Question: Recall from Exercise 3.11 that a positive matrix A > 0 has a dominant eigenvalue and corresponding left eigenvector > 0 and right eigenvector
Recall from Exercise 3.11 that a positive matrix A > 0 has a dominant eigenvalue
and corresponding left eigenvector ω > 0 and right eigenvector ν > 0 (i.e., ωTA = λωT, Aν = λv) which belong to the probability simplex S = {x ∈ Rn : x ≥ 0, 1Tx = 1}. In this exercise, we shall prove that the dominant eigenvalue has an optimization-based characterization, similar in spirit to the “variational” characterization of the eigenvalues of symmetric matrices. Define the function f : S →R++ with values

where aTi is the i-th row of A, and we let 
1.Prove that, for all x ∈ S and A > 0, it holds that
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2. Prove that
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3. Show that f (ν) = λ, and hence conclude that
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which is known as the Collatz-Wielandt formula for the dominant eigenvalue of a positive matrix. This formula actually holds more generally for nonnegative matrices27, but you are not asked to prove this fact.
= P(A) > 0,
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1 2 Multiply the previous inequality on the left by T ... View full answer
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