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Thread: Orthogonal projection

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    Orthogonal projection

    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
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    Quote Originally Posted by champrock View Post
    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}.
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    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
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    Quote Originally Posted by champrock View Post
    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 />
.
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    Quote Originally Posted by champrock View Post
    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}
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    Quote Originally Posted by ThePerfectHacker View Post
    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
    ?
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    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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