i = 0 \forall i$
This is precisely the statement that we needed to prove so we are done.

There are also ways to make it clearer with words what you're doing (that you're going backwards) :

to prove that $\forall i, \alpha_i = 0$, we only need to prove - because we know that the $v_i$ are independant in $V$ - that $\sum \alpha_i v_i = 0$.
Next, because $T$ is injective, to prove that $\sum \alpha_i v_i = 0$ we only need to prove that $T(\sum \alpha_i v_i) = T(0)$ 
Because $T$ is linear [...] ,so we only need to prove that $\sum \alpha_i T(v_i) = 0$
But we have just shown that $\sum \alpha_i T(_i) = 0$ implies $\forall i, \alpha_i = 0$ which is what we needed to prove. 

A: As explained in other answers, you are proving the opposite direction of implication. But, if you observe that $T(x)=T(y)$ is equivalent to $x=y$ when $T$ is one-to-one, you could formulate the correct proof in a way that somewhat resembles your proof.
Indeed, then the following condition 
$$\sum_{i=1}^{n}\alpha_{i}v_{i} = 0$$
holds if and only if
$$T \left( \sum_{i=1}^{n}\alpha_{i}v_{i} \right) = T(0)$$
Since $T ( \sum_{i=1}^{n}\alpha_{i}v_{i})=\sum_{i=1}^{n}\alpha_{i}T(v_i)$ and $T(0)=0$, this is further equivalent to 
$$\sum_{i=1}^{n}\alpha_{i}T(v_i)=0.$$
So this linear combination is zero vector if and only if the first linear combination above is also a zero vector. And this is the case only when all $\alpha_i=0$.
Note that here we have chain of equivalences, so we can also follow implications from bottom toward theh top of the proof.
A: You start by repeating the assumption and expanding its definition, that is ok. So you have $v_1, …, v_n$ such that $∀α_1, …, α_n\ ∑_{i = 1}^n α_n v_n = 0 \implies α_1 = … = α_n = 0$. But then you are doing some operations with some $α_1, … α_n$. What are these? Are these arbitrary such that $∑_{i = 1}^n α_n v_n = 0$? If so, then you have $α_1 = … = α_n = 0$ by the linear independence of $v_1, …, v_n$. But that say nothing about linear independence of $T(v_1), …, T(v_n)$.
Note that you don't use the fact that $T$ is injective in your proof. If you have $T(∑_{i = 1}^n α_i v_i) = T(0)$ and you want to conclude that $T(∑_{i = 1}^n α_i v_i) = 0$, then you just need that $T(0) = 0$, which follows from the linearity of $T$.
If you want to prove the proposition, you have to comply to the following interface: (Having the fixed context of $V$, $W$, $T$, $v_1, … v_n$ with their properies) I will give you $α_1, …, α_n$ such that $∑_{i = 1}^n α_i T(v_i) = 0$, and you have to convince me that $α_1 = … = α_n = 0$. You may formulate a correct proof many different ways and using various logical translations, but it should be always clear to you how to adapt it to my interface.
A: Your proof has issues because you are not consistent in the backwardness. What your prerequisite is is that $\sum \alpha_jv_j=0\Rightarrow \alpha_j=0$, but then you go in the sequence $\sum\alpha_jv_j=0 \Rightarrow ... \Rightarrow \sum\alpha_jT(v_j))=0$. 
That is you prove that $P \Rightarrow Q$ and $P \Rightarrow R$, but that doesn't mean $R\Rightarrow Q$. You have to do the second part the other way around and formulate it so that $P \Leftarrow R$, then it follows that $R \Rightarrow Q$.
The solution is that every next equation must not be shown to follow from the first, but being a prerequisite for the former to be true. In this case that's true, but you formulate it in a way that you don't make that claim in the proof.
For example when you state that you use that $T$ is one-to-one, you should really use that $T$ is linear, because you want to show that $LHS = 0$ implies that $LHS = T(0)$, not the other way around.
A: In your example, the result you want to prove is of the form:

If $A \implies C$, then $B \implies C$.

(Here, statement $A$ is "$\sum_{i=1}^n a_i \mathbf v_i = 0$", $B$ is "$\sum_{i=1}^n a_i T(\mathbf v_i) = 0$", and $C$ is "$a_i = 0\ \forall i$".)
A natural approach to proving such a statement is to first prove that $A \impliedby B$, from which we can then easily derive the desired result.
What's important to note is that proving the other implication, $A \implies B$, does nothing useful for us here!  We really need the implication from $B$ to $A$, not from $A$ to $B$.
Thus, when proving such a lemma, it's natural to start from $B$, and work "backwards" to $A$.
Obviously, here we could also prove the stronger result $A \iff B$, which contains both implications.  But we still only need the reverse part, so this is a strictly stronger result than necessary.  In some cases, it might not even hold, even if $A \impliedby B$ does.
Furthermore, in your own proof, you don't clearly state whether you're claiming equivalence or merely implication between the various intermediate results.  Thus, it looks like you're proving the (useless) implication $A \implies B$, even though each of your steps can in fact be reversed to also show $A \impliedby B$.

It's worth noting that this "reversal" is specific to the type of claim you're trying to prove, where the assumption $A \implies C$ and the desired conclusion $B \implies C$ are both conditional statements with the common consequent $C$.
If you instead wished to prove, say, that if $A \implies B$, then $A \implies C$, then it would indeed be natural to show an implication directly from the assumed consequent $B$ to the desired consequent $C$, given the shared antecedent $A$.
It's only when the consequents are shared but the antecedents differ that you need to go the other way.  And of course, there are plenty of situations where neither of these simple proof strategies will work, and you'll need to try a more complex chain of reasoning.
