# Thread: linear transformation standard matrix

1. ## linear transformation standard matrix

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.

2. ## Re: linear transformation standard matrix

yes your matrix is correct. Since the three vectors you gave, $\vec{v_1}, \vec{v_2}, \vec{v_3}$ from $\mathbb{R}^3$ are linearly independent. They form a basis for $\mathbb{R}^3$.

so any vector in $v \in \mathbb{R}^3$ can be written as $v = c_1 * \vec{v_1} + c_2 * \vec{v_2} + c_3 * \vec{v_3}$.
so $L(v) = L(c_1 * \vec{v_1} + c_2 * \vec{v_2} + c_3 * \vec{v_3}) = c_1*L(\vec{v_1}) + c_2*L(\vec{v_2}) + c_3*L(\vec{v_3}) = \begin{bmatrix} L(v_1) & L(v_2) & L(v_3) \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}$

errr. I think they want the linear transformation written interm of the standard basis for $\mathbb{R}^3$
So since $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & -2 \end{bmatrix}$ is the change of basis matrix from the standard to the basis with ur given vectors.
so $\begin{bmatrix} 0 & 1 & -2 \\ 1 & 1 & 0 \end{bmatrix} * \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & -2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 5 \\ 1 & 2 & 1 \end{bmatrix}$ is the standard matrix for your linear transformation.

3. ## Re: linear transformation standard matrix

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:

$[F]_B = [F]_{B'}P^{-1}$.

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:

$P^{-1} = \begin{bmatrix}1&0&0\\0&\frac{2}{5}&\frac{1}{5}\\0 &\frac{1}{5}&\frac{-2}{5} \end{bmatrix}$

this gives:

$[F]_B = \begin{bmatrix}0&0&1\\1&\frac{2}{5}&\frac{1}{5} \end{bmatrix}$

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).

4. ## Re: linear transformation standard matrix

ok i used the linear combination method to get the standard matrix equal to {[0,0,1],[1,2/5,1/5]} could anybody confirm this? Thanks

5. ## Re: linear transformation standard matrix

Am i also correct in saying the range of F is =span{[0,1],[1,1/5]} and the kernel of F is ={0}?
Thanks

6. ## Re: linear transformation standard matrix

since the range has dimension 2 (and our matrix has RANK 2) the range is all of R2. 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}.

7. ## Re: linear transformation standard matrix

Yes, i see my mistake. Thanks