Projection Transformation on x Axis Parallel to y=2x

Nov 2013
Let $T \colon \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the projection transformation on x axis in parallel to $y=2x$. Find the representing matrix of $T^*$ for the standard basis $B=\left(\left \{1,0 \right \},\left\{0,1\right\}\right)$.

I have a solution which I don't understand.

A basis for $y=2x$ could be $\left \{(1,2) \right\}$. A basis for x axis could be $\left \{(1,0)\right\}$. Therefore A basis for the whole vector would be $$C=\left\{(1,0)\right\},\left\{(1,2)\right\}$$.

I understand why this basis is correct. However:

$e_1=(1,0)=1v_1+0v_2 \Rightarrow [e_1]_B=\left (\begin {matrix} 1\\ 0 \end {matrix}\right)$ - this is quite clear.

$e_2=(0,1)=-\frac{1}{2}v_1+\frac{1}{2}v_2\Rightarrow [e_2]_B=\left (\begin {matrix} -\frac{1}{2} \\ \frac{1}{2} \end {matrix}\right)$ - this is the part I don't understand. Why use $-\frac{1}{2}$ to multiply $v_1$? What am I missing here?

Next, since the proejction transformation is $T(x,y)=(x,0)$ it is easy enough to continue. But there's that part I don't understand. Why is the scalar before the vector $v_1$has to be$
-\frac{1} {2} $


Last edited:


MHF Helper
Sep 2012
Hey nerazzurri10.

You are likely to go from a basis y = x to y = 2x and this involves a re-scaling.

A projection onto the line y = 2x will minimize the orthogonal distance to the line and in two dimensions we have m1*m2 = -1.

This means since m1 = 2 then m2 = -1/2 and you have the equation of the line

y - y1 = -1/2*(x - x1) where you have a known point (x1,y1) and are solving for when both intersect.

This means that you have two equations in two unknowns where:

y = -1/2*x + (y1 + 1/2*x1) and
y = 2x

By solving this system you solve the projection of an arbitrary point onto that line.