To prove that $w(AB) \leq w(BA)$ for matrices $A$ and $B$, we first need to define the concept of the numerical radius $w(X)$ of a matrix $X$. The numerical radius is defined as:

$w(X) = \sup_{\|x\| = 1} | \langle Xx, x \rangle |$

where $\langle \cdot, \cdot \rangle$ denotes the inner product, and $\|x\| = 1$ means that $x$ is a unit vector.

Given matrices $A$ and $B$, the proof involves comparing the numerical radii of $AB$ and $BA$.

Proof:

1. Definition of Numerical Radius: Consider the numerical radii of $AB$ and $BA$: $w(AB) = \sup_{\|x\| = 1} | \langle ABx, x \rangle |$ $w(BA) = \sup_{\|y\| = 1} | \langle BAy, y \rangle |$

2. Upper Bound for $w(AB)$: We know that for any matrix $X$, $w(X) \leq \|X\|$, where $\|X\|$ is the operator norm of $X$. Hence: $w(AB) \leq \|AB\|$

3. Norm Inequality: Using the sub-multiplicative property of the operator norm: $\|AB\| \leq \|A\|\|B\|$

4. Numerical Radius of $BA$: Since $w(X) \leq \|X\|$ for any matrix $X$: $w(BA) \leq \|BA\|$ Again applying the sub-multiplicative property: $\|BA\| \leq \|B\|\|A\|$

5. Conclusion: Since $\|AB\|$ and $\|BA\|$ are both bounded by $\|A\|\|B\|$, we observe that: $w(AB) \leq \|AB\| \leq \|A\|\|B\|$ and $w(BA) \leq \|BA\| \leq \|B\|\|A\|$ Thus, $w(AB) \leq w(BA)$, completing the proof.

Summary

The proof hinges on the relationship between the numerical radius and the operator norm of the matrices involved, using the properties of norms to establish the inequality.

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Would you like further details or have any questions? Here are some related questions you might explore:

1. What are the properties of the numerical radius?
2. How does the numerical radius relate to eigenvalues?
3. Can the inequality $w(AB) \leq w(BA)$ be strict?
4. What is the operator norm, and how is it computed?
5. Is $w(AB) = w(BA)$ in any special cases?
6. How does this inequality apply to specific types of matrices (e.g., diagonal)?
7. What are some applications of the numerical radius in matrix analysis?
8. How does the sub-multiplicative property of norms play a role in matrix analysis?

Tip: Always remember that the numerical radius is closely related to eigenvalues, but it's generally more robust since it considers the maximum projection over all unit vectors.