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 P: 94 1. The problem statement, all variables and given/known data Define the linear transformation $T: R^{3} → R^{3}$ by $T(v)=$ the projection of $v$ onto the vector $w=(1,2,1)$ Find the (standard matrix of $T$) 2. Relevant equations $T: V → W$ 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'$\in$V, c$\in$F For T(v) = Av, the matrix A is called the standard matrix of T projection of v onto w = $(\frac{v \bullet w}{w \bullet w})(w)$ 3. 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 $e_1, e_2, and~ e_3$. Because the transformation takes place in $R^{3}$, I know A will be a 3x3 matrix. Apply T to $e_1$, where proj $e_1$ onto w would be $(1/6, 2/6, 1/6)$ and we can see that $x=1/6, y=1/3, z=1/6$, so should the first column of A be $\begin{pmatrix} 1/6 \\ 1/3 \\ 1/6 \end{pmatrix}$? Then apply T to $e_2$, where proj $e_2$ onto w would be $(1/3, 2/3, 1/3)$ and we can see that $x=1/3, y=2/3, z=1/3$, so the second column of A should be $\begin{pmatrix} 1/3 \\ 2/3 \\ 1/3 \end{pmatrix}$. And again, applying T to $e_3$ proj $e_3$ onto w would be $(1/6, 2/6, 1/6)$ and we can see that $x=1/6, y=1/3, z=1/6$, so should the third column of A be $\begin{pmatrix} 1/6 \\ 1/3 \\ 1/6 \end{pmatrix}$? So should the matrix A be $A=\begin{pmatrix} 1/6 & 1/3 & 1/6 \\ 1/3 & 2/3 & 1/6 \\ 1/6 & 1/3 & 1/6 \end{pmatrix}$? 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!