Carl's Answer to PM #7 33A
It seems like there are a variety of answers for #7. Here is my answer with the work. Hopefully we can reach a consensus. Basically, this problem can be broken into three steps: \\1) Find T for each \beta \\2) Get the arbitrary values in terms of \beta \\3) Factor and substitute First, [T(1,1)]_\gamma = \[ \left( \begin{array}{ccc} 1 \\ 3 \end{array} \right)\] T(1,1) = \[ \left( \begin{array}{ccc} 2 \\ 4 \end{array} \right)\] [T(1,-1)]_\gamma = \[ \left( \begin{array}{ccc} 1 \\ 2 \end{array} \right)\] T(1,1) = \[ \left( \begin{array}{ccc} 3 \\ 2 \end{array} \right)\] Now, solve for a and b in terms of \beta: \[ \left( \begin{array}{ccc} 1 & 1 & | a\\ 1 & -1 & |b \end{array} \right)\] = \[ \left( \begin{array}{ccc} 1 & 0 & | (a+b)/2\\ 0 & -1 & |(a-b)/2 \end{array} \right)\] Finally, T(a,b) = T((a+b)/2 \[ \left( \begin{array}{ccc} 1 \\ 1 \end{array} \right)\] + (a-b)/2 \[ \left( \begin{array}{ccc} 1 \\ -1 \end{array} \right)\] = (a+b)/2 * T\[ \left( \begin{array}{ccc} 1 \\ 1 \end{array} \right)\] + (a-b)/2 * T\[ \left( \begin{array}{ccc} 1 \\ -1 \end{array} \right)\] = (a+b)/2 * \[ \left( \begin{array}{ccc} 2 \\ 4 \end{array} \right)\] + (a-b)/2 * \[ \left( \begin{array}{ccc} 3 \\ 2 \end{array} \right)\] = \[ \left( \begin{array}{ccc} (5a-b)/2 \\ 3a-b \end{array} \right)\]
Sam_Lai: May 4, 2015, 7:44 p.m.
Hi, Carl, thanks for taking the time to outline your steps. I'm a little bit confused with how you managed to get T(1,1) from $[T(1,1)]_\gamma$ when we don't know T. But I disagree with your final answer because if we tried to get $[T]_\gamma^\beta$ from your T(a,b) and the two given bases, we wouldn't get the original matrix given in the question. \\ For example, (1,1) in $\beta$ is (2, 2) under your transformation. In $\gamma$, this is (1,1) and not (1,3) in the first column of the given matrix. This matrix $\begin{pmatrix} (7/2)a + (1/2)b \\ 2a \end{pmatrix}$ seems to do the trick. \\ I managed to get this matrix following these steps. 1) Find the two matrices (P, S) which transform the bases to the standard basis. 2) Invert one of them ($P^{-1}$). 3) $P^{-1}TS = [T]_\gamma^\beta$ Use this to solve for T and then apply arbitrary (a,b) to T.
connorkenny: May 4, 2015, 9:41 p.m.
Are you solving for T or the transformation from gamma to beta? I can't get the answer you got according to what you wrote down. Also, why are you using [T] from gamma to beta and not the other way around?
julianna_burke: May 4, 2015, 10:05 p.m.
Sam_Lai can you explain your step 3. What do you use for the T when you do P^{-1}TS? and why is it [T] from gamma to beta?
TLindberg: May 4, 2015, 11:16 p.m.
Not sure if this is a huge deal when writing down the answer, but writing it in the form: ($\frac{7}{2}$a + $\frac{1}{2}$b, 2a), may be a bit more intuitive because it's mapping a coordinate pair to another coordinate pair in the standard basis.
Sam_Lai: May 5, 2015, 12:42 a.m.
Ahh I'm so sorry. As you already figured out, I do indeed mean [T] from beta to gamma every time I wrote the opposite. As for what I used for T, I used a 2x2 unknown matrix, since we're solving for what it is. I know that it's 2x2 because the surrounding matrices are 2x2. To explain my steps clearer, I defined T to be a transformation acting on the standard basis and P and S are matrices which serve to transform coordinates from beta and gamma bases to the standard basis. // Let's say P is identity matrix from gamma to standard $\begin{pmatrix} 1 & 1 \\ 2 & 0 \end{pmatrix} $ Inverting P will result in $\begin{pmatrix} 0 & 1/2 \\ 1 & -1/2 \end{pmatrix} $ S is the identity matrix from beta to standard $\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} $ Solving $P^{-1}TS=[T]_\beta^\gamma$ with T being $\begin{pmatrix} a & b \\ c & d \end{pmatrix} $ will return the result that I had.
Cullen_Im: May 5, 2015, 5:09 a.m.
A different way to solve for #7 is as follows: Start by setting (a,b) = x(1,1) + y(1,-1) and solving for "x" and "y" in terms of "a" and "b" to find out what direction (a,b) is in terms of the beta bases. You should get (x,y) = ((a+b)/2,(a-b)/2). Then multiply the matrix by (x,y) to get the product (a, (5a+b)/2). Finally, see what the product is with respect to your gamma bases by setting a(1,2) + ((5a+b)/2)(1,0) to get your answer of T(a,b) = ((7a + b)/2, 2a)
Adarsh111: May 6, 2015, 3:15 a.m.
Hey Cullen, thanks for sharing that explanation. It was helpful in understanding the problem - except the last bit - why do we need to see what the product is with respect to the gamma bases? Doesn't the multiplication of the matrix by (x,y) give us the mapping in the Gamma basis? Just confused by this part - clarification would help thanks!
tg1: May 6, 2015, 3:29 a.m.
T(a,b)= a/2 +b/2 [T]beta to gamma= [T(1,1)]gamma = [1 3] written in column form T(1,1)= 1(1,2) + 3(1,0)= (4,2) T(1,-1)= (3,2) T(a,b)= T(x(1,0) + T(y(0,1) take out the y and x T(1,1)-T(1,-1)= T(2,0)= (4,2) - (3,2)= (1,0)*.5=(1/2,0) T(1,-1) use same method as before x(T(1,0) + y(T(0,1) = .5x + .5y
drewgomberg: May 7, 2015, 9:22 p.m.
I think the useful way of thinking of this problem is similar to finding matrices with different and inputs and out puts than the given transformation matrix. First you have to put (a,b) into the basis beta. Then multiply you result by the given matrix and the result is given in gamma. Finally switch from gamma back into the standard basis and your result is shown in terms of (a,b). It is the same as taking the identity matrix from standard to beta multiplying by the transformation matrix beta to gamma then multiplying by the identity from gamma to standard.