Recent content by Myr73

nvm, i have a final tomorrow - i will simply send it in as is, cause i obviously do not understand- and hope its not found to often on the exam! Thank you for your help -

because you said that [T1]=e1xe1

I am not sure, but i thought T2=(e2)x(e2), and i thought e2 would be (0,1,0)

agh ok, and no I was not guessing, the cross product of (0,1,0) and (0,1,0) is also zero...

"If X=Y, it's convenient to choose B=A, and to speak of the matrix representation of T with respect to A instead of with respect to (A,A), or (A,B). The formula for Tij can now be written as Tij=(Tej)i=(Tei),(Tej)" I guess what I did not understand is how to refer it to the matrix of the...

Does [T]= 1 0 0 0 0 0 0 0 0 , make sense? And I will re-read it, I do not really understand much of it, but I will reread it.

yes, I am sorry , a couple of minor errors there. I meant "u" not "U". And yes I ignored the formula with [T], because I did not know what to do with it. But I thought they wanted to know if V=i then find T(u), but I guess not. How am I suppose to compute [T] if v=(1,0,0)?

Ok, well i have answered a and b to the best of my knowledge and ability, is there someone that can help me on part c)? Here is the question again. 1- Question Let V be a fixed vector in R^3. a)Show that the transformation defined by T( u)= v X u is a linear transformation. b) Find the range ot...

anyone?

oh. ok, so are they asking to find the Ax? where x would be the unit square.. so like [10. ] [ 1 0] [ 1 3 ] [ 01 ] ...

1- Let the linear operator on R^2 have the following matrix:A = 1 0 -1 3 What is the area of the figure that results from applying this transformation to the unit square? 2- I am abit confused here, I thought that the matrix for the unit square would be, 0 0 1 0 0 1 1 1... But...

yes, I am aware of that. and I know that to find a subspace of R3 that is orthogonal to v, we would find the orthogonal projection of R3 on W. where w is a subspace of R3. witch would be the Ax, A being the column space of given vectors of v, and x being the least squares solution of Ax=v.

Oh, ok,cool thanks. I will use the symbol bars.. And , I do not know.. I really do not know.

v• (vXu)=0 , v is orthogonal to vXu

oh ok, so how would I write out when v does not equal zero,that the range of T is the plane that is orthogonal to v