- #1
PirateFan308
- 94
- 0
Homework Statement
Define the linear transformation [itex]T: R^{3} → R^{3}[/itex] by [itex]T(v)=[/itex] the projection of [itex]v[/itex] onto the vector [itex]w=(1,2,1)[/itex]
Find the (standard matrix of [itex]T[/itex])
Homework Equations
[itex]T: V → W[/itex] is a function from V to W (which means that for each v in V, there is a T9v) in W such that:
T(v+v')=T(v)+T(v'), T(cv) = cT(v) for all v,v'[itex]\in[/itex]V, c[itex]\in[/itex]F
For T(v) = Av, the matrix A is called the standard matrix of T
projection of v onto w = [itex](\frac{v \bullet w}{w \bullet w})(w)[/itex]
The Attempt at a Solution
I'm having problems understanding Linear Transformations at all, and I'm not really sure if this is at all correct ... I'm thinking I should apply T to [itex]e_1, e_2, and~ e_3[/itex].
Because the transformation takes place in [itex]R^{3}[/itex], I know A will be a 3x3 matrix. Apply T to [itex]e_1[/itex], where proj [itex]e_1[/itex] onto w would be [itex](1/6, 2/6, 1/6)[/itex] and we can see that [itex]x=1/6, y=1/3, z=1/6[/itex], so should the first column of A be [itex] \begin{pmatrix} 1/6 \\ 1/3 \\ 1/6 \end{pmatrix}[/itex]? Then apply T to [itex]e_2[/itex], where proj [itex]e_2[/itex] onto w would be [itex](1/3, 2/3, 1/3)[/itex] and we can see that [itex]x=1/3, y=2/3, z=1/3[/itex], so the second column of A should be [itex] \begin{pmatrix} 1/3 \\ 2/3 \\ 1/3 \end{pmatrix}[/itex]. And again, applying T to [itex]e_3[/itex] proj [itex]e_3[/itex] onto w would be [itex](1/6, 2/6, 1/6)[/itex] and we can see that [itex]x=1/6, y=1/3, z=1/6[/itex], so should the third column of A be [itex] \begin{pmatrix} 1/6 \\ 1/3 \\ 1/6 \end{pmatrix}[/itex]?
So should the matrix A be [itex] A=\begin{pmatrix} 1/6 & 1/3 & 1/6 \\ 1/3 & 2/3 & 1/6 \\ 1/6 & 1/3 & 1/6 \end{pmatrix}[/itex]? This seems odd to have the third row be the same as the first row and the third column the same as the third column.
Is this correct, or am I completely off? Thanks!