Orthogonal projection

**champrock** · September 15th 2009, 05:31 AM

I am not able to formulate the answer for the following question. I think it is one of the very basic questions in linear algebra..please help!

Find the matrix of the orthogonal projection in R2 onto the line x=-2y

**ThePerfectHacker** · September 15th 2009, 06:41 AM

Originally Posted by champrock

I am not able to formulate the answer for the following question. I think it is one of the very basic questions in linear algebra..please help!

Find the matrix of the orthogonal projection in R2 onto the line x=-2y

Let $\theta$ be the angle between this line and the negative x-axis.
Let $\pi_{\text{x}}$ be the projection onto the x-axis.
Let $R_{\theta}$ be the rotation by $\theta$ .
Then the matrix you are looking for is given by $R_{-\theta}\pi_{\text{x}}R_{\theta}$ .

**champrock** · September 15th 2009, 07:44 AM

can u tell me what will be the exact value of theta ? and what is the matrix $\pi_{\text{x}}$ in R2? i am slightly confused as to how to formulate this matrix.

thanks

**Opalg** · September 15th 2009, 09:07 AM

Originally Posted by champrock

Find the matrix of the orthogonal projection in R2 onto the line x=-2y

First, find a unit vector in the range of the projection. This is $\bigl(\tfrac{-2}{\sqrt5},\tfrac1{\sqrt5}\bigr)$ . Then the projection should take the vector (x,y) to $\bigl\langle(x,y),\bigl(\tfrac{-2}{\sqrt5},\tfrac1{\sqrt5}\bigr)\bigr\rangle \bigl(\tfrac{-2}{\sqrt5},\tfrac1{\sqrt5}\bigr) = \bigl(\tfrac45x-\tfrac25y, -\tfrac25x+\tfrac15y\bigr)$ (where the angled brackets denote the inner product). Thus the matrix of the projection is $\begin{bmatrix}\tfrac45&-\tfrac25\\ -\tfrac25&\tfrac15\end{bmatrix}<br />$ .

**ThePerfectHacker** · September 15th 2009, 07:11 PM

Originally Posted by champrock

can u tell me what will be the exact value of theta ? and what is the matrix $\pi_{\text{x}}$ in R2? i am slightly confused as to how to formulate this matrix.

thanks

Notice that $\tan \theta = \frac{1}{2}$ and $\pi_{\text{x}} = \begin{bmatrix} 1&0\\0&0 \end{bmatrix}$

**champrock** · September 17th 2009, 06:00 AM

Originally Posted by ThePerfectHacker

Notice that $\tan \theta = \frac{1}{2}$ and $\pi_{\text{x}} = \begin{bmatrix} 1&0\\0&0 \end{bmatrix}$

wont we need to somehow reduce the length of the vector also? Because the original vector has the length of (x^2 + y^2)^0.5 but the new projected vector will have a length of only (x) ?

Am i missing something or do we need to do it when we are doing similar transformations in R^3 ? for example when we are trying to project a vector in R^3 onto the x-y plane then what will be the matrix like? will be multiply the transformation matrix by
(x^2+y^2)^0.5 / (x^2+y^2 +z^2)^0.5
OR
(x^2+y^2 +z^2)^0.5 /(x^2+y^2)^0.5
?

**HallsofIvy** · September 17th 2009, 08:22 AM

Another way to do this:
(I assume you mean the matrix relative to the standard basis <1, 0> and <0, 1>. A linear transformation only has a basis relative to some basis.)

A vector in the direction of the line x= -2y is v= -2i+ j. The projection of u= i onto that is given by $\frac{u\cdot v}{v\cdot v}v$ which, here, is $\frac{-2}{5}(-2i+ j)= \frac{4}{5}i-\frac{2}{5}j$ .

The projection of w= j onto that is given by $\frac{1}{5}(-2i+ j)= -\frac{2}{5}i+ \frac{1}{5}j$ .

Since applying a linear transformation to the basis vectors in turn gives the columns of the matrix, the matrix corresponding to this projection, in the standard basis, is
$\begin{bmatrix}\frac{4}{5} & -\frac{2}{5} \\ -\frac{2}{5} & \frac{1}{5}\end{bmatrix}$

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