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Ker Positive Definite Matrix

  • Thread starter Mindscrape
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  • #1
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Let [tex]K = A^T C A[/tex], where C>0. Prove that kerK=cokerK=kerA, and rngK = corngK = corng A.

I sort of need a kickstart to get going. I know that each element will be [tex]K_{ij} = v_i^T * C * v_j[/tex], so this is sort of like a Gram matrix, which in turn also means that the matrix is semi-positive definite. I am not quite what sure to do with the A matrix though. Clearly, if the columns of A are linearly independent, then the range of A has dimension m if A is an m x n matrix, and so the kernel for A will have a dimension of n-m. From here, I think I will have to show that the dimension of kerK (will be 0 is C is positive definite), is the same as kerA, and then show that the bases are linearly dependent.
 

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  • #2
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How does this look?

If kerA is all A such that [tex]Ax=0[/tex] then kerK will be [tex]Kx = A^T C A x = 0[/tex] and thus shows that ker A is contained in ker K. Then taking [tex]Kx = 0[/tex] then [tex] 0 = x^T K x = x^T A^T C A x = y^T C y[/tex] where [tex]y = A x[/tex]. Since C>0 then y = 0, and [tex] x \in ker A[/tex].

Now do the same thing for K transpose. [tex]0 = K^T x = (A^T C A)^T x = A^T C^T A x = y^T C^T y [/tex] where [tex]y = Ax[/tex]. If C>0 then C transpose must also be greater than the zero vector, and the portion that matters is the 'y' vector, which is the same as the previous one. This shows that cokerK = kerA, and in turn cokerK = kerK = kerA.
 
  • #3
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Hmm... now for range, should I use fund thm of lin alg to show rngA and rngK have the same dimension, and then I would have to show their bases are linearly dependent, unless there is something tricky to do with the rank.

*Also, the matrix K would not be sort of like a Gram matrix, it is a Gram matrix.
 
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