Change of Basis: Finding PC<->B Matrix for V=R^2 with Given B and C

[snoggerT]

find the change of basis matrix PC<->B from the given ordered basis B to the given ordered basis C of the vector space V:

V=R^2; B={(-5,-3),(4,28)}; C={(6,2),(1,-1)}

The Attempt at a Solution

I'm having a hard time grasping the concept of changing basis. Can someone please explain it to me? my books explanation seems rather abstract and hard to follow.

 

[NateTG]

Can you write an arbitrary vector \vec{v}=&lt;v_x,v_y&gt; in terms of either basis?

 

[Vid]

The change of basis matrix is a matrix that when multiplied by a column B-coordinate vector gives the same vector represented as a column C coordinate vector.

(a b)(-5) = (6)
(c d)(-3) (2)

(a b)(4 ) = (1)
(c d)(28) (-1)

Do the multiplication and solve the relevant systems to find a,b,c,d.

 

[HallsofIvy]

First of all, to "change bases", you have to have two bases! Let's use P2, the set of quadratic polynomials, ax²+ bx+ c, as an example (we can add two polynomials and multiply a polynomial by a real number so that is a vector space over the real numbers). The "standard" basis is x², x, and 1. As long as we understand we are using those "vectors" as basis, in that order, we can use <a, b, c> to represent ax²+ bx+ c: the "x², x, and 1 are "understood".

But (x- 1)²= x²- 2x+ 1, x- 1, and 1 is also a basis. Since the "standard" basis for P2 has 3 vectors, it has dimension 3. This new set also contains 3 "vectors" so as long as they are independent . To show they are independent, look at p(x-1)²+ q(x-1)+ r= 0 (the 0 function: 0 for all x). Since that is 0 for all x, taking x= 1gives r= 0. If that is 0 for all x, it is a constant and so its derivative, 2p(x-1)+ q= 0 for all x. Again taking x= 1, we get q= 0. That is, the original equation was really p(x-1)²= 0 and, taking x= 2 now, p= 0. Since p= q= r= 0, those "vectors" are independent and so a basis.

If I wanted to write ax²+ bx+ c in that new basis, I would have to find p, q, r so that p(x- 1)²+ q(x-1)+ r= ax²+ bx+ c for all x. Do pretty much what we did before: taking x= 1, r= a+ b+ c. Taking the derivative of each side,
2p(x-1)+ q= 2ax+ b, again for all x. Taking x= 1 again, q= 2a+ b. Taking the second derivative of both sides, 2p= 2a so p= a. (Multiplying p(x-1)² and q(x-1) out, collecting coefficients of like and setting corresponding coefficients on both sides equal would give the same thing: p= a, q= 2a+ b, and r= a+ b+ c.) That is, the polynomial represented by <a, b, c> in the first basis (that is, ax²+ bx+ c) is represented by (a, 2a+ b, a+ b+ c) (which is a(x- 1)²+ (2a+ b)(x-1)+ (a+b+c)) in the new basis. If you multiply those out you should see that you get the same thing.

The "change of basis" matrix must convert the <a, b, c> of a vector written in one basis into the components for the same vector as written in a new basis. In the example above, multiplying the basis by <a, b, c> must give <a, 2a+ b, a+ b+ c> . In particular, it must change <1, 0, 0> to <1, 2, 1> (a= 1, b= 0, c= 0 here), change <0, 1, 0> to <0, 1, 1> (a= 0, b= 1, c= 0), and change <0, 0, 1> to <0, 0, 1> (a= 0, b= 0, c= 1). The reason I like to use <1, 0, 0>, <0, 1, 0>, and <0, 0, 1> is because it is easy to see that a matrix multiplied by <1, 0, 0> just gives the numbers in the first column, multiplied by <0, 1, 0> gives the numbers in the second column, multiplied by <0, 0, 1> gives the numbers in the third column. In other words, the "change of basis" matrix must have <1, 2, 1>, <0, 1, 0> and <0, 0, 1> as the first, second and third columns: it is
\left(\begin{array}{ccc} 1 &amp; 2 &amp; 1 \\0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 1\end{array}\right)