The relation between span(In,A,A2, )and it's minimal polynomial

alazhumizhu
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Let A ∈ Mn×n(F )
Why dim span(In, A, A2, A3, . . .) = deg(mA)?? where mA is the minimal polynomial of A.
For span (In,A,A2...)

I can prove its

dimension <= n by CH Theorem

but what's the relation between

dim span(In,A,A2...)and deg(mA)
 
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For example, if the minimal polynomial is x^2+x+1. Then A^2+A+1=0.

Do you see any way to conclude that A^2\in span(A,1)?
 
ybut i don't know why can't I write A
 
ybut i don't know why can't I write A in spanA2I
 
y,but i don't know why can't I write A in span{A2,I}?
I'm sorry about the type..
 
alazhumizhu said:
y,but i don't know why can't I write A in span{A2,I}?
I'm sorry about the type..

That's also true, but I don't see how this fact helps you solve the problem.
 
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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