Co-ordinate transformation in $ 2$ dimensions

Question

Consider a $2$ dimensional $x-y$ co-ordinate system as given below Figure $1$, with unit vectors ${\hat{i}}$ and ${\hat{j}}$ respectively

Figure 1

Now I would like to construct a different co-ordinate axes $x'- y'$ by relating it with the original co-ordinate axis as follows

\begin{equation*} \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \end{equation*}

Such that,

\begin{equation*} x' = x + 2y \end{equation*}

\begin{equation*} y' = 3x + 4y \end{equation*}

Setting $x' = 0$ we can get the equation of the line that gives $y'$ axis i.e.

\begin{equation*} x = -2y \end{equation*}

Setting $y' = 0$ we can get the equation of the line that gives $x'$ axis i.e.

\begin{equation*} x = -\frac{4}{3}y \end{equation*}

The plot for $x'$ and $y'$ axis are super imposed on the $x-y$ axis pictured above shown below in Figure $2$

Figure 2

However if we try to plot the unit vector with respect to the transformation matrix we have,

\begin{equation*} \begin{pmatrix} x' \\ y' \end{pmatrix}=\begin{pmatrix} 1 \\ 3 \end{pmatrix}x + \begin{pmatrix} 2 \\ 4 \end{pmatrix}y \end{equation*}

\begin{equation*} \begin{pmatrix} 1 \\ 3 \end{pmatrix} and \begin{pmatrix} 2 \\ 4\end{pmatrix} the~new~basis~vectors~for~i~ and~ j~ respectively. \end{equation*}

If we plot this new basis vectors(green lines) on figure $2$ as shown below in Figure $3$,

Figure 3

We can now see that the new basis vectors are not parallel to the axes $x' - y'$. Can someone please explain if I have mixed up different concepts. Also, this disagreement does not occur when the transformation matrix chosen is orthogonal.

Thank you.

EDIT $1:$

from the new approach suggested I have plotted the $x-y$ coordinates and the $x'-y'$ coordinates with the basis vector in Figure $4$ below

Figure 4

Answer 1

The way you have transformed the x-y axes to the x'-y'; system is wrong. To do this, x axis in the x'-y' system will be obtained by doing this

since on x axis, y=0, we can substitute this in your transformation ($x'= x+2y$ and $y' = 3x+4y$)

So we get

$x'=x$ and $y'=3x$

So form this your x axes in x'-y' will be $y'=3x'$ and similarly y axis will be $y'=2x'$ And these actually are paralel to $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 4 \end{pmatrix}$

Answer 2

The vectors $$\begin{pmatrix} 1 \\ 3 \end{pmatrix} \quad \text{and}\quad \begin{pmatrix} 2 \\ 4\end{pmatrix}$$ are your original basis vectors (the green vectors in your first figure) expressed in $x',y'$ coordinates.

This is because you have defined your transformation to get the $x',y'$ coordinates this way for the original basis vector $\mathbf i$: $$\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\mathbf i =\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} =\begin{pmatrix} 1 \\ 3 \end{pmatrix},$$ and similarly the $x',y'$ coordinates of the original basis vector $\mathbf j$ come out to $\begin{pmatrix} 2 \\ 4\end{pmatrix}.$

In Figure 3 you take the $x',y'$ coordinates of your original basis vectors and plot them as if they were $x,y$ coordinates. As a result you get something that does not correspond geometrically to any of the vectors you are likely to be interested in--either the original basis or the new basis.

If you want to know how to plot the new basis vectors, you have to convert the vectors with $x',y'$ coordinates $\begin{pmatrix} x' \\ y' \end{pmatrix}=\begin{pmatrix} 1 \\ 0\end{pmatrix}$ and $\begin{pmatrix} x' \\ y' \end{pmatrix}=\begin{pmatrix} 0 \\ 1\end{pmatrix}$ to their $x,y$ coordinates. One way to do this is to find the inverse of your transformation matrix. That's the matrix that lets you do this:

$$\begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x' \\ y' \end{pmatrix}.$$

In fact, any method of finding the $x,y$ coordinates of the new basis vectors algebraically will be equivalent to finding this inverse matrix, because the columns of that matrix are the $x,y$ coordinates of those vectors.

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Update:

In Figure 4 you have illustrated something completely different that you can do with a transformation matrix, that is, you transform the entire plane. Rather than using a new basis to assign new coordinates to points (without actually changing the shape of the things described by those coordinates), you use the old basis to plot the new points to which your transformation maps any figure in the plane.

In particular, the green figure, which used to be two equal segments perpendicular to each other, now has segments of unequal length making an acute angle; moreover, each segment is longer than it originally was. Other figures in the plane also will be distorted.

It may be confusing to see that people use the symbols $x'$ and $y'$ in exercises like this as well as when making a change in basis. You have to read the context in which the symbols are used in order to figure out what they mean, and not assume that $x'$ in one discussion means the same thing it did in another discussion several pages earlier.

By the way, we usually reserve the word "rotation" for cases where everything we're looking at changes its orientation by the same angle and nothing gets larger or smaller. When we talk of rotating a vector, we have in mind a transformation like that, except that we might not be looking at any other vectors to see that they rotate in the same way. You have a transformation here where nearly every line through the origin gets rotated in a different way and every line gets dilated in a different way from most others. Using the word "rotation" to describe it will lead people to misunderstand what you're talking about.