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Math Help - Linear transformation orthogonal projection formula.

  1. #1
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    Unhappy Linear transformation orthogonal projection formula.

    I have an orthonormal basis for the subspace defined by a plane which is [[2/sqrt(5),1/sqrt(5),0],[2/(3*sqrt(5)),-4/(3*sqrt(5)),sqrt(5)/3]] but i need to find a formula for the orthogonal projection onto the plane how would i do this?. I know that the orthogonal projection of a vector is given by (u dot v1)v1+...(u dotvr)vr. How would i find my formula for the plane?

    Thanks
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  2. #2
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    Re: Linear transformation orthogonal projection formula.

    Hey mcleja.

    Hint: The distance between the point and the plane is given by n . p + d where n is the normal and p is the point with plane equation ax + by + cz + d = 0 where the normal is (a,b,c).

    You then move the point along the normal by calculating p' = p - n*(n.p + d) which "backtracks the point" along the normal in the reverse distance.
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  3. #3
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    Re: Linear transformation orthogonal projection formula.

    i have a dislike of "formulas", i prefer to think about what the thing i am doing MEANS.

    let's first find the orthogonal complement of your plane. for such a vector (x,y,z) it has to satisfy:

    2x + y = 0
    2x - 4y + 5z = 0

    substituting y = -2x in the second equation gives us:

    2x + z = 0

    so picking z as a free parameter, we get that the orthogonal complement of your plane is vectors of the form z(-1/2,1,1). normalizing (-1/2,1,1) we obtain (-1/3,2/3,2/3) as a unit basis vector.

    now we have an orthonormal basis for R3: B = {(2/√5,1/√5,0),(2/(3√5),-4/(3√5),√5/3),(-1/3,2/3,2/3)}.

    let call these vectors u1,u2, and u3. the matrix for our projection P in the basis B is easy to write down, it is:

    [P]_B = \begin{bmatrix}1&0&0\\0&1&0\\0&0&0 \end{bmatrix}

    since P(u1) = u1, P(u2) = u2 and P(u3) = 0.

    unfortunately, we need a "change of basis" matrix to get a formula in terms of "standard coordinates". the matrix we WANT is the one that changes "standard to B-coordinates" so we can apply [P]B. but the matrix that is easy to find is "B-coordinates to standard". let's call these two matrices C and C-1. C-1 is the easy one, its columns are just the vectors in B:

    C^{-1} = \begin{bmatrix}\frac{2}{\sqrt{5}}&\frac{2}{3\sqrt{  5}}&\frac{-1}{3}\\ \frac{1}{\sqrt{5}}&\frac{-4}{3\sqrt{5}}&\frac{2}{3}\\0&\frac{\sqrt{5}}{3}& \frac{2}{3} \end{bmatrix}

    however, since we have a change of basis between two orthonormal bases, C is an orthogonal matrix, so C = (C-1)-1 = (CT)-1 = (C-1)T.

    so the matrix we want that defines P in the standard basis is:

    C^{-1}[P]_BC = \begin{bmatrix}\frac{2}{\sqrt{5}}&\frac{2}{3\sqrt{  5}}&\frac{-1}{3}\\ \frac{1}{\sqrt{5}}&\frac{-4}{3\sqrt{5}}&\frac{2}{3}\\0&\frac{\sqrt{5}}{3}& \frac{2}{3} \end{bmatrix} \begin{bmatrix}1&0&0\\0&1&0\\0&0&0 \end{bmatrix} \begin{bmatrix} \frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}&0\\ \frac{2}{3\sqrt{5}}& \frac{-4}{3\sqrt{5}} & \frac{\sqrt{5}}{3}\\ \frac{-1}{3}& \frac{2}{3}& \frac{2}{3} \end{bmatrix}

     = \frac{1}{9}\begin{bmatrix}8&2&2\\2&5&-4\\2&-4&5\end{bmatrix} <---we wind up with a symmetric matrix, as we should.

    so P(x,y,z) = (1/9)(8x+2y+2z,2x+5y-4z,2x-4y+5z).

    let's verify this does 2 things:

    1) leaves the basis vectors for the plane unchanged
    2) send the orthogonal complement of the plane to 0

    for (1): P(2/√5,1/√5,0) = (1/9)(16/√5+2/√5, 4/√5+5/√5, 4/√5-4/√5) = (1/9)(18/√5,9/√5,0) = (2/√5,1/√5,0)

    P(2/(3√5),-4/(3√5),√5/3) = P(2/(3√5),-4/(3√5),5/(3√5)) = (1/9)(16/(3√5)-8/(3√5)+10/(3√5), 4/(3√5)-20/(3√5)-20/(3√5), 4/(3√5)+16/(3√5)+25/(3√5)) = (1/9)(18/(3√5),-36/(3√5),45/(3√5)) = (2/(3√5),-4/(3√5),√5/3)

    for (2): P(-1/3,2/3,2/3) = (1/9)(-8/3+4/3+4/3,-2/3+10/3-8/3,-2/3-8/3+10/3) = (1/9)(0,0,0) = (0,0,0)

    i'm convinced, how about you?
    Last edited by Deveno; January 28th 2013 at 01:56 AM.
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  4. #4
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    Re: Linear transformation orthogonal projection formula.

