# Non-conservatively bound a positive definite matrix

Suppose a matrix $$A_k$$ is a symmetric positive definite real matrix. As $$k$$ increases, $$A_k$$ may change but will always be symmetric positive definite. What I want to do is to have an operation $$f:\mathbb{R}^{n\times n}\rightarrow\mathbb{R}^{n\times n}$$ on this matrix $$A_k$$ such that $$f(A_k)\preceq B$$. Note that when $$A_k$$ and $$B$$ are incomparable, the operation $$f$$ will also make sure that $$f(A_k)\preceq B$$. We can assume that $$B$$ is a simple matrix like $$c I$$, where $$c>0$$.

Is there a non-conservative and (hopefully) efficient way to do this? By non-conservative, I mean I want the matrix $$A_k$$ to change as little as possible.

One possible way to do it is to do a matrix refactorization, $$A_k = Q_k\Lambda_k Q_k^\top$$, and change the eigenvalues in $$\Lambda_k$$ to make it upper bounded by the matrix $$B$$. But apparently that's not efficient.

• Do you care what happens if $A$ and $B$ are incomparable (i.e., neither $A\succeq B$, nor $B\succeq A$?) Jan 15, 2021 at 18:40
• Yes, thanks for mentioning that. If $A_k$ and $B$ are incomparable, after applying the operation $f$, the result should also be $f(A_k)\preceq B$. And also the matrix is a simple matrix like $c I$, where $c>0$. So for triggering the operation, we can find the maximum eigenvalue of $A_k$ and compare it with $c$.
– Evan
Jan 15, 2021 at 18:47