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Non singular matrix M such that MAM^T=F

  1. Mar 26, 2016 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Show that there is a non-singular matrix M such that ##MAM^T = F## for any antisymmetric matrix A where the normal form F is a matrix with 2x2 blocks on its principal diagonal which are either zero or $$\begin{pmatrix} 0 &1 \\ -1&0 \end{pmatrix}$$

    To do so, consider the analogue of the Gram-Schmidt method for antisymmetric matrices using the antisymmetric inner product ##\langle x,y \rangle = x^TA y##

    2. Relevant equations
    Gram Schmidt equations?
    Orthogonality

    3. The attempt at a solution
    I am just really looking for some hints to start. For vectors ##x## and ##y##, $$x^T A y = (x_1 \dots x_n) \begin{pmatrix} A_{11} & A_{12} &..&A_{1n} \\ ..&.. \\ ..& ..&..&A_{nn} \end{pmatrix} \begin{pmatrix} y_1 \\ y_2\\ ..\\y_n \end{pmatrix}$$ and ##A_{ii} = 0, A_{ij} = -A_{ji} ## but I am not seeing how the hint is useful.

    Thanks!
     
  2. jcsd
  3. Mar 27, 2016 #2
    I feel that it is a way or another related to the spectral theorem. Try to work on ##A^2##.
     
  4. Mar 27, 2016 #3

    CAF123

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    In the 2x2 case, ##A^2## is proportional to the identity but this is not true for the 3x3 case and beyond so I am not sure what can be said about ##A^2## in general.

    The product ##x^T A x## is the matrix multiplication of a row of ##M^T## together with ##A## and corresponding transposed row (or column) vector in ##M##. The matrix ##M## is composed of the ##2n## vectors ##x## and amounts to the transition matrix from one basis to another. Is that correct? I guess Gram Schmidt comes into play when we want to work with some arbitrary basis and ##M## is the basis change of matrix that takes us to the basis in which we now have an orthogonal set of basis vectors. I just don't see a way to proceed as of yet.

    Thanks!
     
  5. Mar 27, 2016 #4
    Let's say that ##A## is a real matrix.
    If ##A## is anti-symmetric (## A^T = -A##), then ##A^2## is symmetric. The spectral theorem says it is diagonalizable in an orthonormal basis.
    Could the diagonal matrix be a representation of ##F^2## in another orthonormal basis ? I don't know, it is just a suggestion, this is how I would start.
    If yes, you will get an equality that looks like ##(M A M^T)^2 = F^2##, with ##M## being a product of 2 orthogonal matrices, therefore invertible. But still, there are these squares you'll have to get rid of.
     
  6. Mar 27, 2016 #5

    CAF123

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    Ok, so given that ##A^2## is always symmetric it can always be diagonalised in a basis spanned by its (normalised) eigenvectors. So, something like $$RA^2R^T = \text{diag}(\lambda_1, \dots \lambda_{2n})$$ Then multiply by ##P## on the left and by ##P^T## on the right so that $$PR A^2 R^T P^T = P\text{diag}(\lambda_1, \dots \lambda_{2n}) P^T \overset{!}{=} F^2.$$ If ##M = PR## then get ##(MAM^T)^2 = F^2##

    So somehow I need to prove that it is possible to go from a diagonal form ##D = \text{diag}(\lambda_1 \dots \lambda_{2n})## to ##F^2##? ##F^2## is so specific though about the elements on its block diagonal. And to resolve the sign where I square root both sides i thought about considering ##M=\text{Id}## to get the sign right for det but this depends on how much I'm allowed to assume about ##\text{det F}## (I can see that it is only 1 or 0 though)

    Does the hint about Gram Schmidt help me at all and in particular using the quantity ##x^T A y##?
     
  7. Mar 28, 2016 #6

    CAF123

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    I got the following solution from another source:

    We suppose that ##A## is written in the canonical basis. Consider the product defined by ##\langle x,y\rangle=x^TAy##, the bilinear antisymmetric product. There exists a basis ##e_1,...e_{2n}## such that this product in this basis is ##e_{n+1}\wedge e_{1}+...+e_{2n}\wedge e_{n}##. The matrix of the bilinear for in this basis is ##\pmatrix{0 & 1 \cr -1 & 0}##.

    Let ##M## be the transtion matrix from the basis to ##e_1,...,e_{2n}## to the canonical basis ##M^TAM =\pmatrix{ 0 &1\cr -1 & 0}##

    I didn't completely understand this proof - What does the term ##e_{n+1}\wedge e_{1}+...+e_{2n}\wedge e_{n}## evaluate to and why is the bilinear in this basis in which the inner product is equal to that given by $$\begin{pmatrix} 0 & 1 \\ -1&0 \end{pmatrix}?$$
     
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