    Im still not too sure about the why the vector (-1/3,2/3,2/3) is in the orthonormal basis why is it not just B = {(2/√5,1/√5,0),(2/(3√5),-4/(3√5),√5/3)}?

    thanks.
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  5. #5
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    Re: Linear transformation orthogonal projection formula.

    your vectors have 3 coordinates, your plane lives in 3-dimensional space. 2 vectors is not big enough for a basis of all of R3. finding the orthogonal complement is the same thing as finding the "normal vector to the plane" (except we don't pick a direction).

    so the basic idea is:

    a) find a basis in which our projection has a simple matrix. in three dimensions this is:

    spanning vector 1 + spanning vector 2 + normal vector 3

    b) find the change-of-basis matrix from the standard basis to the basis of the form: {basis of the plane} + {basis of the orthogonal complement}. the inverse of this matrix is usually what we find first (new basis to standard). then, if we have taken the time to make our plane+complement basis orthonormal, we can find the matrix we want by taking the transpose of the inverse matrix we have.

    c) line up all 3 matrices we have now, and multiply them together. if the change of basis matrix from new basis to standard basis is C, this is:

    CTMC, where M is the matrix we found in part (a).

    Cv takes the vector in the standard basis, to its form in the "plane" basis.

    MCv projects the vector Cv onto the plane, straight down (orthogonally).

    CTMCv takes the projected vector MCv back to the "old coordinates".
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  6. #6
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    Re: Linear transformation orthogonal projection formula.

    Quote Originally Posted by mcleja View Post
    I have an orthonormal basis for the subspace defined by a plane which is [[2/sqrt(5),1/sqrt(5),0],[2/(3*sqrt(5)),-4/(3*sqrt(5)),sqrt(5)/3]] but i need to find a formula for the orthogonal projection onto the plane how would i do this?. I know that the orthogonal projection of a vector is given by (u dot v1)v1+...(u dotvr)vr. How would i find my formula for the plane?

    Thanks
    You’re given two orthonormal vectors in the plan e1 and e2. Projection of v onto the plane is (v.e1)e1 + (v.e2)e2.

    A simple sketch makes this clear.

    For the equation of the plane,
    Normal n to the plane is e1Xe2
    if r = xi + yj + zk is any point in the plane:
    r.n = 0 is the equation of the plane
    Last edited by Hartlw; January 28th 2013 at 09:48 AM. Reason: X should be . in last line
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  7. #7
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    Re: Linear transformation orthogonal projection formula.

    if one actually carries out Hartlw's procedure, this gives (i will use u1 and u2, since e1 and e2 are often reserved for the standard basis vectors (1,0,0), (0,1,0)), for v= (x,y,z):

    (\mathbf{v}\cdot\mathbf{u}_1)\mathbf{u}_1 + (\mathbf{v}\cdot\mathbf{u_2})\mathbf{u}_2

    =\left((x,y,z)\cdot\left(\frac{2}{\sqrt{5}},\frac{  1}{\sqrt{5}},0\right)\right)\left(\frac{2}{\sqrt{5  }},\frac{1}{\sqrt{5}},0\right) + \left((x,y,z)\cdot\left(\frac{2}{3\sqrt{5}},\frac{-4}{3\sqrt{5}},\frac{\sqrt{5}}{3}\right)\right) \left( \frac{2}{3\sqrt{5}},\frac{-4}{3\sqrt{5}},\frac{\sqrt{5}}{3} \right)

    = \left(\frac{2}{\sqrt{5}}x + \frac{1}{\sqrt{5}}y\right) \left(\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}},0 \right) + \left( \frac{2}{3\sqrt{5}}x - \frac{4}{3\sqrt{5}}y + \frac{\sqrt{5}}{3}z\right) \left(\frac{2}{3\sqrt{5}},\frac{-4}{3\sqrt{5}},\frac{\sqrt{5}}{3} \right)

    = \left(\frac{4}{5}x + \frac{2}{5}y, \frac{2}{5}}x + \frac{1}{5}y, 0 \right) + \left( \frac{4}{45}x - \frac{8}{45}y + \frac{2}{9}z, \frac{-8}{45}x + \frac{16}{45}y - \frac{4}{9}z,\frac{2}{9}x - \frac{4}{9}y + \frac{5}{9}z \right)

     = \frac{1}{9}(8x+2y+2z,2x+5y-4z,2x-4y+5z)

    which is the same answer we obtained above. this procedure is also perfectly fine, and if you prefer to use it, and find it easier to remember, by all means do so.
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  8. #8
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    Re: Linear transformation orthogonal projection formula.

    ok, i understand now. Thanks everybody!
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