I have a linear transformation satisfying F(1,0,0) = (0,1), F(0,2,1) = (1,1) and F(0,1,−2) = (−2,0) is the standard matrix {[0,1,-2],[1,1,0]} ? or if not how do i find the stardard matrix correctly?
Thanks.
yes your matrix is correct. Since the three vectors you gave, from are linearly independent. They form a basis for .
so any vector in can be written as .
so
errr. I think they want the linear transformation written interm of the standard basis for
So since is the change of basis matrix from the standard to the basis with ur given vectors.
so is the standard matrix for your linear transformation.
no, that is the matrix in the basis B' = {(1,0,0),(0,2,1),(0,1,-2)}.
as jakncoke points out, you need to employ a change of basis similarity transform, to obtain the "standard matrix". if the change of basis matrix is P (from B' to B, the standard basis B = {(1,0,0),(0,1,0),(0,0,1)})
the matrix you want is:
.
the matrix you have is [F]_{B'}. P is given by jakncoke above.
you COULD compute the inverse using the adjugate, but i always found that to be tedious and time-consuming.
we know that (1,0,0)--->[1,0,0]_{B'}, so we just need to see how to express (0,1,0) and (0,0,1) as linear combinations of (0,2,1) and (0,1,-2).
a little fooling around gives:
(0,1,0) = (2/5)(0,2,1) + (1/5)(0,1,-2)
(0,0,1) = (1/5)(0,2,1) + (-2/5)(0,1,-2) so:
this gives:
as the correct matrix for F in the standard basis. we can verify this like so:
F(1,0,0) = (0,1) since this is the first column of F.
F(0,2,1) = (0+0+1,0+4/5+1/5) = (1,1)
F(0,1,-2) = (0+0-2,0+2/5-2/5) = (-2,0).
since the range has dimension 2 (and our matrix has RANK 2) the range is all of R^{2}. F is onto.
by the rank-nullity theorem, this means the dimension of the kernel is 1, so there HAS to be more in the kernel than just {0}.