# Homework Help: Linear Algebra - Change of Basis

1. Sep 13, 2008

### kehler

1. The problem statement, all variables and given/known data
Let B & C be the following subsets of R^2
B= {[3 1] , [2 2]} (the vectors should be in columns instead of rows)
C= {[1 0] , [5 4]}

Let T: R^2 -> R^2 be the linear transformation whose matrix with respect to the basis B is
[2 1]
[1 5] (the brackets should be joint, it's a 2 x 2 matrix)

Find the matrix T with respect to C. Check your answer by finding the determinant and the trace of each matrix.

3. The attempt at a solution
I found the change of basis matrix from B to C to be
[(7/4) (-2/4)]
[(1/4) ( 2/4) ] (again a 2x2 matrix)

I did this by multiplying the change of basis matrix from the standard basis to C, with the change of basis matrix from the B basis to the standard matrix like this:
[1 (-5/4)] [3 2]
[0 (1/4)] [1 2]

I then multiplied the given matrix by the change of basis matrix from B to C to get
[3 (-3/4)]
[1 (11/4)]

I thought my answer was correct but the trace for the matrix I got is 8.25 whilst the trace of the original matrix is 7. They should be the same, right??
Can anyone see where I went wrong? I'm not sure if I got the change of basis matrices right :S. The columns of the set B should form the change of basis matrix from B to the standard basis, shouldn't they?

Any help would be appreciated :)

2. Sep 13, 2008

### gabbagabbahey

You've actually done the exact opposite here:

The matrix for going from the standard basis (S) to C is actually:

$$\overleftrightarrow{P}_{C \leftarrow S} \leftrightarrow \left( \begin{array}{cc} 1 & 5 \\ 0 & 4 \end{array} \right)$$

While the matrix for going from B to S is actually:

$$\overleftrightarrow{P}_{S \leftarrow B} \leftrightarrow \left( \begin{array}{cc} \frac{1}{2} & \frac{-1}{2} \\ \frac{-1}{4} & \frac{3}{4} \end{array} \right)$$

You can check these by operating on each of the basis vectors in S (i.e. {[1,0],[0,1]}) with
$$\overleftrightarrow{P}_{C \leftarrow S}$$ , to make sure you get the basis vectors of C; and by operating on each of the basis vectors in B with
$$\overleftrightarrow{P}_{S \leftarrow B}$$ , to make sure you get the corresponding basis vectors in S.

You can then easily obtain the change of basis matrix from B to C by using :

$$\overleftrightarrow{P}_{C \leftarrow B} = {\overleftrightarrow{P}_{C \leftarrow S}}{\overleftrightarrow{P}_{S \leftarrow B}}$$

Last edited: Sep 13, 2008
3. Sep 14, 2008

### muso07

Are you in MATH1116 at ANU? If you're not then my lecturer's been doing some hard core plagiarism. :P

Sorry I don't actually have an answer for you... I got the same as you but the trace that I calculated was 5.75...

:S

4. Sep 14, 2008

### gabbagabbahey

Muso, the answer for $$\overleftrightarrow{T_c}$$ should be:

$$\overleftrightarrow{T_c} {\leftrightarrow} \left( \begin{array}{cc} \frac{83}{4} & \frac{-277}{16} \\ 17 & \frac{-55}{4} \end{array} \right)$$

If you show me your work, I can tell you where you went wrong.

5. Sep 14, 2008

### muso07

Okay, I got P(C<--B)=
[7/4 -1/2]
[1/4 1/2] (2x2 matrix)

so [T]C=P(C<--B)[T]B=
[7/4 -1/2][2 1]=
[1/4 1/2][1 5]

[3 -3/4]
[1 11/4]
Much appreciated. :)

6. Sep 14, 2008

### gabbagabbahey

Okay, your first error is that your $$\overleftrightarrow{P}_{C \leftarrow B}$$ Matrix is incorrect.

$$\overleftrightarrow{T_C} =\overleftrightarrow{P}_{C \leftarrow B} \cdot \overleftrightarrow{T_B} \cdot \overleftrightarrow{P}_{C \leftarrow B}^{-1}$$

NOT just

$$\overleftrightarrow{T_C} =\overleftrightarrow{P}_{C \leftarrow B}} \cdot \overleftrightarrow{T_B}$$

Why don't you show me how you got your $$\overleftrightarrow{P}_{C \leftarrow B}$$?

