its my first question here. I'm re-self-studying linear algebra from different sources and one of them is Linear Algebra and Its applications by g.strang 4th ed. . While i have studied a bunch of material i still don't grasp some basics and i really struggle. Page 4-5 gives two approaches for the geometry of linear equations:

my problem is the second graph. I get it, the operations, how we use them as a transformation to create the vectors and that we have to guess a solution to get to the (1,5). I also get that he is trying to make a point for later on, on Gauss elimination. But how/why can he take the coefficients of each equation for example for the y part \begin{bmatrix}-1\\1\end{bmatrix} and use it as x,y coordinates to map it to the graph? How can -1y and 1y coefficients turn into a (x,y) / ( -1,1) vector? It looks like to me that we got two coeffients and turned them from y,y to x,y.I don't get how that concept works. Do i make sense? I've searched a lot but i may lack knowledge of the correct terms, so even if the question is novice level/not worth answering please provide me with some keywords.And of course correct me for any mistakes. Much appreciated


2 Answers 2


Confusion can arise by calling the unknowns $x$ and $y$. In the picture on the left, they are actual $x,y$ coordinates of a point with geometrical significance. In the picture on the right, $x,y$ are called $x$ and $y$, but they're not connected to $x$ and $y$ coordinates anymore. I think it's conceptually better to separate the left picture and right picture completely, and when trying to understand the picture on the right, don't even call the unknowns $x$ and $y$, because they're not going to be $x$ and $y$ coordinates of anything in our second picture.

Take vectors $$u=\begin{pmatrix} p \\ q \end{pmatrix}$$ $$v=\begin{pmatrix}r\\s\end{pmatrix},$$ $$w=\begin{pmatrix} \alpha \\ \beta\end{pmatrix}.$$

Suppose we want to know how to express $w$ as a linear combination of $u,v$. We take unknown constants $a,b$ and ask: How do we achieve $$au+bv=w?$$ This is coordinate-wise equality, so we write out what this means. We want $$au+bv = a\begin{pmatrix} p \\ q \end{pmatrix}+b\begin{pmatrix}r\\ s\end{pmatrix} = \begin{pmatrix}ap+br \\ aq+bs\end{pmatrix},$$ and we want this to be equal to $\begin{pmatrix} \alpha \\ \beta\end{pmatrix}$. So this is equivalent to solving the system of equations $$ap+br=\alpha$$ $$aq+bs=\beta.$$

Instead of calling the coefficients $a,b$, if we call them $x,y$, we can see the relationship you're asking about. However, this translation is a little difficult precisely because we have pre-existing interpretations of what $x,y$ should be. We think of $p,s,\beta$ as the $y$-coordinates of the vectors if we plot them in $\mathbb{R}^2$ (which is what the picture on the right is doing), but this has no connection to the fact that the variables are called $x,y$. The fact that $x,y$ are actually $x,y$ coordinates in some meaningful plot is reflected in the plot on the left in your link.

In general, if ${\bf X}$ is a matrix with columns $x_1,\ldots, x_p$ and if $\beta_1,\ldots, \beta_p$ are numbers, the linear combination $$\sum_{i=1}^p \beta_i x_i$$ is exactly the matrix product $${\bf X}\begin{pmatrix} \beta_1 \\ \beta_2 \\ \vdots \\ \beta_p\end{pmatrix}.$$ If we denote $$\begin{pmatrix} \beta_1 \\ \beta_2 \\ \vdots \\ \beta_p\end{pmatrix}$$ as $\beta$, then asking whether or not $${\bf X}\beta=w$$ has a solution is equivalent to asking whether $w$ is in the column space of ${\bf X}$, which is the span of the columns of ${\bf X}$

  • $\begingroup$ Thank you for your answer and explaing it further beyond. My brain still cannot fully digest it but i understand it, it's way more clear now. $\endgroup$
    – MatVe
    Mar 1 at 14:59
  • 1
    $\begingroup$ @MatVe - in a coordinate plane, the axes are named (x-axis, y-axis) by convention (i.e., that's how we typically name them), but those names don't really mean anything in an of themselves. It could just as well be (apple axis, orange axis) or (a-axis, b-axis) or (horizontal axis, vertical axis) or (x1-axis, x2-axis), it doesn't really matter as long as you have a way to distinguish the two. For the diagram on the right, the x & y variables in the equation have nothing to do with the coordinate axes (if you look closely, the axes aren't even labeled using x and y, unlike in the first diagram). $\endgroup$ Mar 1 at 23:54

The author suggests two methods of solving the linear equations. One is standard algebra class elimination, while the other looks at combining the vectors in some way to get to a goal vector.

$2x - y = 1$

$x + y = 5$

Above are the equations that we have.

He gets those column vectors with the properties of vector addition and scalar multiplication.

$x\begin{bmatrix}2\\1\end{bmatrix} + y \begin{bmatrix}-1\\1\end{bmatrix} = \begin{bmatrix}2x\\1x\end{bmatrix} + \begin{bmatrix}-y\\y\end{bmatrix} = \begin{bmatrix}2x-y\\x+y\end{bmatrix} = \begin{bmatrix}1\\5\end{bmatrix}$

Therefore the first element is $2x-y$ which corresponds with 1, and the second element is $x+y$ corresponding with 5.

That is how he got the coefficients.


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