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Math Help - Orthogonal projection on the line y=x?

  1. #1
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    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.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by chocaholic View Post
    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.

    Thanks in advance.
    .
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  3. #3
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    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.
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  4. #4
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    Quote Originally Posted by chocaholic View Post
    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.
    .
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  5. #5
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    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?
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  6. #6
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    Quote Originally Posted by chocaholic View Post
    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
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  7. #7
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    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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