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Math Help - Matrix of Linear Trans. in basis i, j, k

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    Matrix of Linear Trans. in basis i, j, k

    Let \phi: R^3 \rightarrow R^3 be the projection of the three-dimensional space onto the plane x + y + z = 0. Find the matrix of this linear transformation in the standard basis i, j, k.

    [Hint: Find the coordinates of the normal vector N to the plane. Then use the fact that the projection of a basis vector m (m = i, j, or k) onto the plane is m - a where a is the projection of m onto the direction of N.]
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    Re: Matrix of Linear Trans. in basis i, j, k

    Hint 2: A normal vector to the plane ax + by + cz = d is (a, b, c).

    Hint 3: The projection of m onto the direction of N is (m, N) / |N| (here (m, N) is the scalar, or dot, product).
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    Re: Matrix of Linear Trans. in basis i, j, k

    Quote Originally Posted by emakarov View Post
    Hint 2: A normal vector to the plane ax + by + cz = d is (a, b, c).

    Hint 3: The projection of m onto the direction of N is (m, N) / |N| (here (m, N) is the scalar, or dot, product).
    So from your hints I would say that the coordinates of the normal vector N to the plane is (1,1,1) right? So N = (1 1 1)? I am not sure I understand what m is?
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    Re: Matrix of Linear Trans. in basis i, j, k

    Quote Originally Posted by vidomagru View Post
    So from your hints I would say that the coordinates of the normal vector N to the plane is (1,1,1) right? So N = (1 1 1)?
    Yes.

    Quote Originally Posted by vidomagru View Post
    I am not sure I understand what m is?
    Have you read the hint in your own post?
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    Re: Matrix of Linear Trans. in basis i, j, k

    Quote Originally Posted by emakarov View Post
    Yes.

    Have you read the hint in your own post?
    I really do not understand what you are getting at here.

    m = i, j, or k for i = (1,0,0), j = (0,1,0), k = (0,0,1) and N = (1,1,1), but you N is not a n x n matrix so how would you take the determinant?
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    Re: Matrix of Linear Trans. in basis i, j, k

    Quote Originally Posted by vidomagru View Post
    I really do not understand what you are getting at here.
    Even though you are saying this, you answered one of your own questions (about what m is), so you are making progress.

    Quote Originally Posted by vidomagru View Post
    N is not a n x n matrix so how would you take the determinant?
    Vertical bars denote many things in mathematics. Here, |N| is the length of vector N.

    You may wish to review the material about projections and scalar (dot) product.
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    Re: Matrix of Linear Trans. in basis i, j, k

    I am sorry, hint 3 from post #2 should be as follows.

    Hint 3: The projection of m onto the direction of N is ((m, N) / |N|2) N, where (m, N) is the scalar, or dot, product.

    Previously I said that the projection was (m, N) / |N|, but this is the scalar projection, not a vector. To get the vector projection onto the direction of N, the scalar projection must be multiplied by the unit vector in the direction of N, which is N / |N|. This article in Wikipedia describes projections well.
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    Re: Matrix of Linear Trans. in basis i, j, k

    Quote Originally Posted by emakarov View Post
    I am sorry, hint 3 from post #2 should be as follows.

    Hint 3: The projection of m onto the direction of N is ((m, N) / |N|2) N, where (m, N) is the scalar, or dot, product.

    Previously I said that the projection was (m, N) / |N|, but this is the scalar projection, not a vector. To get the vector projection onto the direction of N, the scalar projection must be multiplied by the unit vector in the direction of N, which is N / |N|. This article in Wikipedia describes projections well.
    Ok, yes in the research I was just doing I had seen that it should be ((m, N) / |N|2) N. So here is what I have done:
    a = proj_N(m) = (m, N) / |N|2) N = m, N) / 3) N.
    Now for m = i = (1,0,0) we have a_1 = \dfrac{1}{3}.

    Now for m = j = (0,1,0) we have a_2 = \dfrac{1}{3}.

    Now for m = k = (0,0,1) we have a_3 = \dfrac{1}{3}.

    so a = (\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}). From here we should do m - a. Do I do (i, j, k) - (\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}) to give me a 3x3 matrix?
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    Re: Matrix of Linear Trans. in basis i, j, k

    I'll drop the bold font.

    Quote Originally Posted by vidomagru View Post
    a = proj_N(m) = (m, N) / |N|2) N = m, N) / 3) N.
    I agree, but what is the type of the left- and right-hand sides? It's a vector, not a number. Getting a vector was the reason the scalar projection (m, N) / |N| was multiplied by the unit vector N / |N|.

