# Thread: Orthogonal projection on the line y=x?

1. ## Orthogonal projection on the line y=x?

Hi I need help with a linear algebra problem please.

Let T:R^2->R^2 be the orthogonal projection on the line y=x.

(a) Find a formula for T(x,y)
I don't know where to start on this one because I don't know how to define the transformation.If anyone could explain the transformation and process to find the formula it would be greatly apprerciated.

(b)Use the formula in (a) to show that T^2=T.
Here I just want to know whether T^2 is the composition of T with T?

(c) Find the eigenvalues an bases of the eigenspaces of T.
Here I just need to know how do I set up the matrix whose eigenvalues I must find?

Any input would be greatly appreciated.

2. Originally Posted by chocaholic
Hi I need help with a linear algebra problem please.

Let T:R^2->R^2 be the orthogonal projection on the line y=x.

(a) Find a formula for T(x,y)
I don't know where to start on this one because I don't know how to define the transformation.If anyone could explain the transformation and process to find the formula it would be greatly apprerciated.

To orthogonally project a general vector $\binom {x}{y}$ into a line with basis $\binom{a}{b}$ you must map $\binom{x}{y}\mapsto \frac{\binom {x}{y}\cdot \binom{a}{b}}{a^2+b^2}\,\binom{a}{b}$ .

In your case, you can choose $\binom{a}{b}=\binom{1}{1}$ and get the general form from the above. Of course, the dot denotes the usual inner product in $\mathbb{R}^2$

(b)Use the formula in (a) to show that T^2=T.
Here I just want to know whether T^2 is the composition of T with T?

Yes, and when you'll get the form for $T$ it'll be obvious that $T^2=T$

Tonio

(c) Find the eigenvalues an bases of the eigenspaces of T.
Here I just need to know how do I set up the matrix whose eigenvalues I must find?

Any input would be greatly appreciated.

.

3. Hi again.

I understand the orthogonal projection part as that is the formula, but why can (a,b) be (1,1)? Since we're working in R^2 shouldn't we have 2 basis vectors like (1,0) and (0,1)? And once we have the new vector/s is/are that T? How does this relate to x=y? unless (1,1) is a basis of x=y if so then how do we confirm this?

Thanks for the help.

4. Originally Posted by chocaholic
Hi again.

I understand the orthogonal projection part as that is the formula, but why can (a,b) be (1,1)?

It has to be a basis for the subspace on which we're going to project, and that's why (1,0) and (0,1) don't work.

Tonio

Since we're working in R^2 shouldn't we have 2 basis vectors like (1,0) and (0,1)? And once we have the new vector/s is/are that T? How does this relate to x=y? unless (1,1) is a basis of x=y if so then how do we confirm this?

Thanks for the help.
.

5. Ok so I understand why it's (1,1) and finding the orthogonal projection results in the formula:
T(x,y)=T((x+y)/2,(x+y)/2)
Is this correct? How do you set up a matrix of this formula?

6. Originally Posted by chocaholic
Ok so I understand why it's (1,1) and finding the orthogonal projection results in the formula:
T(x,y)=T((x+y)/2,(x+y)/2)
Is this correct? How do you set up a matrix of this formula?

The formula is $T\binom{x}{y}=\binom {\frac{x+y}{2}}{\frac{x+y}{2}}$ , and then $T=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}\\{}\\\fra c{1}{2}&\frac{1}{2}\end{pmatrix}$

Tonio

7. O I get it because its 1/2x+1/2y and 1/2 is the coefficient.
Thanks a million. This really cleared things up for me!

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### the orthogonal projection on the plane defined by equation y=x

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