Linear Algebra Inverse Generalization

gabriels-horn
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Homework Statement


Show that if T is a square (matrix) and if T^4 is the zero matrix then (I-T)^{-1} = I+T+T^2+T^3. Generalize.

The Attempt at a Solution


To be honest I don't even know where to begin. Supposedly this is a simple generalization.
 
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Well, what is (I+T+T^2+T^3)(I-T)?
 
gabbagabbahey said:
Well, what is (I+T+T^2+T^3)(I-T)?

Is it...I?
 
confirm it by multplying out. matrix addition & multiplication are distrubutive.
 
lanedance said:
confirm it by multplying out. matrix addition & multiplication are distrubutive.

Yeah, I know. It was a joke; of course its I. Everything in Linear is I.
 
gabriels-horn said:
Is it...I?

Multiplying it out should give you I-T^4. So, if T^4=0,, then you get the identity matrix and hence (1+T+T^2+T^3+T^4)=(I-T)^{-1}.

Now, can you generalize this statement?
 
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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