Orthogonal projection onto a vector with matrix transformation

Question

So I have this question here which says:

$a)$ Find the standard matrix of the linear operator $T:R^2\rightarrow R^2$ given by the orthogonal projection onto the vector $(1,-2)$.

$b)$ Given the linear transformation $T:R^2\rightarrow R^3$ such that

$$T(1,0)=(1,-2,3), T(1,1)=(0,1,4)$$

Find $T(2,3)$.

For part $a)$, I used my standard basis vectors to find the orthogonal projection. Thus, I went:

$T(e_1)=(1,0)-\frac{(1,0)\cdot(1,-2)}{\|(1,-2)\|}(1,-2)$ $=(\frac{4}{5},\frac{2}{5})$

Similarly, I get that $T(e_2)=(\frac{2}{5},\frac{1}{5})$

So thus, my standard matrix is:

$\left( \begin{array}{cc} \frac{4}{5} & \frac{2}{5}\\ \frac{2}{5} & \frac{1}{5} \\ \end{array} \right) $

Apparently this isn't right. Why? Is there something I did wrong with my work?

For part $b)$, I know I need $T(1,0)$ and $T(0,1)$ however, I only know what $T(1,0)$ is. Can I find $T(0,1)$ by going $T(1,1)-T(1,0)$?

Some guidance would be appreciated. Thanks!

Answer 1

a) The projection of $v$ on the vector $w$ is $\displaystyle\frac{v.w}{\lVert w\rVert^2}w$. So, the projection of $(1,0)$ is $\displaystyle\left(\frac15,-\frac25\right)$ and the projection of $(0,1)$ is $\displaystyle\left(-\frac25,\frac45\right)$. So, the matrix is$$\begin{bmatrix}\frac15&-\frac25\\-\frac25&\frac45\end{bmatrix}.$$

b) Note that $(2,3)=3(1,1)-(1,0)$. Therefore, $T(2,3)=3T(1,1)-T(1,0)$.

Answer 2

The formula fir the orthogonal projection of $e_1$ along $(1,-2) $ as you have written is incorrect (a length should be squared, check).

In part two, since $T$ is linear, $T (0,1)=T ((1,1)-(1,0))=T (1,1)-T (1,0)$.