Clarifying my understanding of Linear Algebra I don't think I fully understand matrix and I need help
E.g. Consider $$A\hat x=\hat b$$ where
$$A=\begin{bmatrix}
        1 & -2 \\
        3 & -4  \\
        \end{bmatrix},\hat x=\begin{pmatrix}
        1  \\
        2   \\
        \end{pmatrix}, \hat b=\begin{pmatrix}
        -3  \\
        -5   \\
        \end{pmatrix}$$
$\hat x$ is based on the row space of A right while $\hat b$ is based on the column space of A right? If this is true, does that mean that I can represent $\hat x$ with the linear combination of the vectors in the row space of A? 
Like this $$\begin{pmatrix}
        1  \\
        2   \\
        \end{pmatrix}=-5\begin{pmatrix}
        1  \\
        -2   \\
        \end{pmatrix}+2\begin{pmatrix}
        3  \\
        -4   \\
        \end{pmatrix}$$
Likewise for $\hat b$, can I represent it with the linear combination of the vectors in the column space of A?(This one I know is definitely true)
$$\begin{pmatrix}
        -3  \\
        -5   \\
        \end{pmatrix}=1\begin{pmatrix}
        1  \\
        3   \\
        \end{pmatrix}+2\begin{pmatrix}
        -2  \\
        -4   \\
        \end{pmatrix}$$
These linear combinations are really just similar to how we assign a reading on each of the axes that vector is in. 
But since the value of the vector is dependent in each of these coordinates, does it have anything that allows it to be independent of coordinates? Will that be its magnitude? $\hat x$ and $\hat b$ should be describing the same vector since the matrix only just transforms it from one coordinate to another. But the magnitude of $\hat x$ is different of that $\hat b$. Am I missing something out?
Edited: My reason(my own interpretation and I don't know whether this is true or not) why $\hat x$ is based the row space of A is because $\hat x =A^{-1}\hat b$. The row space of A is the same as the column space of $A^{-1}$ other than the scalar factor. 
If $\hat x $ is not based on the row space of A, then its basis is based on what?
 A: Your last equation
$$\begin{pmatrix}
        -3  \\
        -5   \\
        \end{pmatrix}=1\begin{pmatrix}
        1  \\
        3   \\
        \end{pmatrix}+2\begin{pmatrix}
        -2  \\
        -4   \\
        \end{pmatrix}$$
is right, and, indeed, it shows that $b$ (column) is a linear combination of the columns of $A$. Hence, in $A x = b$ , it's true that $b$ belongs to the column space of $A$. 
The other assertion ($x$ is based on the row space of $A$) makes no sense to me. $x$ is arbitrary, and does not need to belong to any "space" of $A$. Take for example $A$ as the all-zeroes matrix and $x=(1, 2)´$. 
What we can say about the row space of $A$ is that if we write $c=xA$ (with $c$ and $x$ being row vectors), then, yes, $c$ belongs to the row space of $A$.
Concerning the last paragraph, I don't fully understand it. It's true that matrix multiplication can be thought as a change of coordinates. But the vector will not have the same lengths in different coordinates (unless the transform is unitary, say a rotation).
A: Sorry I may have misunderstood your question but I'll give it a shot.
if I understand correctly, you're asking is there an method of representing a vector independent from it's co-ordinates?
i mean we can describe a vector v as the product of it's magnitude in the direction of its unit vector $$\hat{v} = v/\|v\|$$ so $\textbf{v}=\hat{v}\|v\|$ 
if that helps
i mean from your example $\hat{x}$ is the solution to your system $A\hat{x}=\hat{b}$ what i would argue was originally asked was
find the solution which satisfies $x-2y=-3$ and $3x-4y=-5$ which we can represent as the matrix A and then test to see if there exists a unique/infinite/no solutions.
in this case $A\sim I$ So it has Rank(A)=2, which is the same as the Rank(M) of the augmented Matrix M and given that Rank(A)=Rank(M)=n=2 where n is the number of equations you're dealing with there exists a unique solution. which you've found to be x=1 y=2.
Any linear combination of the vectors which make up the matrix A will also have the same solution set.
Does that help? (if I've gotten something wrong please someone correct me...)
oh i'll add something just incase that helps.
A vector space is made up by the basis which Span's that space. then all vectors in that space is made up as linear combinations of that basis. so in Euclidean Space we use i,j,k..so when we write a vector u=1i+2j+3k u is a linear combination of those 3 individual vectors..
