I'm not understanding what a(sub i)s means in linear algebra. I'm just jumping into linear algebra without many other mathematics experience (Algebra 1, 2, and Geometry)
Also, what is n-vector?
[Math] What does $a_i$ mean in linear algebra
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Related Solutions
I'm not going to try and answer all of your questions because it's really a very broad question. I will at least give you a start and then the best thing you can do is either take another course or open a book and learn some things, coming back to ask more specific questions along the way.
Surely you are familiar with equations like $5x = 6$ where we want to find all solutions $x$ where $x$ is in some field, say in this case the real numbers. In this case every nonzero element has an inverse so you can multiply by $1/5$ to get $x = 5/6$.
Although linear algebra isn't really just about solving equations, that's where it starts. It's called linear because we only want to solve equations that are linear in the unknown variable. The simplest case would be something like
$$ x + y = 4, 2x - y = -1 $$
Actually we can just write this in matrix form. If you remember how to multiply a matrix, then we can write this system as $Ax = b$:
$$ \begin{pmatrix} 1 & 1\\ 2 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4\\ -1 \end{pmatrix} $$
The reason why we multiply matrices is because we want to solve $Ax = b$ by multiplying by the inverse of $A$ to get $x = A^{-1}b$. Of course, not all matrices have inverses. So the set of all $n\times n$ matrices, with addition and multiplication is a ring but not a field. A ring is sort of like a field but now we remove the requirement where inverses exist for all nonzero elements. Also matrix multiplication is not commutative: $AB$ is not necessarily equal to $BA$.
In the above case $A$ does have an inverse, and you can multiply on the left by $A^{-1}$ (see if you can find it) to get the solution to this system of equations.
Thus we have found: multiplication of matrices helps us solve equations.
However, we are only beginning because finding the inverse of a matrix is tricky, so we study the different ways to represent matrices and calculate with matrices in order to more efficiently move them around. This is a bit vague but intentionally so since there is so much mathematics going on in the background which you need to learn.
Linear algebra is really about vector spaces. To appreciate the idea of a vector space you should first get some experience with abstraction by doing hundreds of problems. A vector space is just a set of elements together with addition and scalar multiplication that satisfy certain axioms. It turns out that matrices correspond to maps between vector spaces in a chosen basis of that space. This may not make too much sense to you now, but the important point is that putting matrices in different forms corresponds to changing the basis of the vector space in different ways.
The reason why we like to use vector spaces is because then we can concentrate on the algebraic properties of vector spaces without having to worry about specific numbers or equations, which then can be applied to all sorts of problems which have little do with solving equations.
The best thing you can do to understand linear algebra is to take a course/read a book and just start solving problems. It is impossible to really understand what it is about first and then practice doing it. The understanding comes with the practice.
Isomorphisms are defined in many different contexts; but, they all share a common thread.
Given two objects $G$ and $H$ (which are of the same type; maybe groups, or rings, or vector spaces... etc.), an isomorphism from $G$ to $H$ is a bijection $\phi:G\rightarrow H$ which, in some sense, respects the structure of the objects. In other words, they basically identify the two objects as actually being the same object, after renaming of the elements.
In the example that you mention (vector spaces), an isomorphism between $V$ and $W$ is a bijection $\phi:V\rightarrow W$ which respects scalar multiplication, in that $\phi(\alpha\vec{v})=\alpha\phi(\vec{v})$ for all $\vec{v}\in V$ and $\alpha\in K$, and also respects addition in that $\phi(\vec{v}+\vec{u})=\phi(\vec{v})+\phi(\vec{u})$ for all $\vec{v},\vec{u}\in V$. (Here, we've assumed that $V$ and $W$ are both vector spaces over the same base field $K$.)
Best Answer
The $a_i$s represent the elements in the vector. For example if you have $\begin{align} a &= \begin{bmatrix} a_{1} \\ a_{2} \\ \vdots \\ a_{n} \end{bmatrix} \end{align}$ where $a_1,...,a_n$ are in $\mathbb{R}$ then it could look something like $\begin{align} a &= \begin{bmatrix} 1 \\ 2 \\ \vdots \\ 10 \end{bmatrix} \end{align}$ so $a_1 = 1, a_2 = 2, a_n = 10$
And an $n$-vector is just saying that there are $n$ elements (essentially telling you the size of the matrix). Hope this helps!