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Hyperplanes of ##M_n(\mathbb{C})##

  1. Feb 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that any hyperplane of ##M_n(\mathbb{C})## contains at least an invertible matrix

    2. Relevant equations
    Let ##H## be an hyperplane of ##M_n(\mathbb{C})##.

    3. The attempt at a solution

    By contradiction, assume that for any matrix ##A\in H##, ##A## is not invertible.
    Therefore 0 is an eigenvalue of A, and there exists a basis of ##M_{n,1}(\mathbb{C})## in which at least the first column of ##A## is 0.
    This can be translated to : there is a surjection ##\phi## between the vector space ##U = \{ M\in M_n(\mathbb{C}) : \forall i = 1...n\ m_{i1} = 0 \} ## and hyperplane ##H##.

    My problem to obtain a contradiction is that ##\phi## is not linear. If it was I would use the rank theorem and
    ## n^2 - 1 = \text{dim}(H) = \text{rk}(\phi) \le \text{dim}(U) = n^2 -n ##
    which is a contradiction as soon as ##n\ge 2##.

    How would you solve this ?
     
  2. jcsd
  3. Feb 13, 2016 #2

    Krylov

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    To me it is not immediately clear that there exists a basis that accomplishes this simultaneously for all ##A \in H##.
    I would perhaps start by noting that there exists a nonzero ##c \in \mathbb{C}^{n \times n}## such that ##H## is linearly isomorphic to
    $$
    \left\{a \in \mathbb{C}^{n \times n}\,:\,\sum_{j = 1}^n\sum_{i=1}^nc_{ij}\overline{a_{ij}} = 0 \right\}
    $$
    (Every hyperplane can be written as the kernel of some non-zero functional.) Since ##c## is non-zero, you could then solve for one of the entries of ##a## while keeping the freedom to choose the other ##n^2 - 1## entries at will.

    Does this help?
     
    Last edited: Feb 13, 2016
  4. Feb 13, 2016 #3
    We can write ##M_n(\mathbb{C}) = H \bigoplus \mathbb{C} M_0 ##, where ##H ## is the kernel of a linear form ##\phi : M_n(\mathbb{C}) \to \mathbb{C} ##, and ##\phi(M_0) \neq 0##

    Assuming that ##H## doesn't contain any invertible matrix, the family ##(M_i)_{i\in I}## of all invertible matrices has the form : ##M_i = A_i + \lambda_i M_0##, with ##\lambda_i \neq 0 ##.

    But that means that for all ##i\in I##, ##\phi(\lambda_i^{-1} M_i) = \phi(M_0) \neq 0 ##. This implies that for all ##i\in I##, ## \lambda_i^{-1} M_i \in \mathbb{C}M_0##

    But the restriction of ##\phi## to ##\mathbb{C}M_0## is a bijection onto ##\mathbb{C}##. Since the ##(\phi(\lambda_i ^{-1} M_i))_{i\in I}## are all equal to ##\phi(M_0)##, that means by injectivity that all invertible matrices should be proportional 2 by 2. This is absurd.

    Is it correct?
     
  5. Feb 13, 2016 #4

    fresh_42

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    It implies ## \lambda_i^{-1} M_i = \lambda_i^{-1} A_i + M_0 ∉ H##, i.e. its projection on ##\mathbb{C}M_0## equals ##1## and not that it is included in ##\mathbb{C}M_0##.
    I think that you somehow need to use the fact that ##GL_n(ℂ) ⊂ M_n(ℂ)## is a dense subset, i.e. you probably need to wiggle a little bit on a point ##M## in ##H## to find an invertible point ##M+εM'## without leaving ##H##.

    Edit: IMO induction on n seems to be the easiest way.
     
    Last edited: Feb 13, 2016
  6. Feb 13, 2016 #5
    Doomed ! Once again :woot:
     
  7. Feb 14, 2016 #6

    Samy_A

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    (bolding mine)
    Without loss of generality, you can assume ##a_{11}## is that one entry.
     
  8. Feb 14, 2016 #7
    I think I have it :

    Let ##(E_i)_{i = 1...n^2-1}## be a basis of ##H##. Then the matrices ## M_\lambda = E_i + \lambda E_j ## are in ##H##.
    But ##P(\lambda) = \det M_\lambda ## is a polynomial of ##\mathbb{C}_n[X]##.
    Therefore it has at most ##n## zeros and there is a complex ##\lambda_0## for which ##P(\lambda_0)\neq 0##.
    Meaning that ##M_{\lambda_0} \in H## is invertible.

    Is it good now?
     
  9. Feb 14, 2016 #8

    Samy_A

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    ##E_1=\begin{pmatrix}
    1 & 0\\
    0 & 0
    \end{pmatrix}
    ##, ##E_2=\begin{pmatrix}
    0& 1\\
    0 & 0
    \end{pmatrix}
    ##
    What is the polynomial ##P(\lambda)##?
     
  10. Feb 14, 2016 #9
    Hmmm I get it, but really, I'm starting to dry up with this problem !
     
  11. Feb 14, 2016 #10

    Samy_A

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    Look again at @Krylov 's tip and its implications.
     
  12. Feb 14, 2016 #11
    I did not understand what he said
     
  13. Feb 14, 2016 #12

    Samy_A

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    Since H is a hyperplane, the n² matrix elements of matrices in H satisfy a linear equation. Solve this equation for one the elements (without loss of generality, say ##a_{11}##): that element is then a linear combination of all the others. To get a matrix in H, you then have n²-1 degrees of freedom, meaning you can chose n²-1 matrix elements freely. The last one will then be a linear combination of these freely chosen n²-1 elements.
     
  14. Feb 14, 2016 #13
    I'm sorry, but this is above my head, I don't follow you.
    If you know how to solve it I'll be glad to read your solution. But it has already taken me too much time and I need to move on.
    Really sorry.

    Thread closed, unless someone posts a solution
     
  15. Feb 14, 2016 #14

    Samy_A

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    Is ##A=\begin{pmatrix}
    x & 0 & 0 & 0 & 1\\
    0 & 0 & 0 & 1 & 0\\
    0 & 0 & 1 & 0 & 0\\
    0 & 1 & 0 & 0 & 0\\
    1 & 0 & 0 & 0 & 0
    \end{pmatrix}## invertible?
     
  16. Feb 14, 2016 #15

    Krylov

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    Don't be afraid to ask me next time. @Samy_A, thank you for clarifying.
    I hope you are willing to let it rest a bit (maybe for a few days) and then reconsider with a fresh mind. Sometimes it takes a while to solve an exercise, don't give up on it.
     
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