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Question 6.4.30: If λmax is the largest eigenvalue of a symmetric matrix A, n...

If {\lambda}_{max} is the largest eigenvalue of a symmetric matrix A, no diagonal entry can be larger than {\lambda}_{max}. What is the first entry {a}_{11} of A = Q\wedge {Q}^{T}? Show why {a}_{11} \le  {\lambda}_{max}.

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{a}_{11} is \left[ \begin{matrix} {q}_{11} & \cdots &{q}_{1n} \end{matrix} \right] \left[ \begin{matrix} {\lambda}_{1}\bar{{q}_{11}} & \cdots &{\lambda}_{n}\bar{{q}_{1n}}\end{matrix} \right]^{T} \le  {\lambda}_{max} (\left|{q}_{11} \right|^{2} + \cdots + \left|{q}_{1n} \right|^{2})= {\lambda}_{max} .

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