Question

Prove that the spectral radius function does not satisfy the triangle inequality by finding matrices $A, B$ such that $\rho(A+B)>\rho(A)+\rho(B)$.

   Prove that the spectral radius function does not satisfy the triangle inequality by finding matrices $A, B$ such that $\rho(A+B)>\rho(A)+\rho(B)$.

Applied Linear Algebra (Undergraduate Texts in Mathematics)

Applied Linear Algebra (Undergraduate Texts in Mathematics)

Peter J. Olver,… 2nd Edition

Chapter 9, Problem 35

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** The spectral radius, denoted as \(\rho(A)\), of a matrix \(A\) is the largest absolute value of its eigenvalues. The triangle inequality for the spectral radius would state that \(\rho(A + B) \leq \rho(A) + \rho(B)\) for any matrices \(A\) and \(B\). We need to Show more…

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Prove that the spectral radius function does not satisfy the triangle inequality by finding matrices $A, B$ such that $\rho(A+B)>\rho(A)+\rho(B)$.

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