# What are the conditions on $\text{tr}(AB) \leq \text{tr(A)} \text{tr(B)}$ to be true?

Let $$A$$ and $$B$$ be two arbitrary matrix with proper dimension for multiplication.

Consider this trace inequlaty which is trace of multiplication of two matrices versus their individual traces

$$\text{tr}(AB) \leq \text{tr(A)} \text{tr(B)}$$

1- Do we have result for rectangular matrix that satisfy this inequality?

2- If they were square matrices what are the conditions?

3- Is there any specific name for this inequality?

• How do you define the trace of a matrix which is not a square? In case you take $A=B$ this equivalent to $$\sum_{i=1}^n \lambda_i^2 \leq \left(\sum_{i=1}^n \lambda_i \right)^2$$ – Severin Schraven Jan 4 '19 at 17:39
• If we have a $m\times n$-matrix, and $k:=\min(m,n)$, the definition $$tr(A)=\sum_{j=1}^k a_{jj}$$ would make sense. – Peter Jan 4 '19 at 17:43
• @Peter You are right, I was thinking about something which is coordinate-free. – Severin Schraven Jan 4 '19 at 17:45
• @Severin Schraven: peter answered that. But to get into the problem, let they be square first and focus on 2 and 3. – Saeed Jan 4 '19 at 17:46

Can't think of anything deep, but if both $$A$$ and $$B$$ are positive semidefinite, the inequality is true: when $$a=\operatorname{tr}(A)$$, we have $$A\preceq aI$$ and hence $$\operatorname{tr}(AB) =\operatorname{tr}(B^{1/2}AB^{1/2}) \le\operatorname{tr}(B^{1/2}(aI)B^{1/2}) =\operatorname{tr}(A)\operatorname{tr}(B).$$ This also follows from (and hence is weaker than) von Neumann's trace inequality, which in this context says that $$\operatorname{tr}(AB)\le\sum_i\lambda_i(A)\lambda_i(B)$$ when the eigenvalues of $$A$$ and $$B$$ are arranged in the same (ascending or descending) order.
• Can you explain what $A^{1/2}$ is? When they are psd we have $A=U_A\Lambda_AU_A^T$ and $B=U_B\Lambda_BU_B^T$. So $\text{tr}(U_A\Lambda_AU_A^TU_B\Lambda_BU_B^T)=\text{tr}(U_B^TU_A\Lambda_AU_A^TU_B\Lambda_B)$, how can I get what you have? – Saeed Jan 4 '19 at 17:54
• What's that $F$ norm? – Yanko Jan 4 '19 at 17:57
• @Yanko $\|\cdot\|_F$ denotes Frobenius norm. – user1551 Jan 4 '19 at 18:00