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#### Fernando Revilla

##### Well-known member
MHB Math Helper
Here is the question:

I'm supposed to show that this transformation from R^2 to R^2 is linear by showing that it is a matrix transformation.

P projects a vector onto the line y=x

Here is a link to the question:

Projection and linear transformation? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Consider $(x_0.y_0)\in\mathbb{R}^2$. The perpendicular line to $r:y=x$ passing through $(x_0,y_0)$ is $s:y-y_0=-(x-x_0)$. The intersection point between $r$ and $s$ is:
$$\left \{ \begin{matrix}x-y=0\\x+y=x_0+y_0\end{matrix}\right.\Leftrightarrow\ldots\Leftrightarrow(x,y)=\left(\dfrac{x_0+y_0}{2},\dfrac{x_0+y_0}{2}\right)$$
That is, if $p:\mathbb{R}^2\to \mathbb{R}^2$ projects $(x_0,y_0)$ onto $r:y=x$ then,
$$p\begin{pmatrix}{x_0}\\{y_0}\end{pmatrix}= \dfrac{1}{2} \begin{pmatrix}{x_0+y_0}\\{x_0+y_0}\end{pmatrix}= \dfrac{1}{2} \begin{pmatrix}{1}&{1}\\{1}&{1}\end{pmatrix}\begin{pmatrix}{x_0}\\{y_0}\end{pmatrix}=A \begin{pmatrix}{x_0}\\{y_0}\end{pmatrix}$$
Now, we easily prove that $p$ is a linear map. For all $\lambda,\mu\in\mathbb{R}$ and for all $v=(x_0,y_0)^t$, $v'=(x'_0,y'_0)^t$ in $\mathbb{R}^2$:
$$p(\lambda v+\mu v')=A(\lambda v+\mu v')=\lambda Av+\mu Av'=\lambda p(v)+\mu p(v')$$