7. Sep 14, 2008

### muso07

I used the [c1 c2 | b1 b2] ~ [ I | P(C<--B)] formula

so I got

[1 5 | 3 2] ~
[0 4 | 1 2]

[1 0 | 7/4 -1/2]
[0 1 | 1/4 1/2]

So P(C<--B)=
[7/4 -1/2]
[1/4 1/2]

Thanks

8. Sep 14, 2008

### gabbagabbahey

Hmmm... I've never seen that formula before, but I can derive a correct version of it for you:

$$\left( \begin{array}{cc} \vec{c_1} ,& \vec{c_2} \end{array} \right) = \overleftrightarrow{P}_{C \leftarrow B} \cdot \left( \begin{array}{cc} \vec{b_1} ,& \vec{b_2} \end{array} \right)$$
since, by definition, the change of basis matrix will change the basis vectors of B into those of C.

And so,

$$\left( \begin{array}{cc} \vec{c_1} ,& \vec{c_2} \end{array} \right) \cdot \left( \begin{array}{cc} \vec{b_1} ,& \vec{b_2} \end{array} \right)^{-1} = \overleftrightarrow{P}_{C \leftarrow B} \cdot \left( \begin{array}{cc} \vec{b_1} ,& \vec{b_2} \end{array} \right) \cdot \left( \begin{array}{cc} \vec{b_1} ,& \vec{b_2} \end{array} \right)^{-1} =\overleftrightarrow{P}_{C \leftarrow B}$$

Try this formula, and don't forget to take the inverse.

9. Sep 14, 2008

### muso07

Thank you! I think my textbook is wrong... :S

10. Sep 14, 2008

### gabbagabbahey

Your welcome, wht do you get for $$\overleftrightarrow{P}_{C \leftarrow B}$$ now?

11. Sep 14, 2008

### muso07

I got
[-3/4 13/4]
[-1 3]

Is that right?

12. Sep 14, 2008

### gabbagabbahey

Yup, did you get the right $$\overleftrightarrow{T_c}$$ now?

13. Sep 14, 2008

### muso07

Yep. Or at least the traces and determinants agree now. :)

14. Sep 14, 2008

### kehler

Yes I am. Lol. Are u done with the assignment? I'm still stuck on a couple of questions :(

Thanks, gabbagabbahey :)

15. Sep 14, 2008

### gabbagabbahey

WOW! what are the odds of that?!:rofl: And you're welcome

16. Sep 14, 2008

### muso07

I've done everything but the first and last questions. Need help with anything?

17. Sep 14, 2008

### kehler

I'm stuck on the last one too! I just posted it actually...
And question 3 and 5. I've tried expanding q3 but I'm not getting anywhere :(. For q5, I'm having a bit of trouble trying to explain b and c.

U need help with the first question?
I actually found the geometric series by trial and error. Then I just showed that the sum of my series from n=10 to infinity is less than 10^-6.

Last edited: Sep 14, 2008
18. Sep 14, 2008

### muso07

yeah my tutor did an example for the first one and basically he used 1/n! from 10 to infinity but that's not a geometric series, so... :S

With 3 expand and factor out x^4 on the denominator and numerator. You should get a limit of -10/3.

And with 5, solve for r^3-9r^2-12r+20=0, you should get r=-2, 10, 1. Then yk is the solution with initial values as given in the question. yk=a(-2)^k+b(10)^k+c(1)^k and solve for a, b, c

19. Sep 14, 2008

### kehler

Ahh thanks :).. I finally got ques 3. Expanding functions is such a pain!
Try using fractions for both the 'a' and 'r' in your geometric series.