Dimension Theorem. Understanding the proof.

Question

I'm currently rereading through Linear Algebra by Friedberg. The outline of his proof for the dimension theorem is as follows:

1. Extend a basis ${v_1, ... , v_k}$ of $N(T)$ to a basis ${v_1, ... , v_n}$ of vector space $V$.

2. Show $\mathrm{span}( T(v_{k+1}), ... , T(v_n) ) = R(T)$

3. Show $T(v_{k+1}), ... , T(v_n)$ are linearly independent.

I understand every single step. My only issue is the last lines of the proof says that this argument also shows that $T(v_{k+1}), ... , T(v_n)$ are all distinct.

My main question is how does this argument show that? I have a feeling this has to do something with the vectors being linearly independent. But I'm uneasy about it.

1. Doesn't the very question of asking if $T(v_k+1) , ... , T(v_n)$ are linearly independent assume that they are in fact distinct? That is, we are assuming they are distinct from the very start. For instance, we wouldn't ask if $(1,0,0)$, $(1,0,0)$, and $(0,1,0)$ are linearly independent.

2. How does this all tie into the definition of linear independence. And where is my understanding of it faulty concerning this question at hand?

I've tried to reason as follows.

If our vectors $T(v_{k+1}) , ... T(v_n)$ are not distinct. Then the task of showing that they are linearly independent makes no sense. This is because linear independence only concerns distinct vectors.

Edit: Here are the definitions of linear dependence and independence given from the text:

A subset $S$ of a vector space $V$ is called linearly dependent if there exists a finite number of distinct vectors $u_1, ... ,u_n$ in $S$ and scalar $a_1, ... ,a_n$ all not zero such that $a_1 u_1+ ... + a_n u_n = 0$ (zero vector)

Linear independence is given as the negation of the above statement. That is, $S$ is linearly independent if it is not linearly dependent.

Answer 1

Why are linearly independent vectors distinct? Suppose they're not, and that $v_i=v_j$ for some distinct $i,j$. Then $$1\cdot v_i + (-1)\cdot v_j = 0$$ That is a nontrivial linear dependence relation, contradicting the assumed independence. Done.

It's not that linear independence only concerns distinct vectors - it's that vectors that aren't distinct can't be linearly independent.

Answer 2

I'm answering a question Gabe asked in a comment, but which seems to be at the heart of Gabe's misunderstanding, and for good reason. Gabe asks:

[...] doesn't lilnear dependence only consider the case where you can write a non-trivial linear combination of distinct vectors. That is, the above isn't valid. You've written them as a linear combination of two nondistinct ones.

Here's an underlying question: what is a linear combination? Depending on your text, it'll say something like "An expression of the form $$ c_1 v_1 + \ldots + c_n v_n, $$ where the $c_i$ are scalars and $\{v_1, \ldots, v_n\}$ is a set of vectors in the vector space $V$."

The problem with this is that if all the vectors are equal, then the set contains only one vector, and ... it's not clear what's meant.

Also, do we mean the expression? Or do we mean its value? If we mean the expression, just what is that mathematically? What set does it lie in? (The value of the expression lies in $V$, but the expression? That's an altogether different kind of object.)

So here's perhaps a better version of that definition (but one I've never seen, despite having taught linear algebra multiple times):

If $v_1, \ldots, v_n$ is a finite sequence of vectors in a vector space $V$ over a field $F$, then a "linear combination" of these vectors is a finite sequence $c_1, \ldots, c_n$ of elements of $F$; the value of this linear combination is the vector $$ c_1 v_1 + \ldots + c_n v_n,$$ which is a vector in $V$.

Note that this sequence of vectors might contain multiple copies of the same vector, or contain the zero vector, or ...just about anything. If $v$ is some nonzero vector in a vs V over $\Bbb R$, then the sequence might be $$ v, v, -v $$ and then the coefficients could be $1,1,0$, in which case the value would be $$ 1v + 1v + 0(-v) = 2v $$ or they might be $1, 0, 1$, in which case the value would be the zero-vector, or they might be $1, -1, 0$, in which case the value would again be the zero-vector. But all three, despite having the vector $v$ repeated, would be valid linear combination values of the sequence $v, v, -v$. (I used a sequence because sequences are allowed to have repeated elements. Indeed, formally defining a sequence is an interesting exercise in itself, but not the right topic for this answer...)

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In actual practice, we get sloppy. When we write a set of vectors in some order, like $\{v, w, v+w\}$, we pretend that the order matters, even though sets are unordered -- we mangle together our typography with the underlying meaning. As it happens, this never confuses anyone except people trying to learn it for the first time, when it can be completely baffling. That's probably why you never see the definition I gave above --- it's valid, but somewhat complicated, and tough to apply without a lot of fiddling, so no one wants to do all that, and perhaps they even forget that to make logical sense, it might be necessary. So we live with the less formal definitions, and move on.

(One exception: people who try to write theorem provers or proof-assistant programs. They actually have to suffer with these things and iron out the informality. It's probably a pretty worthwhile thing to do now and then.)