Is using Linear Algebra this way valid? When can we convert one question into a Linear Algebra question? Before I start, I am NOT trying to prove Fermat's Last Theorem using linear algebra, I am just using it as an example. Anyways...
Given the form of FLT,
$$x^n + y^n = z^n$$
and then some minor algebraic manipulation,
$$x^n + y^n - z^n=0$$
Find vectors $u$ and $v$ that generate a dot product equal to the left side, like
$$u = <x, y, z>$$
$$v = <x^{n-1}, y^{n-1}, -z^{n-1}>$$
Then it is clear that 
$$u \cdot v = x^n + y^n - z^n = 0$$
So now, given this is valid (I certainly have my doubts), we have transformed FLT to say, given $x,y,x,n \in\mathbb{N}$ and $n > 2$, $u$ and $v$ cannot be orthogonal (their dot product cannot be $0$). 
Since both vectors are in $\mathbb{R}^3$, they are orthogonal to their cross product that I will call $u \times v = w$. If $u$ is to be orthogonal to $v$, which are both orthogonal to $w$, then all 3 have ought to form an orthogonal basis. Thus we can fix a matrix
$$\begin{bmatrix}
v_1 & u_1 & w_1\\ 
v_2 & u_2 & w_2\\ 
v_3 & u_3 & w_3
\end{bmatrix}$$
and ask "is this matrix ever orthogonal for $x,y,z,n \in \mathbb{N}$?" Which also asks the question, when is this 
$$det\begin{bmatrix}
v_1 & u_1 & w_1\\ 
v_2 & u_2 & w_2\\ 
v_3 & u_3 & w_3
\end{bmatrix}=\pm1$$
ever true? Amongst other things that have ought to be true for a matrix to be orthogonal, so we could do something like apply Sylvester's Theorem. Nonetheless, it doesn't seem like we have asked a more "simplified" question equal to FLT. It seems more so that we have converted the question to terms of linear algebra, and so we get all the great algorithms that linear algebra is equipped with, so it may be helpful. I am not sure if this is valid at all, though. I am not sure exactly what is wrong with saying the dot product of 2 vectors is equivalent to the statement of FLT, but I am sure there is something preventing me from doing so. More or less I am asking for more intuition on when I can (and cannot) attempt to convert a problem into terms of linear algebra. 
 A: 
Nonetheless, it doesn't seem like we have asked a more "simplified" question equal to FLT. It seems more so that we have converted the question to terms of linear algebra, and so we get all the great algorithms that linear algebra is equipped with, so it may be helpful.

This is exactly right. Transforming a problem from one subfield of mathematics to another is valid as long as all steps in your transformation are. The interesting question is whether it is helpful. There are some spectacular examples of situations where reformulating things in terms of linear algebra saves the day in an otherwise hopeless looking situation. I'll give two below. Then there are other situations where the translation just transforms one hopeless problem into another one without any real progress being made. I think you are asking for intuition about how to tell the two apart? I think I can say a little bit about that. First the examples.
Has anybody ever told you how to solve differential equations of the form $y''(t) = ay'(t) + by(t) + c$, with initial conditions $y'(0) = p$, $y(0) = q$? If not, please find someone to explain it to you because it is a really great story. But either way, for now let's pretend we don't know it and approach the thing naively.
We start with making a sketch. We know the value of $y$ at $t = 0$ (it is $q$) and since we know that $y'$ is $p$ at $t = 0$ we get some sense of the direction in which the function is moving and guess that at $t =1$, the value of $y$ will be roughly $q + p$. Also using the equation we get some guess of what $y'$ will roughly be at $t  = 1$ and we can use that to get a rough idea where the function will be at $t = 2$ etc etc etc. In the end we end up with a very sloppy, cubist approximation of the graph of the true function $y$ which will nevertheless give us an idea what kind of behaviour we can expect for the true $y$ we are looking for.
Long story short: it is a complete horror-show. Some sine-wave whose amplitude and frequency themselves fluctuate sine-like over time. Surely there is no way that the solution to this equation fits neatly into our little brains, let alone that we will ever find it... But even if we do we can be 100% absolutely sure that this function will not be linear, so linear algebra seems useless here, right?
WRONG: by translating everything into linear algebra in the right way it turns out that we can get a very short and understandable expression for $y$ that fits into our brains perfectly. Truly a miracle of exactly the kind you were referring to when talking about 'great algorithms of linear algebra'.
Ok, a miracle, but I'm not gonna type it out because it is long and well known. Here is another example which I will describe in a bit more detail. 

