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Is it true that for any two matrices $A \preceq B \iff C^T A C \preceq C^T B C$ for any conformable matrix $C$?

This related question proves ($\Rightarrow$) We have $x^TC^T(B-A)C x = (Cx)^T(B-A)C x \ge 0$ for any conformable vector $x$ so that $C^T(B-A)C \ge 0.$

What about the other direction?

  • What do you mean by "any conformable matrix"? If you're allowed to have C=I (the identity matrix) then the answer is yes (trivially). On the other hand if you restrict to C = 0 (the zero matrix) the answer is no (trivially). – ors Mar 26 '21 at 12:49
  • Any non-trivial matrix (different from zero and identity) such that the dimension of the matrix multiplication agree –  Mar 26 '21 at 12:53
  • see https://en.wikipedia.org/wiki/Conformable_matrix –  Mar 26 '21 at 12:53

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The converse is clearly false. Consider $C=0$ for instance.

For another counterexample we have $\pmatrix{1&0}\pmatrix{1&0\\ 0&-1}\pmatrix{1\\ 0}\succeq0$ but $\pmatrix{1&0\\ 0&-1}$ is indefinite.

user1551
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