# Nonnegative matrices and singular values

I would like to prove (or prove it is not true with a counter example) the following result:

Let $A$, $B$ be two squares matrices of size $n\times n$ with positive entries. If $A \leq B$, then $\sum_{i = 1}^n \sigma_i(A) \leq \sum_{i = 1}^n \sigma_i(B)$.

The sign $\leq$ between matrices is meant elementwise and the notation $\sigma_i(A)$ is used to denote the i-th singular value of $A$. The Perron-Frobenius theorem only provides information on the spectral radius, while the above statement involves all singular values.

• Probably I'm missing something trivial, so let me ask. Your definition of $\leq$ implies that $\mathrm{trace}(A) \leq \mathrm{trace}(B)$, so it seems to me that what you are asking is trivially true. Where am I wrong? Feb 19 '18 at 9:42
• @FrancescoPolizzi: The matrices $A$ and $B$ need not be symmetric, so the sums of the singular values are $\operatorname{trace}(\sqrt{A^*A})$ and $\operatorname{trace}(\sqrt{B^*B})$ rather than $\operatorname{trace}(A)$ and $\operatorname{trace}(B)$ . Feb 19 '18 at 9:46
• @Jochen Glueck: Ah, right, I was missing that point. Thanks! Feb 19 '18 at 9:47

Let $B = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ and let $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ (my favourite counterexample matrix).
The singular values of $B$ are $2$ and $0$. A short computation shows that the singular values of $A$ are $(\frac{3 + \sqrt{5}}{2})^{1/2}$ and $(\frac{3 - \sqrt{5}}{2})^{1/2}$; the sum of those two values is strictly larger than $2$ (approximately $2.236$, but admittedly I used a calculator to check this...)
• Since science needs more confirmatory results and replication: I think this is entirely correct. Besides hand-calculation, e.g. the following sage-code can be used to confirm the singular vales: A = matrix(CDF, [[1,1],[1,1]]);B = matrix(CDF, [[1,1],[0,1]]); [A.SVD() , B.SVD() , N( ((3+sqrt(5))/2)^0.5 ) , N( ((3-sqrt(5))/2)^0.5 ) ]. The last statement is easily seen by squaring and rearranging to be equivalent to $3^2−(\sqrt{5})^2>1$, which is true. Feb 19 '18 at 12:28