Cauchy-Schwarz inequality applied to Trace of two products $\mathbf{Tr}(A'B)$ has the form

$$ \mathbf{Tr}(A'B) \leq \sqrt{\mathbf{Tr}(A'A)} \sqrt{\mathbf{Tr}(B'B)} $$

I saw many places where people use this inequality. But did not see a formal proof. Is it difficult to prove ? Anyone can give a simple proof ?


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$Tr(A-tB)'(A-tB) \geq 0$ for all $t$ real. Expand this, use the fact that $Tr(M')=Tr(M)$ (so $Tr(A'B)=Tr(B'A))$ and minimize the left side over $t$. You will get the inequality you want. (This is the standard proof of C-S inequality)


The Cauchy-Schwarz inequality is valid for any inner product, so you just need to show $\operatorname{\textbf{Tr}}A'B$ is an inner product. It's clearly bilinear (or sesquilinear if by $'$ you meant a complex adjoint), with $$\operatorname{\textbf{Tr}}A'A=\sum_i (A'A)_{i}=\sum_{ij}A'_{ij}A_{ji}.$$Depending on whether you're working with the real or complex case, this quantity is either $\sum_{ij}A_{ji}^2$ or $\sum_{ij}|A_{ji}|^2$. Either way it's non-negative, completing the proof.

  • 4
    $\begingroup$ The point here is that while $\mathbf{Tr}(A,B)$ looks complicated, it is really nothing more than a plain vanilla inner product with n^2-element vectors, so there's nothing special in proving C-S here. $\endgroup$ – einpoklum Nov 20 '18 at 19:13

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