    Quote Originally Posted by vidomagru View Post
    Now for m = i = (1,0,0) we have a_1 = \dfrac{1}{3}.
    But here I understand that a₁ is a number. Which formula did you use to get a number?

    Quote Originally Posted by vidomagru View Post
    so a = (\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}).
    You used three vectors i, j and k to get one? Where did you get such formula?

    ((i, N) / |N|) N is the projection of i onto N (the result is a vector), and similarly for j and k. You subtract that projection from the corresponding original vector (i, j or k), and you get the projection of that vector onto the plane.
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    Re: Matrix of Linear Trans. in basis i, j, k

    Quote Originally Posted by emakarov View Post
    I'll drop the bold font.

    I agree, but what is the type of the left- and right-hand sides? It's a vector, not a number. Getting a vector was the reason the scalar projection (m, N) / |N| was multiplied by the unit vector N / |N|.

    But here I understand that a₁ is a number. Which formula did you use to get a number?

    You used three vectors i, j and k to get one? Where did you get such formula?

    ((i, N) / |N|) N is the projection of i onto N (the result is a vector), and similarly for j and k. You subtract that projection from the corresponding original vector (i, j or k), and you get the projection of that vector onto the plane.
    From ((i, N) / |N|) N one has ((1,0,0) \cdot(1,1,1)/3)N = \dfrac{1}{3}N = (\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}). So for m = i one has i - a = (\dfrac{2}{3},\dfrac{-1}{3},\dfrac{-1}{3})? Then for m = j one has j - a = (\dfrac{-1}{3},\dfrac{2}{3},\dfrac{-1}{3})? And finally for m = k one has k - a = (\dfrac{-1}{3},\dfrac{-1}{3},\dfrac{2}{3})? How does that translate into the matrix of linear transformation?
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    Re: Matrix of Linear Trans. in basis i, j, k

    Quote Originally Posted by vidomagru View Post
    From ((i, N) / |N|) N one has ((1,0,0) \cdot(1,1,1)/3)N = \dfrac{1}{3}N = (\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}). So for m = i one has i - a = (\dfrac{2}{3},\dfrac{-1}{3},\dfrac{-1}{3})? Then for m = j one has j - a = (\dfrac{-1}{3},\dfrac{2}{3},\dfrac{-1}{3})? And finally for m = k one has k - a = (\dfrac{-1}{3},\dfrac{-1}{3},\dfrac{2}{3})?
    Correct.

    Quote Originally Posted by vidomagru View Post
    How does that translate into the matrix of linear transformation?
    The matrix consists of the coordinates of images of basis vectors written as columns.
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    Re: Matrix of Linear Trans. in basis i, j, k

    Quote Originally Posted by emakarov View Post
    Correct.

    The matrix consists of the coordinates of images of basis vectors written as columns.
    So just \begin{bmatrix} \dfrac{2}{3}&\dfrac{-1}{3}&\dfrac{-1}{3}\\ \\ \dfrac{-1}{3}&\dfrac{2}{3}&\dfrac{-1}{3}\\ \\ \dfrac{-1}{3}&\dfrac{-1}{3}&\dfrac{2}{3} \end{bmatrix}?
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    Re: Matrix of Linear Trans. in basis i, j, k

    I \{u_1, u_2, ..., u_m\} is a basis for U and \{v_1, v_2, ..., v_n\} is a basis for V, A is a linear transformation from U to V,
    then the "matrix representation" of A in those two bases requires calculating [tex]A(u_i)[tex] and writing as a linear combination of the v_i vectors. The coefficents of v_i in A(u_j) form the i^{th} column of matrix.

    So what are (1, 0, 0), (0, 1, 0), and (0, 0, 1) projected to?
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    Re: Matrix of Linear Trans. in basis i, j, k

    Quote Originally Posted by vidomagru View Post
    So just \begin{bmatrix} \dfrac{2}{3}&\dfrac{-1}{3}&\dfrac{-1}{3}\\ \\ \dfrac{-1}{3}&\dfrac{2}{3}&\dfrac{-1}{3}\\ \\ \dfrac{-1}{3}&\dfrac{-1}{3}&\dfrac{2}{3} \end{bmatrix}?
    Yes.
    Last edited by emakarov; October 22nd 2013 at 04:54 PM.
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