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
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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