Relation between eigenvalues of $A$ and $B=(f(a_{i,j}))$, where $f$ is concave/convex Let $A$ be an $n\times n$ real and symmetric matrix with unit diagonal. Furthermore, let the entries $0\leq  a_{ij} \leq 1, i\neq j$.

Now, define a function $f:[0,1] \mapsto [0,1]$ which is concave. For instance we can take $f(x)=2\sin\left(\frac{\pi x}{6}\right)$ which is the transformation of rank correlation coefficients to linear correlation coefficients in a Gaussian copula.

This defines  $B=(f(a_{i,j}))$, which is still symmetric with unit diagonal and $b_{ij}\geq a_{ij}$.

I have three related questions: 

1) What is the relationship between the eigenvalues of $A$ and $B$ in this case? Does a general result exist when $f$ is concave or convex?

2) Imagine that $A$ has $t$ negative eigenvalues, how about $B$? 

3) If $B$ is positive semi-definit is $A$?

The question arised when i implemented an algortihm to find the nearest (rank) correlation matrix to $A$ (where a requirement is non-negative eigenvalues) which worked as intended. Afterwards, I converted the rank-coefficients to linear correlation coefficients and this correlation matrix had some negative eigenvalues. This made me curious about the relation between the eigenvalues as the function $f$ possesses very nice properties (monotone, concave, infinitely differentiable). Perhaps a general result exists in this case?
Some quick observations. Presumably $n&gt;1$.

(2) Since your $A$ always has nonnegative entries and your $f$ is a function on $[0,1]$, $B$ must have a nonnegative trace. It follows that $B$ must possess at least one non-negative eigenvalue, but the number of non-negative entries may vary wildly.

Example: let $A$ be indefinite, then:


when $f(x)=x$, $B=A$ is indefinite;
when $f(x)=1$, all eigenvalues of $B=ee^T$ (where $e$ is the all-one vector) are non-negative.


(3) No. In the above example with $f(x)=1$, $B$ is positive semidefinite but $A$ is indefinite.
Some quick observations. Presumably $n&gt;1$.

(2) Since your $A$ always has nonnegative entries and your $f$ is a function on $[0,1]$, $B$ must have a nonnegative trace. It follows that $B$ must possess at least one non-negative eigenvalue, but the number of non-negative entries may vary wildly.

Example: let $A$ be indefinite, then:


when $f(x)=x$, $B=A$ is indefinite;
when $f(x)=1$, all eigenvalues of $B=ee^T$ (where $e$ is the all-one vector) are non-negative.


(3) No. In the above example with $f(x)=1$, $B$ is positive semidefinite but $A$ is indefinite.