anonymous
  • anonymous
Anyone good with changing basis!!!! I neeeeeed Help! :(
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
logs?
anonymous
  • anonymous
no changing basis of a transformation matrix.
anonymous
  • 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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • 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
  • anonymous
(a) First column: 1, 0 Second column: 0, 0
anonymous
  • 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
  • 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
  • anonymous
(where {e1,e2} is the canonical standard basis for R2)
anonymous
  • anonymous
but how to i change the basis to be working in the standard basis instead of basis B?
anonymous
  • anonymous
So you have to ask yourself: How do I write (1,0) in terms of linear combinations of v1 and v2?
anonymous
  • anonymous
Note that v1+2v2 = (3,0), so 1/3*v1+2/3*v2=e1
anonymous
  • anonymous
You do the same for e2.
anonymous
  • anonymous
Then you can calculate T(e1), T(e2) in terms of the standard basis and write down the matrix for (b).
anonymous
  • anonymous
Btw. this is equivalent to finding the inverse matrix for the matrix with columns v1, v2.
anonymous
  • anonymous
Do you understand?
anonymous
  • 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
  • anonymous
a basis for B? What do you mean by that? B IS a basis.
anonymous
  • 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
  • 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
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.