Someone asked me why the quotients of successive Fibonacci numbers converge to the golden ratio. This does not sound like a linear algebra question. I mean of the four operations addition, subtraction, multiplication and division (i.e. taking ratios) the last one seems the one that is least related to linear algebra.
Still, linear algebra is useful here, as a way to simplify things after which other types of mathematics can take over again. The key insight here is that the set of all sequences that satisfy the Fibonacci recurrence is a vector space. Take any linear combination of two such sequences and the resulting sequence (of real or complex numbers) satisfies the same recurrence. Even more beautiful: even though the members of these space are sequences and hence look like infinitely long vectors, the space is only two dimensional! After all: each sequence is determined by only its first two terms. So the mysterious space of all these sequences is no more complicated than $\mathbb{R}^2$! (Or $\mathbb{C}^2$ if you allow complex numbers.) 
Not only is this reassuring on a psychological level, it also means (as you say) that we can use all the great algorithms of linear algebra. Here I want to use one of the most basic ones (pun intended): write everything with respect to an as-simple-as-possible basis and magically the whole world looks simpler too! 
We know from linear algebra that pretty much any two sequences would span the space. So if we can take two that look really simple, that is bound to help us. What would simple mean in this case? Remembering that our original motivation was understanding the ratio between successive terms in a sequence, the very very most helpful thing that could happen to us is if we could find two sequences where this ratio stays constant. That is: sequences of the form $1, x, x^2, x^3, x^4, ...$. 
Is it possible for such a sequence to satisfy the Fibonacci recurrence? This is not a question of linear algebra, but we can still solve it. (Using tools from quadratic algebra if that is even a word.) We conclude that, yes, there are exactly two such $x$: the golden ratio $\phi$ and the number $\epsilon := 1 - \phi$ which I denote by $\epsilon$ to stress that it is small, by which I mean concretely that $|\epsilon| < 1$. 
Now from the point of view of the original question we might think 'o.k., nice that there are Fibonacci-like sequences for which we can compute the ratio, but these ratios are not even the same in these two examples. What does that tell us about the actual Fibonacci sequence?' 
However from the perspective of linear algebra the natural reaction is something different: 'Yay! Two solutions!' That means that the sequences $1, \phi, \phi^2, \phi^3, ...$ and $1, \epsilon, \epsilon^2, \epsilon^3, ...$ form a basis of the space of all Fibonacci-like sequences and hence there are some real scalars $a$ and $b$ such that the $n$th Fibonacci number can be computed as:
$$F_n = a \phi^n + b \epsilon^n$$
You can compute $a$ and $b$ if you want but this formula already gives you the answer: if $n$ gets bigger so does $a\phi^n$ while $b\epsilon^n$ dies out. Some reasoning from analysis (so not linear algebra) then helps you establish that the limit of the ratios between $F_{n + 1}$ and $F_n$ then equals the limit of the ratios of $a \phi^{n+1}$ and $a \phi^n$, thus $\phi$.

O.k., so here are two highly successful examples of solving things by translating everything to linear algebra, at least as a step in a bigger process. There are also situations where this doesn't seem helpful at all. What is the difference? What do the two successful applications have in common that we can use as some sort of intuition when else linear algebra might be successful?
I don't have a full answer to that and would like to hear other people's thoughts, but here is one thing:

In both cases we could recognize something as an object from the world of linear algebra, and this observation captured really some of the structure of the problem.

In the Fibonacci case the crucial insight was recognizing the (bigger) set of all sequences satisfying the recurrence as a linear space, clearly an object from the world of linear algebra. It also captures a large part of the problem because it talks about the recurrence and the recurrence pretty much determines everything there is to know about the Fibonacci sequence.
In the differential equation case we can similarly note that the space of all functions is a linear space. This is crucial in the rest of the argument that I didn't type, but it does not by itself capture much of the structure of the problem. After all 'being a function' is not very impressive given how much functions float around in mathematics. So we need a second insight, and that is that, on the vector space of functions, differentiation is a linear map! And consequently, rewriting the equation as $y'' - ay' - by = c$, the map that sends function (vector) $y$ to the left hand side of our differential equation is a linear map as well. Now we have linear algebra objects capturing a large part of our problem!
Finally in your FLT example you certainly (and cleverly) recognized an object from linear algebra and hence fulfilled the first part of my rule of thumb above. However when it comes to the second part, i.e. if the inner product really captures a large part of the structure of the problem, I have my doubt.
