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I Understanding rectangular matrices

  1. Oct 3, 2016 #1
    So if I have a system of equations:
    $$x_1+x_2+x_3=1$$
    and $$x_4+x_5+x_6=1$$

    Then they can be put into a matrix representation
    \begin{equation*}
    \begin{bmatrix}
    1 & 1 & 1 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & 1 & 1\\
    \end{bmatrix}
    \begin{bmatrix}
    x_1 \\
    x_2 \\
    x_3 \\
    x_4 \\
    x_5 \\
    x_6 \\
    \end{bmatrix}
    =
    \begin{bmatrix}
    1 \\
    1 \\
    \end{bmatrix}
    \end{equation*}




    So I know that columns in a matrix represent vectors. Is it true that in this matrix we therefore have 6 2D vectors?

    Also it looks like there are only 2 vectors, and 3 each of them.

    It's just surprising to me that there are 6 variables and only 2D vectors.

    If I imagine them to be 2D vectors in x-y plane, then they are also mutually perpendicular. So eventually, the equations I mentioned above, although they look like planes, are they just really the x and y axes?

    I guess I'm confused in going from the vector interpretation to the equation interpretation of the matrix.

    I would appreciate help in understanding how to interpret the matrix form
     
  2. jcsd
  3. Oct 3, 2016 #2

    Erland

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    Does this originate in some textbook problem? Could you then please state it from the book....

    It is not clear to me which 2D-vectors you are talking about. Also, your two equations are independent in the sense that they can be solved separately, and I see no reason to put them together in a system of equations.
     
  4. Oct 3, 2016 #3
    It's from a linear programming question in standard form. It's a part of a worksheet we got, but I didn't want help with the whole question (so I didn't put it in the homework help forum), just understanding the meaning of the rectangular matrix. I've attached the question here for context.

    By 2D vectors I mean
    \begin{pmatrix}
    1\\
    0
    \end{pmatrix}
    and
    \begin{pmatrix}
    0\\
    1
    \end{pmatrix}

    There are 3 each of them in the matrix.
     

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  5. Oct 3, 2016 #4

    BvU

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    You are mapping 6 variables (so a 6 dimensional space) onto 2 numbers (so a 2 dimensional space).
    The 6 2D vectors you mention are the images of the 6 unit vectors in the 6D domain space.
    So ##(1,0,0,0,0,0,0)\rightarrow(1,0)## etc.
    These images are clearly not perpendicular: ##(0,1,0,0,0,0,0)\rightarrow(1,0)## also, etc.

    With 6 DImensions and 2 constraints (your 2 equations) you have four degrees of freedom.
    6D vectors that satisfy your equations look like e.g. ##(x-1, x_2, 1-x_1-x_2,x_4,x_5,1-x_4-x_5)## so they are definitely not the axes.
     
    Last edited: Oct 4, 2016
  6. Oct 3, 2016 #5
    Can you elaborate a bit more? In a regular Ax=b formulation, where A is, say, a 3x3 matrix, the columns are vectors in 3D space. the elements of x are 'weighting factors' which multiply the columns (vectors) and the result b is the vector sum.

    So we have

    \begin{equation*}
    \begin{pmatrix}
    1 & 2 & 3\\
    5 & 7 & 8 \\
    3 & 1 & 12\\
    \end{pmatrix}
    \begin{pmatrix}
    x\\
    y \\
    z\\
    \end{pmatrix}
    =
    \begin{pmatrix}
    1x+2y+3z\\
    5x+7y+8z \\
    3x+1y+12z\\
    \end{pmatrix}
    =
    x\begin{pmatrix}
    1\\
    5 \\
    3\\
    \end{pmatrix}
    +
    y\begin{pmatrix}
    2\\
    7 \\
    1\\
    \end{pmatrix}
    +
    z\begin{pmatrix}
    3\\
    8 \\
    12\\
    \end{pmatrix}
    \end{equation*}

    Why is this not the same for my example? Why are the columns in my rectangular matrix not 2D vectors?
     
  7. Oct 3, 2016 #6

    BvU

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    Sounds weird when I just wrote
    thereby calling them vectors, isn't it ?

    Exactly analogous to your 3 x 3 example, where ##
    \begin{pmatrix}
    1\\
    5 \\
    3\\
    \end{pmatrix}## is the image of unit vector ##
    \begin{pmatrix}
    1\\
    0 \\
    0\\
    \end{pmatrix}##
     
  8. Oct 4, 2016 #7
    Ok so thinking about this a bit more, I'm getting most of it. You say that since ##(1,0,0,0,0,0,0)\rightarrow(1,0)##
    and also that ##(0,1,0,0,0,0,0)\rightarrow(1,0)##,
    The 2D vectors are not perpendicular? I'm just struggling with the idea of why the 2D vectors in their 2D system aren't perpendicular. As I said, it's just easiest for me to imagine them as the x and y axes. Thanks for your help!
     
  9. Oct 4, 2016 #8

    BvU

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    Yes, ##(1,0)## and ##(1,0)## are clearly one and the same and therefore not perpendicular. You can't find more than two mutually perpendicular vectors in 2D. As it happens the images of ##(1,0,0,0,0,0)## and ##(0,0,0,1,0,0)## are perpendicular, but you can't say: "the 6 unit vectors in the 6D domain of this matrix have images in the 2D range space that are perpendicular"
     
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