Solving for positive semidefiniteness Given a real matrix M, is there a matrix function f(M) such that $f(M)-M$ is guaranteed to be positive semidefinite, other than the idea of multiplying $M$ with its transpose and apart from the trivial case where $f(M)=M$? If so, what would $f(.)$ be?
 A: Expanding on user Helms Answer: Let $M$ be diagonalizable and $M=A\Lambda A^{-1}$ for some invertible $A$ and $\Lambda$ as $M$'s eigenvalues. Let $f(M)$ be some polynomial function of $M$. Then
\begin{align}
f(M)-M=A(f(\Lambda)-\Lambda)A^{-1}
\end{align}
Thus positive definiteness of $f(M)-M$ will only come from the same of $f(\Lambda)-\Lambda$ which for a given eigenvalue $\lambda$ of $M$ corresponds to $f(\lambda)-\lambda > 0$. Thus, every polynomial function $f(x)$ such that $g(x)=f(x)-x$ is positive for $x=\lambda_i,~\forall i=1,2,..,N$ is a candidate for your function. Here $\lambda_i$ denotes eigenvalues of $M$. 
A: The following is useful, if the matrix is diagonalizable, and the idea is applied to the eigenvalues. Then ...         
...any function $f(x)$ shall do, which has that $f(x)>x$ . The most simple one is $f(x)=x+1$ . so if $f(M) = M + I$ where $I$ is the identity should suffice. 
You could also use $f(M) = M^2 + 2*I$ or $f(M) = \cosh(M)+ I$ where $\cosh(.)$ is the power series, which takes also matrices as arguments.
A: There are infinitely many such functions, such as $M+\sum_{i=0}^m\sum_{k=0}^na_k(g_i(M)^Tg_i(M))^k$ with arbitrary matrix functions $g_i$ and arbitrary nonnegative coefficients $a_k$, but I cannot recall any well known ones.
