The factorization norm, sometimes also called $\gamma\_2$ norm play an important role in (quantum) communication complexity and is defined for a $n\times n$ matrix $A$ by:


$\gamma\_2(A) = \max || A \circ uv^t||\_{\mathrm{tr}}$ where the maximization runs over all unit vectors $u$ and $v$ ($||u||=||v||=1$)


We can find many equivalent definitions such as: $\gamma\_2(A) = \min \lambda$ such that $(A)\_{ij} = \langle u\_i | v\_j\rangle$ and $\forall i,j$ we have $ ||u\_i||\leq \lambda$ and $||v\_j|| \leq \lambda$.


And the trace norm is defined by $||A||\_{\mathrm{tr}}=\mathrm{tr}\sqrt{A^\dagger A}$.



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> These two norms are equivalent, so there exists a constant $C\_n$ such that $||A||\_{\mathrm{tr}} \geq C\_n\gamma\_2(A)$. What is the value of $C\_n$?
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Having played with a few examples I conjecture that $C\_n=1$. Also note that the reverse inequality can be easily obtained: $||A||\_\mathrm{tr} \leq n\cdot \gamma\_2(a)$.


In particular, I am interested by the case where $A$ is definite positive. In this case the trace norm is simply the trace of $A$, and I can prove that $\gamma\_2(A) \geq \sqrt{tr(A)}$.



Equivalence constant between factorization norm and trace norm
You are right, the best constant is $1$. In fact, the stronger inequality $\gamma\_2(A) \leq \|A\|\_{\infty}$ is also true (and is stronger since $\|A\|\_{\infty}\leq \| A\|\_{1}$).


For simplicity I denote $\|\cdot\|\_p$ the Schatten $p$-norm. ($p=1$ corresponds to the trace norm, $p=2$ the Hilbert-Schmidt norm and $p=\infty$ the operator norm).


For a proof, one has to show that $\|A \circ u v^t\|\_{1} \leq \|A\|\_{\infty}$ if $u,v$ are unit vectors in the euclidean space $\mathbb C^n$. (it could be recalled that the notation $\circ$ is the Hadamard product, or product entrywise). Denote by $D\_u,D\_v$ the diagonal matrices with diagonal entries the coordinates of $u$ and $v$, so that $A \circ u v^t = D\_u A D\_v$. By Hölder's inequality $\|D\_u A D\_v\|\_1 \leq \|D\_u\|\_2 \|A\|\_\infty\|D\_v\|\_2 = \|A\|\_\infty$.


