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What subspace of 3x3 matrices is spanned by rank 1 matrices

  1. Oct 6, 2015 #1
    So that's the question in the text.

    I having some issues I think with actually just comprehending what the question is asking me for.

    The texts answer is: all 3x3 matrices.

    My answer and reasoning is:

    the basis of the subspace of all rank 1 matrices is made up of the basis elements

    [tex]\begin{bmatrix}1 & 1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix},\begin{bmatrix}0 & 0 & 0\\ 1 & 1 & 1\\ 0 & 0 & 0\end{bmatrix},\begin{bmatrix}0 & 0 & 0\\ 0 & 0 & 0\\ 1 & 1 & 1\end{bmatrix},\begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 0\\ 1 & 0 & 0\end{bmatrix},\begin{bmatrix}0 & 1 & 0 \\ 0 & 1 & 0\\ 0 & 1 & 0\end{bmatrix},\begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & 1\\ 0 & 0 & 1\end{bmatrix}[/tex]

    I figure these are the minimum elements you need to create any and all rank 1 matrices. By linearly combining these matrices you can make all rank 1 matrices...why do the rank 1 matrices also span the space of all 3x3 matrices?
     
  2. jcsd
  3. Oct 6, 2015 #2

    RUber

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    What about the rank 1 matrices:
    ##\begin{bmatrix} 1&0&0\\0&0&0\\0&0&0 \end{bmatrix},\begin{bmatrix} 0&0&0\\0&1&0\\0&0&0 \end{bmatrix},\begin{bmatrix} 0&0&0\\0&0&0\\0&0&1 \end{bmatrix}.##
    Sum the three of these, what do you get?
    This matrix is clearly in the space of all 3x3 matrices, since any 3x3 matrix multiplied by it will still be in the space of 3x3 matrices.
     
  4. Oct 7, 2015 #3

    mathwonk

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    I guess I misunderstood the question, as I thought you wanted to describe the set of rank one matrices, which of course does not consitute a linear subspace, but rather a cone, (minus the vertex zero). Since such matrices are determined by their image which is spanned by one vector, plus the linear function mapping each source vector to a multiple of that vector, they are all products of form v.wperp, where v,w are column vectors. I.e. dotting with w takes a vector to a number,a nd then multiplying that number by v, takes it to a vector in the line spanned by v. However we can multiply through by t and 1/t and get the same map. So this cone seems to have dimension 2n-1, or 5, inside the 9 dimensional space of 3x3 matrices.

    However if you really just want to know what matrices can be written as linear combinations of matrices of rank one, well, that is easily shown to all 3x3 matrices, using the hints in the previous post, i.e. the usual basis of all 3x3 matrices contains only rank one matrices. Intuitively it is also unlikely that the non linear cone just described would lie in any hyperplane.
     
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