Linear Algebra Eigenspace Question

Nexttime35
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Homework Statement


Let T: C∞(R)→C∞(R) be given by T(f) = f'''' where T sends a function to the fourth derivative.

a) Find a basis for the 0-eigenspace.
b) Find a basis for the 1-eigenspace.


The Attempt at a Solution



I just want to verify my thought process for this problem. For a), finding the basis for the 0-eigenspace, essentially I needed to find a basis for the vectors v in V such that T(v) = 0v .

So, would the basis for this 0-eigenspace be all polynomials in P3? If you solve the fourth derivative of any polynomial in P3, you will get 0.

As for b), when finding the basis for the 1-eigenspace, we need to find a basis for the vectors v in V such that T(v) = 1v, or that after solving the fourth derivative, you get a function that is equal to 1? Is this the correct logic? So would the basis for the 1-eigenspace be any polynomial in P4?

Thanks much for your help.
 
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The 1 eigenspace is vectors that map to themselves. Not to the vector 1.
 
If by ##C^∞## you mean the space of all infinitely differentiable functions, then there are a lot more than polynomials around.

Let ##f \in C^∞##. Look at the power series: ##f(x) = ∑_{i=0}^{∞} a_i x^i##. If the fourth derivative of ##f## is 0, then you have that ##a_i = 0## for ##i \geq 4##. Thus the choice of ##a_0, a_1, a_2, a_3## determines ##f## in the 0-eigenspace. Using this, can you come up with a basis? (It will have 4 functions in it).
 
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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