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- anonymous

Anyone good with changing basis!!!! I neeeeeed Help! :(

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- anonymous

Anyone good with changing basis!!!! I neeeeeed Help! :(

- jamiebookeater

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- anonymous

logs?

- anonymous

no changing basis of a transformation matrix.

- anonymous

the transformation matrix is made up of the basis vectors as columns, transformation is applied by multiplying with the transformation matrix, where is the part that's troubling you?

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- anonymous

my problem is here:
Suppose that vectors v1 = (1, 2), v2 = (2,−1), and that the basis B is B=v1, v2 . (this is a list)
Let T be the linear transformation from R2 to R2 given by T(v1) = v1 and T(v2) = 0.
(a) Write down the matrix for T in the new basis B. (You should be able to do this
directly from the definition of T).
(b) Use this to write down the matrix for T in the standard basis.

- anonymous

(a) First column: 1, 0
Second column: 0, 0

- anonymous

The first column has to contain the coefficients of T(v1) in the new basis {v1,v2}.
The second column contains the coefficients of T(v2) in the new basis {v1,v2}.

- anonymous

Now, for (b) you have to express T(e1) in terms of the basis {e1,e2} and that will be your first column, and the second accordingly for T(e2).

- anonymous

(where {e1,e2} is the canonical standard basis for R2)

- anonymous

but how to i change the basis to be working in the standard basis instead of basis B?

- anonymous

So you have to ask yourself: How do I write (1,0) in terms of linear combinations of v1 and v2?

- anonymous

Note that v1+2v2 = (3,0), so 1/3*v1+2/3*v2=e1

- anonymous

You do the same for e2.

- anonymous

Then you can calculate T(e1), T(e2) in terms of the standard basis and write down the matrix for (b).

- anonymous

Btw. this is equivalent to finding the inverse matrix for the matrix with columns v1, v2.

- anonymous

Do you understand?

- anonymous

i understand for the most part, until applying the transformation, becuase if i'm applying that transformation aren't i still working in a basis for B?

- anonymous

a basis for B? What do you mean by that? B IS a basis.

- anonymous

When you are asked to find a matrix of a linear transformation T of Rn to Rn with respect to a certain basis {b_1,b_2,...,b_n}, then this just means you are supposed to find the coefficients of the images T(b_i) in terms of the given basis. So you are supposed to express the images of the basis vectors in terms of the basis.

- anonymous

And then these coefficients are written down in a nicely ordered way, which is called a matrix. I.e. the i-th column contains in the j-th row the coefficient in front of b_j when you write T(b_i) in terms of the basis {b_1,...,b_n}.

- anonymous

i know but thats what i mean, if i apply the transformation matrix to the standard basis vectors arent we still working in that basis? when for part b what we're supposed to do is put the transformation matrix into standard basis for instead of B basis

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