Let $A$ and $B$ be two $p\times p$ real symmetric matrices such that $A\succeq B$, meaning that $A-B$ is non-negative definite. I want to prove these two results.
First result: $\lambda_j(A)\geq \lambda_j(B)$, $j=1,...,p$, where $\lambda_j(\cdot)$ is the $j$ eigenvalue of the respective matrix.
Second result: $\log(A)\succeq \log(B)$, where $\log(X)=P\text{diag}(\log\lambda_j(X))P^{\top}$ and $X$ is a $p\times p$ symmetric matrix with spectral decomposition $X=P\text{diag}(\lambda_j(X))P^{\top}$.
I think that the two results must be straightforward. My complication is that $A$ and $B$ have (in general) different spectral decompositions, say $$A=PDP^{\top}$$ and $$B=QMQ^{\top}.$$ It will be helpful if, for example, I can say something about $P$ and $Q$, like $P^{\top}Q=I$, which I think it could be false in general.
For the first result, it suffices to use the Courant-Fischer theorem. In particular, we have a proof to the effect of $$ \lambda_j(A) = \min\{\max_x\{x^TAx : \cdots\}\} = \min\{\max_x\{x^TBx + x^T(A-B)x : \cdots\}\}\\ \leq \min\{\max_{x}\{x^TBx: \cdots\}\} = \lambda_j(B).\\ $$ Your second result will only hold if we have $B \succ 0$. In this case, your statement is equivalent to saying that the matrix-logarithm is "operator monotone" over the positive definite matrices. For a reference on this, I recommend Bhatia's Matrix Analysis or Horn and Johnson's Topics in Matrix Analysis.