I am reading a paper that tries to find an upper bound for a phrase and for this purpose, it uses something like the following:
$$ I - A \succeq 0 \quad \text{so} \quad A \preceq I \quad \text{then} \quad x^TABx \leq x^TBx $$
I can't find the proof of this term, maybe it is obvious, but I do not get the intuition or any proof for it. I would appreciate any hint for the proof.
Some information about $A$ and $B$ which might be helpful:
- $A$, $B$ and $AB$ are symmetric
- Both $A$ and $B$ are positive semi-definite
If $B$ is positive semidefinite, there's an orthonormal basis $(e_i)$ of its eigenvectors with eigenvalues $\lambda_i\ge 0$, and we can verify the statement for $x=e_i$: $${e_i}^TABe_i=\lambda_i\,{e_i}^TAe_i\le \lambda_i\,{e_i}^Te_i={e_i}^TABe_i\,.$$
Now if $A$ also commutes with $B$ ($AB=BA$) then they share a common orthonormal eigenbasis, and thus the above inequalities extend to any vector $x$ because all the mixed terms ${e_i}^TABe_j$ become zero with $i\ne j$.