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, $\displaystyle \vec{v_1}, \vec{v_2}, \vec{v_3}$ from $\displaystyle \mathbb{R}^3 $ are linearly independent. They form a basis for $\displaystyle \mathbb{R}^3$.
so any vector in $\displaystyle v \in \mathbb{R}^3 $ can be written as $\displaystyle v = c_1 * \vec{v_1} + c_2 * \vec{v_2} + c_3 * \vec{v_3} $.
so $\displaystyle L(v) = L(c_1 * \vec{v_1} + c_2 * \vec{v_2} + c_3 * \vec{v_3}) = c_1*L(\vec{v_1}) + c_2*L(\vec{v_2}) + c_3*L(\vec{v_3}) = \begin{bmatrix} L(v_1) & L(v_2) & L(v_3) \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} $
errr. I think they want the linear transformation written interm of the standard basis for $\displaystyle \mathbb{R}^3 $
So since $\displaystyle \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & -2 \end{bmatrix} $ is the change of basis matrix from the standard to the basis with ur given vectors.
so $\displaystyle \begin{bmatrix} 0 & 1 & -2 \\ 1 & 1 & 0 \end{bmatrix} * \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & -2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 5 \\ 1 & 2 & 1 \end{bmatrix}$ 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:
$\displaystyle [F]_B = [F]_{B'}P^{-1}$.
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:
$\displaystyle P^{-1} = \begin{bmatrix}1&0&0\\0&\frac{2}{5}&\frac{1}{5}\\0 &\frac{1}{5}&\frac{-2}{5} \end{bmatrix}$
this gives:
$\displaystyle [F]_B = \begin{bmatrix}0&0&1\\1&\frac{2}{5}&\frac{1}{5} \end{bmatrix}$
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}.