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For a given positive semi-definite matrix $A$, expressed using singular value decomposition as: $A=UD^2U'$ (subject to the orthonormality conditions), any positive semi-definite matrix $X$ that satisfies $A - X \succeq 0$ can be expressed in terms of $X= U Z^2 U'$ subject to the constraint that $\forall i .D_{ii}^2 - Z_{ii}^2 \geq 0$.

Please bear in mind, the eigenvalues of $X$, i.e. $Z_{ii}^2$ can appear in any order, as long as they satisfy the above constraint.

I have done some numerical examples and it seems to support the statement, but I don't have a proof for it yet.

I really appreciate a counterexample to the statement or a proof.

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Notice that $Z$ does not need to be diagonal. Thus, $Z_{ii}^2$ need not be an eigenvalue of $X$. In fact $A$ and $X$ are simultaneous diagonalizable if and only if $AX = XA$.

Now the proof:

We have $$A - X = U D^2 U' - U Z^2 U' = U (D^2 - Z^2) U' \succeq 0$$ if and only if $$ D^2 - Z^2 \succeq 0 $$ as $U$ is unitary. Now, let $e_i$ be the unit vector with $1$ at the $i$-th component and otherwise $0$. Then, we have $$ D^2_{ii} - Z^2_{ii} = e_i'(D^2 - Z^2) e_i \ge 0. $$

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  • $\begingroup$ Does this mean that there is no positive semi-definite matrix $X$ that satisfies $A-X \succeq 0$ but cannot be expressed as $X=UZ^2U'$ where $U$ is the unitary matrix of SVD of $A$? $\endgroup$
    – Amir
    Commented Aug 18, 2015 at 23:57
  • $\begingroup$ double negation is evil :) For every $X\succeq 0$ we have $U'XU \succeq 0$. However, it needs not be diagonal. If $AX = XA$, then $U'XU$ is diagonal. $\endgroup$
    – user251257
    Commented Aug 19, 2015 at 0:00
  • $\begingroup$ My bad ;D Thanks alot. $\endgroup$
    – Amir
    Commented Aug 19, 2015 at 0:02

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