1. ## Change of base

Hi,

I've just started Linear Algebra
I have one question i'd like some help with.

The line l $Oe_{1}e_{2$ is described by the equation $2x_{1} + 3x_{2} = 0$

Find an equation for the line in $Oe´_{1}e´_{2}$
if $~~\begin{cases} e´_{1} = \frac{1}{13}(5e_{1} + 12e_{2}) \\
e´_{2} =\frac{1}{13}(-12e_{1} + 5e_{2}) \end{cases}$

2. suppose we knew how to transform e'-coordinates into e-coordinates, by using some matrix A. it stands to reason that A^-1 would translate e-coordinates to e'-coordinates.

now the matrix that sends e'-coordinates to e-coordinates is very easy to write down. in e'-coordinates, e1' is (1,0). and for ANY 2x2 matrix A, A(1,0) is the first column of the matrix. similarly, in e'-coordinates, e2' is (0,1), and A(0,1) is the second column of the matrix A. so the first column of A is A(e1') = (5/13, 12/13) (in e-coordinates). and the second column of A is e2' in e-coordinates, or (-12/13,5/13) so A =

[5/13 -12/13]
[12/13 ..5/13]

by a standard computation, A^-1 =

[ 5/13 12/13]
[-12/13 5/13]

so in e' coordinates, e1 = (5/13,-12/13) = (5/13)e1' - (12/13)e2', while e2 = (12/13,5/13) = (12/13)e1' + (5/13)e2'.

in e'-coordinates the points (x1,x2) that lie on the line now have coordinates ((5/13)x1+(12/13)x2, (5/13)x2 - (12/13)x1)), so the equation for our line becomes:

2(5/13)x1 + 2(12/13)x2 + 3(5/13)x2 - 3(12/13)x1 = 0, or upon simplfying:

-2x1 + 3x2 = 0

The key to this question says it's $46x_{1} - 9x_{2}$

4. $2x_1+ 3x_2= 0$ is the same as saying $(x_1e_1+ x_2e_2)\cdot(2e_1+ 3e_2)= 0$. That is, that the line is perpendicular to the vector $2e_1+ 3e_2$

Given that $e_1'= \frac{5}{13}e_1+ \frac{12}{13}e_2$ and $e_2'= -\frac{12}{13}e_1+ \frac{5}{13}e_2$, then the matrix having those coefficients as columns,
$\begin{pmatrix}\frac{5}{13} & -\frac{12}{13} \\ \frac{12}{13} & \frac{5}{13}\end{pmatrix}$
will change a vector in $e_1'$ and $e_2'$ components to the same vector in $e_1$ and $e_2$ components. What we want to do is change the vector $2e_1+ 3e_2$ to $e_1'$ and $e_2'$ components. We can do that by using the inverse matrix which happens to be
$\begin{pmatrix}\frac{5}{13} & \frac{12}{13} \\ -\frac{12}{13} & \frac{5}{13}\end{pmatrix}$

So we are saying that our vector, in $x_1'e_1'+ x_2'e_2'$ components must be perpendicular to
$\begin{pmatrix}\frac{5}{13} & \frac{12}{13} \\ -\frac{12}{13} & \frac{5}{13}\end{pmatrix}\begin{pmatrix}2 \\ 3\end{pmatrix}= \begin{pmatrix}\frac{46}{13} \\ \frac{-9}{13}\end{pmatrix}$

That is, $\left(\frac{46}{13}e_1'- \frac{9}{13}e_2'\right)\cdot\left(x_1'e_1'+ x_2'e_1')= \frac{46}{13}x_1'- \frac{9}{13}x_2'= 0$

Now, multiply both sides of that equation by 13.

5. Essentially the same thing but slightly different point of view:
solve $e_1'= \frac{5}{13}e_1+ \frac{12}{13}e_2$, $e_2'= -\frac{12}{13}e_1+ \frac{5}{13}e_2$ for $e_1$ and $e_2$ in terms of $e_1'$ and $e_2'$. (That is equivalent to finding the inverse matrix above.) Once you have done that, replace $e_1$ and $e_2$ in $2e_1+ 3e_2$ by their expression in terms of $e_1'$ and $e_2'$ to find the vector our $x_1'e_1'+ x_2'e_2'$ must be perpendicular to.

6. Thanks