# Ker Positive Definite Matrix

Let $$K = A^T C A$$, 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 $$K_{ij} = v_i^T * C * v_j$$, 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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How does this look?

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

Now do the same thing for K transpose. $$0 = K^T x = (A^T C A)^T x = A^T C^T A x = y^T C^T y$$ where $$y = Ax$$. 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.

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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