MHB IADPCFEVER's question at Yahoo Answers (projection and linear transformation)

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The discussion focuses on demonstrating that a transformation projecting vectors from R^2 onto the line y=x is linear by expressing it as a matrix transformation. The transformation is defined mathematically, showing that the projection of a vector (x_0, y_0) results in the point ((x_0 + y_0)/2, (x_0 + y_0)/2). This projection can be represented using a matrix A, which confirms the linearity of the transformation. The proof involves verifying that the transformation satisfies the properties of linearity for any scalar multiples and vector additions. The conclusion establishes that the projection is indeed a linear map.
Fernando Revilla
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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

How do I go about?

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.
 
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Hello IADPCFEVER,

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 $\mathbb{R}^2$:
$$p(\lambda v+\mu v')=A(\lambda v+\mu v')=\lambda Av+\mu Av'=\lambda p(v)+\mu p(v')$$
 
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