Results 1 to 6 of 6

Math Help - Construct a matrix

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    24

    Construct a matrix

    v1 = (2 -1 2 1) v2 = (0 -1 1 -3) These vectors are column vectors.



    Find a 2x4 matrix A, with linearly independent rows, such that Av1 = 0 and Av2 = 0.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    A basis of <v_1,v_2>^{\perp} with the usual inner product provides the rows of A .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,444
    Thanks
    1863
    Or, find a, b, c, d, e, f, g, h such that
    \begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}2 \\ -1 \\ 2 \\ 1\end{bmatrix}= \begin{bmatrix} 2a- b+ 2c+ d \\ 2e- f+ 2g+ h\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}
    and
    \begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}0 \\ -1 \\ 1 \\ -3\end{bmatrix}= \begin{bmatrix} -b+ c- 3d \\ -f+ g- 3h\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}

    That gives you four equations to solve for 8 values. Of course, there will be an infinite number of such matrices just as there are an infinite number of possible bases for the orthogonal space FernandoRevilla suggests.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by HallsofIvy View Post
    Or, find a, b, c, d, e, f, g, h such that
    \begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}2 \\ -1 \\ 2 \\ 1\end{bmatrix}= \begin{bmatrix} 2a- b+ 2c+ d \\ 2e- f+ 2g+ h\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}
    and
    \begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}0 \\ -1 \\ 1 \\ -3\end{bmatrix}= \begin{bmatrix} -b+ c- 3d \\ -f+ g- 3h\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}

    That gives you four equations to solve for 8 values. Of course, there will be an infinite number of such matrices just as there are an infinite number of possible bases for the orthogonal space FernandoRevilla suggests.

    You are right, but we only need two equations and four unknowns.

    <v_1,v_2>^{\perp}\equiv\begin{Bmatrix} 2x_1-x_2+2x_3+x_3=0\\{}\quad -x_2+x_3-3x_4=0 \end{matrix}

    has dimension 2, so two linearly independent solutions of the system are "your" (a,b,c,d) and (e,f,g,h) , that is, we needn't to denote twice the unknowns.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2011
    Posts
    24
    Thanks so much for the help!

    I was just wondering what does the thing the looks like an upside down T by the (v1, v2) mean?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by skittle View Post
    I was just wondering what does the thing the looks like an upside down T by the (v1, v2) mean?

    The subspace of \mathbb{R}^4 whose elements x=(x_1,x_2,x_3,x_4) are orthogonal to v_1 and v_2. Equivalently:

    \begin{Bmatrix}(x_1,x_2,x_3,x_4)\cdot (2,-1,2,1)=0\\(x_1,x_2,x_3,x_4)\cdot (0,-1,1,-3)=0\end{matrix}\Leftrightarrow\begin{Bmatrix}2x_1-x_2+2x_3+x_4=0\\{}\quad\; \;-x_2+x_3-3x_4=0\end{matrix}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mathematica, how to construct a tridiagonal matrix?
    Posted in the Math Software Forum
    Replies: 3
    Last Post: August 23rd 2010, 03:38 AM
  2. Construct a histogram
    Posted in the Statistics Forum
    Replies: 1
    Last Post: June 7th 2010, 04:50 PM
  3. Construct an angle
    Posted in the Geometry Forum
    Replies: 8
    Last Post: June 6th 2010, 09:42 AM
  4. How to construct triangle
    Posted in the Geometry Forum
    Replies: 3
    Last Post: April 22nd 2010, 04:10 PM
  5. Construct equation.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: December 17th 2009, 08:52 AM

Search Tags


/mathhelpforum @mathhelpforum