I know this isn't a hardcore math problem but I feel like I've developed anxiety over math. I realized I can't just use theorems that I don't have intuition for: it just feels like I'm manipulating stuff I don't understand. And so I feel very anxious because it just doesn't make sense. But this is the important part: even when I logically follow every step, if I don't have the intuition, it makes zero sense.
First, let me say my OCD/anxiety has gotten better. For the past three months, I was questioning what it means for equations to be equivalent: $x+y =z$ iff $y = z-x$. Yes, I know it's super simple and you may be wondering how I could be confused by something so easy, but I guess the best explanation is this: imagine saying out loud the same word over and over and over again; at a certain point, you're going to get confused and say a different word. Another example is trying to read a book, but getting stuck on a sentence, and rereading that one sentence so much that you're reading word by word, which as everyone knows leads to more confusion since we typically read words by clusters. and you're also angry because you know that this one tiny sentence is stopping you from reading the rest of the book.
I've gotten over this anxiety, I think. This was a very extreme form of anxiety in which I was questioning everything I know since like age 4; I think it stems from a desire to study math for a long time – have it potentially be my job – and this pressure led to me wanting to understand everything, leading me to seriously overthink every single step, things I'd just take for granted and pay no attention to in the past. some examples of overthinking are the equivalent equations thing; why is x – (-y) = x+y. like yes, it's because of the axioms, but it also happens to coincide exactly with the "real world's" physical understanding of negative * negative – people say math was created to model the physical world, but – – = + isn't a way to model the physical world; it comes from the axioms, which is its own mathematical world; you get the gist – i'd get these random thoughts.
Now my anxiety stems from things that are logically sound but don't make intuitive sense. (This is an improvement!) Let me explain: the proof for the rank nullity theorem makes sense. If we have $T : V \mapsto W$, we can extend the basis of $null(T)$ to create a basis for $V$. Then, we apply $T$ to both sides of an arbitrary vector, and the basis vectors in $null(T)$ go to $0 \in W$ as the rest go to the nonzero parts of $W$, loosely speaking. And then so we have the rank-nullity theorem, which was possible because our basis for $V$ was split between the union of $null(T)$ and other stuff that corresponds to the basis vectors of $Im(T)$. This intuitively makes sense – every vector in v must go somewhere when $T$ is acted upon it and our choice of basis demonstrates the split very clearly – and so I follow it.
(Before I go on, let me explain one anxiety I had: this proof relied on our choice of basis for $V$. I know this isn't true but how do we know there's not a bbasis / counterexample out there that proves the rank-nullity theorem false. like yes, the rank-nullity theorem is true, but could it also not be true. of course not, but this is the anxiety I'm talking about)
But then we have the theorem: a map to a smaller dimensional space is not injective, and what's frequently used is the rank-nullity theorem.
dim $null(T)$ = dim $V$ – dim $range(T)$ $\geq $ dim $V$ – dim $W > 0$.
Yes, I completely accept the rank-nullity theorem, but my brain cannot go around accepting this, despite the fact I completely follow every steps because I don't feel an intuition. It feels like we're just rearranging the rank nullity theorem and just doing some substitution. There's no physical mapping in this proof the way there is for rank-nullity, and I have no intuition I know I $should$ accept this since it's completely logically sound but I just can't (because of my anxiety). Also, the rank-nullity theorem relied on our choice of basis, but here it doesn't, so I'm just like: we can use this?
As I begin my mathematics journey, I think I just need to accept proofs that are logically sound but are weird to me. In a sense, my brain is very underdeveloped, so ofc everything feels weird, since it's new. like, logically, you can't have a false and true statement (e.gl rank nullity anxiety I voiced in the parentheses a few paragraphs up).
Anyway, I just wanted everyone's thoughts
Edit: Figured out what exactly I'm anxious about: I think the main source of anxiety I have is that the choice of basis was specific: we needed it to include the basis vectors of $Null(T)$. Is there a way to think about this for an arbitrary basis? I know $T(v)$ must go to 0 or not, and there's dimensions for each of these spaces – hence whatever is "outputted" has "dimension" $dim(NullT) + dim(ImT)$ but how do I relate that to dimV for an arbitrary basis for V? I put "outputted" in quotes since the dimension of the Nullity isn't part of the dimension of the output. The proof makes complete sense but I have anxiety that this only works for this specific basis because I can't imagine how the proof would work if we had a different basis. So I guess the only way to relieve my anxiety is to essentially prove the rank-nullity theorem without that specific basis or to be convinced that if it holds for that specific basis, it holds for all
Best Answer
To help with the intuition behind that proof, the rank is the size (In terms of dimension) of the image of the space. Whatever dimensions of the input space don't go into the image need to be dumped to 0, thus are in the null space.
The key concept here is that the dimension of a space is invariant, no matter the choice of basis, all bases have the same cardinality. Once you have that and you know a linear transformation is uniquely defined by what it does to the basis elements, the rest of the intuition follows.
The intuition of the proof of larger dimension to smaller dimension is not injective: Take any basis you want of the larger set. The image of each element under that map CANNOT be linearly independent, or we would have a linearly independent set in our codomain that has more elements than the dimension! So since the rank is the dimension of the image, we must have the rest be in the nullity, which means we have a nullity of dimension larger than 0. Thus there's a nonzero vector that goes to 0. But 0 also goes to 0, so it's not injective.
Does that help? Questioning your intuition on everything in math isn't a bad thing, especially if you go into a field like constructive mathematics where we lose very common things like the law of the excluded middle.