# Matrix composition question

1. Aug 19, 2011

### JamesGoh

1. The problem statement, all variables and given/known data

Suppose that T : R2 → R2 is linear and has matrix

-2 1
5 2

with respect to the standard basis S of R2.

B = 1 1
5 6

(B is another poorly constructed matrix)

What is the matrix of T with respect to B?

2. Relevant equations

T$_{C,B}$ = (T$_{C,S}$)$^{-1}$I$_{B,S}$

3. The attempt at a solution

Please see the pdf called q6b

Also to see the question better presented look at problems1.pdf. Go to Problem sheet 3 question 6b

#### Attached Files:

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• ###### problems1.pdf
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Last edited: Aug 20, 2011
2. Aug 20, 2011

### JamesGoh

sorry guys, I should point out that

the answer to the question should be

3 7
0 -3

which I don't seem to be getting

3. Aug 20, 2011

### ehild

That is wrong. A matrix in a new basis B is IBS-1T IBS

ehild

4. Aug 20, 2011

### JamesGoh

When you say T, do you mean T with respect to the standard basis ?

5. Aug 20, 2011

### JamesGoh

In terms of basis reference, does your formula do the following (in terms of matrix composition)

1. You create the identity map of matrix B with respect to the standard basis

2. You multiply the B identity map with T to create a new matrix with respect
to the standard basis ?

3. You multiply the result of step 2 with the inverse of the identity map in step 1 to get the answer with respect to B

6. Aug 20, 2011

### ehild

With respect to the standard basis T=Ts, and Tb=IBS-1Ts IBS is the matrix of the linear transformation in the new basis.

Think: The new basic vectors b1, b2 are B times the standard basic vectors. Edit: With "B"I denoted the matrix with columns equal to the basic vectors b1 and b2. And you get beck the standard basic vectors by multiplying b1, b2 by B-1.
T transforms the basic vectors of B into vectors Tb1 and Tb2, defined in the standard basis. You need to express these vectors in terms of the new basis, so apply B-1 to them.

Last edited: Aug 20, 2011
7. Aug 20, 2011

### ehild

You explained it very well. Yes. that is what I wanted to do, but I could not express myself so well.

ehild

8. Aug 20, 2011

### HallsofIvy

Just a word about notation: It does not really make sense to ask about the matrix of a linear transformation "with respect to" another matrix. In your attached pdf files, B is NOT a matrix, it is two vectors, a new basis for $R^2$.

Another way to find the matrix of a linear transformation with respect to a given (ordered) basis is: Apply the linear transformation to the each basis vector in turn. Write the result as a linear combination of the basis vectors. The coefficients give the columns of the matrix.

Last edited by a moderator: Aug 20, 2011
9. Aug 20, 2011

### ehild

Thanks, HallsofIvy. I know that I cannot mix vectors with transformations...I just used that the matrix that transforms the standard basis into a new basis has columns equal to the new basic vectors.
I hope, I am right... Being a physicist, I use Maths a bit sloppy way.

10. Aug 20, 2011

### Ray Vickson

If the standard basis is e1=[1 0]^t and e2=[0 1]^t (^t = transpose) and b1=[1 5]^t, b2 = [1 6]^t, you can express e1 and e2 as linear combinations of b1 and b2, just by solving the equations b1=e1+5*e2 and e2=e1+6*e2 for e1 and e2; you can do this just as though the e's and b's were real variables instead of vectors---the algebra does not care what they are. A vector v = x1*b1 + x2*b2 can thus be written in terms of basis {e1,e2}, say as y1*e1+y2*e2, with y1,y2 = known linear combinations of x1,x2. Applying the matrix A = [[-2 1],[5 2]]to v = y1*e1+y2*e2 gives (-2y1+y2)*e1+(5y1+2y2)*e2. Now back-substitute for the yi in terms of the xi and for the ej in terms of the bj. You should get the result Tv = (3x1+7x2)*b1+(-3x2)*b2. Therefore, the matrix representation of T in the basis {b1,b2} is [[3 7],[0 -3]]. Once you have grasped these concepts through some simple examples, done step-by-laborious-step, then you can more successfully see where the matrix formulae come from and what they mean.

RGV