Norm of an Operator: Show llTll = max ldl

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Homework Statement



Let D be a nxn diagonal matrix and T:Rn -> Rn be the linear operator associated with D. ie., Tx = Dx for all x in Rn. Show that:

llTll = max ldl

where d1, ..., dn are the entries on the diagonal of D

Homework Equations



the smallest M for which llTxll <= M*llxll is the norm of T

The Attempt at a Solution



i have shown that llTll <= max ldl which was relatively straight forward

im struggling to guess a y such that: llTyll >= maxldl * llyll

which would allow me to conclude that llTll >= max ldl and hence llTll = max ldl

any hints in the right direction is appreciated
 
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actually i think y = ek where dk = max |d| works
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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