suppose we work a slightly different problem. suppose, as before, that we have new basis vectors rotated 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

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:

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,

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

.

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

.

we no longer have , but that’s ok. we just compute :

and apply it to z:

getting the vector

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 and be the original basis vectors.

that the original components of z are

means precisely that z is

.

similarly, our new basis vectors b1 and b2 are

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

= x

i.e.

= A

(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

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

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

then i split that into two equations.

i solve the two equations for Y1 and Y2, getting

,

which is the same answer.

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

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

,

both P and Q are transition matrices. (in particular, 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

= A

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

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