I'm proceeding through MIT OCW's 18.06 class in Linear Algebra, and I've reached a sticking point on the first lecture - I was wondering if someone can offer some clarification on a specific point for me.
My confusion is on the explanation behind 'how' you get away with creating a column vector out of the coefficients of the same variable in a system of equations, and then use those same coefficients to become 'motion' along a different coordinate axes.
For instance, in the system of equations below,
\begin{matrix} 1x && + && 2y && = && 2 \\ -3x && + && 4y && = && 5 \\ \end{matrix}
Professor Strang comes up with three separate column vectors. For x, it is: \begin{bmatrix} 1 \\ -3 \\ \end{bmatrix} and for y, it is:
\begin{bmatrix} 2 \\ 4 \\ \end{bmatrix} Similarly, he derives the (2,5) vector as the answer vector which we then take linear combinations of the previous two to determine a solution - I fully grasp all the mechanics of how this works, but DO NOT understand how he can use the 'x' vector (with components 1 & -3) to draw a vector of '1' unit in the x direction and '-3' units in the y direction using nothing but coefficients that came from the x variable! This is the crux of my confusion.
I would have an easier time understanding what was going on if the actual mechanics of creating the two vectors was to take the coefficients of each equation and put those into column vectors. For instance, if it were as follows:
X = \begin{bmatrix}1 \\ 2 \\ \end{bmatrix} and Y = \begin{bmatrix}-3 \\ 4\\ \end{bmatrix}
IF it were this way, it would make sense to me because it would correspond to the 'movement' each equation produced in each coordinate system - I also realize that this 'breaks' the idea of 'x' and 'y' because in this second example (in which I realize my understanding is wrong) I've arbitrarily assigned the 'labels' x and y to the separate vectors.
Can someone offer an explanation?
Thanks,
Luke
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$\begingroup$Without having worked through the course myself, this appears to be more a product of confusing notation than of a deeply rooted misunderstanding.
Instead, considering the following system:
$$ 1a + 2b = 2 \\ −3a + 4b = 5 $$
The goal of creating vectors here is that we want to be able to write the set of equations above as $aA + bB = C$ for some column vectors $A, B, C$. If you think about it in those terms, then it becomes more straightforward why you need to define:
$$ A=\begin{bmatrix}1 \\ -3\end{bmatrix} B=\begin{bmatrix}2 \\ 4\end{bmatrix} C=\begin{bmatrix}2 \\ 5\end{bmatrix} $$
Now we can re-write this system as: $aA + bB = C$.
$$ \begin{bmatrix}1a \\ -3a\end{bmatrix} + \begin{bmatrix}2b \\ 4b\end{bmatrix} = \begin{bmatrix}2 \\ 5\end{bmatrix} $$
It is not so much that these vectors represent vectors in the traditional plane, but rather as a concise way to represent the system of equations. Ultimately, people will begin to express these not as equations in vectors, but as augmented matrices, which nicely summarize the system of equations.
$$ \left[\begin{array}{cc|c} 1 & 2 & 2 \\ -3 & 4 & 5 \end{array}\right] $$
It is my opinion that suspending parallels to the $\mathbb{R}^2$ plane as soon as possible when studying linear algebra tends to be a worthwhile endeavour. It can be useful when drawing parallels, but ultimately hampers the ability to think more broadly about the concepts.
$\endgroup$ 6 $\begingroup$The following content is taken from the link. The content has been reproduced below, with slight changes to make it suitable for the answer (I could have simply mentioned the link, just to make sure the availability of answer in the absence link, the content has been reproduced).
An invertible system of two linear equations admits several geometric interpretations.
In the row picture (when the slider is in the left-most position), each equation is visualized as a line. The goal is to find the coordinates of the intersection point P.
In the column picture (when the slider is in the left-most position), each column of coefficients is a vector on a plane. The goal is to find a linear relationship among those vectors. It's easier to view the picture if you move the slider all the way to the right.
There are no such things as "absolute coordinates" in Linear Algebra! From the users perspective, the row and column pictures are simply different visual interpretation of the same geometric phenomenon: change of coordinates.
Figure 1 - Row Picture Visualisation
Figure 2 - Column Picture Visualisation
(Sorry, for hyperlinking the pictures, instead of embedding them in the answer. As my reputation is below 10, stackexchange.com did not allowed me to embed them in the answer. I will do my best for my next answer, plz adjust for now ...)
Thank you very much...
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