# Proof for Cauchy-Schwarz inequality for Trace [closed]

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 ?

## closed as off-topic by user21820, Namaste, KReiser, user10354138, CesareoNov 21 '18 at 2:11

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## 2 Answers

$$Tr(A-tB)'(A-tB) \geq 0$$ 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.

• 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. – einpoklum Nov 20 '18 at 19:13