Finding Matrix P for B & R Basis in R^2

[UrbanXrisis]

Let B be the basis of R^2 consisting of the vectors

$$\left(\begin{array}{c}3 & 1 \end{array}\right)$$ and $$\left(\begin{array}{c}-1 & 3 \end{array}\right)$$

Let R be the basis consisting of

$$\left(\begin{array}{c}2 & 3 \end{array}\right)$$ and $$\left(\begin{array}{c}1 & 2 \end{array}\right)$$

find a matrix P such that $$[x]_R= P [x]_B$$ for all x in R^2[/tex]

the answer should be a 2x2 matrix but I don't see how that is possible since [x] is only a column vector. I'm not sure how to solve this problem. any ideas?

 

[Euclid]

A nx2 times a 2x1 gives a nx1, so really P has to be a 2x2 matrix.

Start by finding the components of each basis in terms of the other. A motivating question may be something like what is $$P(1, 0)^T$$? Fool around with this idea to find the components of P.

 

[UrbanXrisis]

I'm not sure what is meant by "Start by finding the components of each basis in terms of the other."

I understand that "A nx2 times a 2x1 gives a nx1, so really P has to be a 2x2 matrix." but what about the fact that [x] can be ANY numbers. [x] is not even dependent of the basis so I don't know why they would even give the basis.

So if [x] can be anything, then so can P, so how do I find a specific 2x2 matrix?

 

[MathematicalPhysicist]

Let B be the basis of R^2 consisting of the vectors

$$\left(\begin{array}{c}3 & 1 \end{array}\right)$$ and $$\left(\begin{array}{c}-1 & 3 \end{array}\right)$$

Let R be the basis consisting of

$$\left(\begin{array}{c}2 & 3 \end{array}\right)$$ and $$\left(\begin{array}{c}1 & 2 \end{array}\right)$$

find a matrix P such that $$[x]_R= P [x]_B$$ for all x in R^2[/tex]

the answer should be a 2x2 matrix but I don't see how that is possible since [x] is only a column vector. I'm not sure how to solve this problem. any ideas?

your matrix could be regarded as a 'changing basis matix'.
in other words you need to let one of them to be a linear combination of the other basis:
2=a1*3+a2*1
3=b1*3+b2*1
1=a1*(-1)+a2*3
2=b1*(-1)+a2*3
and your P would be the matrix:
a1 a2
b1 b2

 

[UrbanXrisis]

your matrix could be regarded as a 'changing basis matix'.
in other words you need to let one of them to be a linear combination of the other basis:
2=a1*3+a2*1
3=b1*3+b2*1
1=a1*(-1)+a2*3
2=b1*(-1)+a2*3
and your P would be the matrix:
a1 a2
b1 b2

I tried that already, that does not give the correct answer. Also, I believe you mean:
2=b1*(-1)+b2*3

this gives you
a1=.5
a2=.5
a3=.7
a4=.9

which is not the correct answer

 

[MathematicalPhysicist]

your textbook has different answers than these?

 

[UrbanXrisis]

yes, i entered them for answers on the web and it tells me that they are wrong

 

[UrbanXrisis]

ok, i did the exact same thing for the other matrix and got the right answer, thanks for the help!

 

[MathematicalPhysicist]

which other matrix are you referring to?

 

[UrbanXrisis]

3=a1*2+a2*3
-1=a1*(1)+a2*2
1=b1*(2)+a2*3
3=b1*1+b2*2

 

[MathematicalPhysicist]

but this gives you the solution of this:
[x]_B=P[x]_R