a 2D change of basis: a non-orthonormal basis

suppose we work a slightly different problem. suppose, as before, that we have new basis vectors rotated $30{}^{\circ}$ CW from the old ones, but suppose the new x basis vector is of length 2, and the new y basis vector is of length 1/2. that is, the new basis vectors are given as

$(\sqrt{3},\ -1)$

$(\frac{1}{4},\ \frac{\sqrt{3}}{4})$

here’s what the new basis looks like wrt the old one:

again, we arrange the basis vectors in rows to get a new attitude matrix A:

$A = \left(\begin{array}{cc} \sqrt{3}& -1\\ \frac{1}{4}& \frac{\sqrt{3}}{4}\end{array}\right)$

because the new basis vectors are not of length 1, this is not an orthogonal matrix. (i hate to remind you of this, but i should. because we changed the lengths but not the directions, the new basis vectors are orthogonal to each other, but not orthonormal. i know the terminology is awkward, but it’s also universal. the matrix is orthogonal iff the vectors are an orthonormal basis; orthogonal vectors do not suffice for the matrix to be orthogonal)

we still have that the transition matrix maps new components to old,

$z = P \ y$

and that the transition matrix is the transpose of the attitude matrix,

$P = A^T$.

and we want y, so we use the same equation as for a rotation…

$y = P^{-1}\ z = A^{-T}\ z$.

we no longer have $A^{-T} = A$, but that’s ok. we just compute $A^{-T}$:

$A^{-T} = \left(\begin{array}{cc} \frac{\sqrt{3}}{4}&-\frac{1}{4}\\ 1&\sqrt{3}\end{array}\right)$

and apply it to z:

$\left(\begin{array}{c} \sqrt{2}\\ \sqrt{2}\end{array}\right)$

getting the vector

$(\frac{-1+\sqrt{3}}{2 \sqrt{2}},\ \sqrt{2}+\sqrt{6})$

we draw z and y:

our new (red) vector y is way over against the y axis. what’s happened? our new x basis vector is twice as long, so the new x component is half-size; similarly, the new y component is twice as big. the outcome is that our new vector y has been compressed against the y axis and streched along it.

can we check this somehow? well, we can write it out including the basis vectors explicitly. let $e_1$ and $e_2$ be the original basis vectors.
that the original components of z are

$(\sqrt{2},\ \sqrt{2})$

means precisely that z is

$z = \sqrt{2} \ e_1+\sqrt{2} \ e_2$.

similarly, our new basis vectors b1 and b2 are

$b1 = \sqrt{3} \ e_1- \ e_2$

$b2 = \frac{e_1}{4}+\ \frac{\sqrt{3}\ e_2}{4}$

incidentally, this is where the attitude matrix A comes from. if we write symbolic columns of e’s and b’s, the b vectors are given by

$\left(\begin{array}{c} b_1\\ b_2\end{array}\right)$ = $\left(\begin{array}{cc} \sqrt{3}& -1\\ \frac{1}{4}& \frac{\sqrt{3}}{4}\end{array}\right)$ x $\left(\begin{array}{c} e_1\\ e_2\end{array}\right)$

i.e.

$\left(\begin{array}{c} b_1\\ b_2\end{array}\right)$ = A $\left(\begin{array}{c} e_1\\ e_2\end{array}\right)$

(strictly, i suppose those are not matrix equations because the b’s and and e’s are not numbers, but it’s so convenient to write the equations that way that after a while we get used to it.)

we want to find new components (Y1, Y2) of y where y is of the form

$\text{Y1} \ b_1+\text{Y2} \ b_2$

but y and z are the same vector. it’s the components that are different. we set y = z.

$\sqrt{2} \ e_1+\sqrt{2} \ e_2==\text{Y1} \ b_1+\text{Y2} \ b_2$

replace the b’s by their definitions wrt the e’s.

$\sqrt{2} \ e_1+\sqrt{2} \ e_2 = \text{Y1} \sqrt{3} \ e_1-\ e_2+\text{Y2} \ \frac{e_1}{4}+\frac{\sqrt{3} \ e_2}{4}$

then i split that into two equations.

$\sqrt{2} \ e_1==\sqrt{3} \ \text{Y1} \ e_1+\frac{\text{Y2} \ e_1}{4}$

$\sqrt{2} \ e_2==-\text{Y1} \ e_2+\frac{1}{4} \sqrt{3} \ \text{Y2} \ e_2$

i solve the two equations for Y1 and Y2, getting

$(Y1,Y2) = (\frac{-1+\sqrt{3}}{2 \sqrt{2}},\ \sqrt{2}+\sqrt{6})$,

finally, why do we care about the transition matrix P and the attitude matrix A? we care about P because in the eigendecomposition

$D = P^{-1}\ X \ P$

which diagonalizes X, we have to know that P is a transition matrix. further, in the matrix change-of-basis equation,

$Y = Q^{-1}\ X \ P$,

both P and Q are transition matrices. (in particular, $Q^{-1}$ is not a transition matrix.)
and i personally care about A because very often when i need a new basis, i know what i need it to be – which means i can write the attitude matrix right away. more to the point, very often when i need a new basis, what i’ve been given is, in fact, the equations

$\left(\begin{array}{c} b_1\\ b_2\end{array}\right)$ = A $\left(\begin{array}{c} e_1\\ e_2\end{array}\right)$

so i’m used to reading off the elements of A.