Question

Coordinates with respect to basis? Let B_1={u_1,u_2,u_3} and B_2 = {v_1,v_2,v_3} be the bases for R^3 which u_1=(-3,0,-3) u_2=(-3,2,-1) u_3=(1,6,-1) v_1=(-6,-6,0) v_2=(-2, -6, 4) and v_3=(-2,-3,7). Find the transition matrix P_B1---> _B2

Answer 1

Don't know how to approach this question

Answer 2

hmm...im trying to see how i would do this in my mind, i dont want to look in my book....

Answer 3

alright. lets say we have a vector x (a,b,c) with respect to the first basis B_1.
That means that the vector in the standard basis (1,0,0) (0,1,0) (0,0,1) would be:
\[\vec v = a\vec u_1+b\vec u_2+c\vec u_33\]
If you want a matrix equation to simulate that, you would make a matrix out of the columns of B_1. So that means so far we have:
\[A\vec x = \vec v\]
Now that we have the coordinates with respect to the standard basis, we can convert them to the other basis.

Answer 4

Okay I sort of get what you're saying

Answer 5

To convert them to the other basis, we would set up a matrix (call it B) with columns using the second basis vectors (B_2), and look for a solution to:

\[B\vec y = \vec v\]

the answer we are looking for is the vector y, which will tell us what linear combination of the B_2 vectors we need. Since the columns of B form a basis, it is invertible. So multiplying by its inverse we get:
\[B\vec y = \vec v \iff \vec y = B^{-1}\vec v \iff \vec y = B^{-1}A\vec x \]
Thus the change of basis matrix would be:
\[B^{-1}A\]
where A is formed by making the first basis vectors into columns, and B^-1 is formed by getting the inverse of the matrix formed from by making the second basis vectors into columns.

Answer 6

Now, im going to test this out...i dont know if its correct <.< it makes sense in my head lolol

Answer 7

okay just for the first part though, I don't know how you'd find a b and c if you don't know v

Answer 8

if we were interested in actually converting a vector from one basis to the other, we would need to know a, b, and c. Since this question just asks for the matrix that will do the job, we dont need that info.
I'll do calculations on paper and post in a sec.

Answer 9

would you happen to have the answer to this problem for verification?

Answer 10

Man I'm so confused but maybe seeing it will make more sense. Yes I do, [0.75 0.75 1/12; -0.75 -17/12 -17/12; 0 2/3 2/3]

Answer 11

yeah, i think im on the right track, as soon as im done with the tedious calculations i'll post lol

Answer 12

Of course you'd be.. I wish I was a math genius so I wouldn't have to study so much. Thanks again

Answer 13

ok, in the first picture, thats just putting the first basis vectors into a matrix, and taking the second basis vectors, putting them in matrix, and finding the inverse of the matrix (which could be easy or not easy, unfortunately)

The second picture is just multiplying the two matrices out, which gives the change of basis matrix.

Im going to spend a little time thinking of a more efficient method.

Answer 14

Okay well is that the standard way of finding a transition matrix from one basis to another?

Answer 15

i dont think it is. Its just something that came to mind lol >.< i just came up with a better way to do it.

You want to create the augmented matrix

[ B | A ]

and row reduce it.

You will end up with:

[ I | P ]

where P is your change of basis matrix from B_1 (represented by A), to B_2 (represented by B)

Answer 16

I think that's the method they have in my book but it's worded so weirdly didn't understand it until you restated it

Answer 17

oh okay I get it now thanks for your help

Answer 18

but just wondering does coordinates have any relation to orthogonal projections?

Answer 19

We were learning that then bam we went into this and I got so lost

Answer 20

if your basis happens to be an orthogonal basis (better yes, an orthonormal basis), then finding out its coordinates is really really easy.

Answer 21

yet*

Answer 22

Oh okay by using that w dot with all the vectors method?

Answer 23

lets say you have a vector x in the standard basis, and you want its coordinates in the new basis:
\[B =\left\{ q_1, q_2,\ldots , q_n\right\}\]
where B is an orthogonal basis.
Because B is a basis, we know that the vector x is some linear combination of those q's:
\[\vec x = c_1q_1+c_2q_2+c_3q_3+\ldots +c_nq_n\]

Now, lets say i wanted the first coordinate, c1. To get rid of all the other junk, i do a dot product (sometimes called inner product) with q1:
\[x*q_1 = c_1q_1*q_1+c_2q_2*q_1+\ldots +c_nq_n*q_1\]
Because its an orthogonal basis, each q is perpendicular to every other q, leaving their dot products 0:
\[x*q_1 = c_1q_1*q_1+0+0+\ldots+0\]

Answer 24

Thats why orthogonal basis are nice, dot products are generally easier to compute.

Answer 25

how do you check to make sure b is an orthogonal basis, that question I asked earlier apparently it wasn't? do you dot each vector together and you should get 0?

Answer 26

yes, thats exactly how you would check :)

Answer 27

so if it's not then you'd have to make a matrix and row reduce

Answer 28

and find the unique solutions?

Answer 29

yep. thats correct

Answer 30

oh snap that make sense, haha thanks a lot!!