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July 9th, 2016, 07:46 PM   #1
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Find eigenvector and eigenvalue of T:P2(R)->P2(R) Tf=f''+f'

My first thought on solving this is defining $f=a_2x^2+a_1x+a_0$

Now using $T(f)=f''+f'$

$T(a_2x^2+a_1x+a_0)=2a_2x+a_1+2a_2=\lambda a_2x^2+ \lambda a_1x+ \lambda a_0$

After that I get the next system of equations:
$$\lambda a_2=0$$
$$\lambda a_1=2a_2$$
$$\lambda a_0=a_1+2a_2$$
Which gives me $\lambda =0$, after this I substitute $\lambda$ in $f=\lambda a_2x^2+ \lambda a_1x+ \lambda a_0=0$.
Can I say my eigenvector and eigenvalue $=0$ ?
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July 10th, 2016, 04:31 AM   #2
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No. The definition of "eigenvalue" for operator A is that an eigenvalue is a value of $\displaystyle \lambda$ such that $\displaystyle Av= \lambda v$ for some non-zero ("non-trivial") vector, v. Further, any number times an eigenvector is also an eigenvector corresponding to the same eigenvalue- there are, necessarily, an infinite number of eigenvectors.

Your calculations are correct, your interpretation is flawed. From the first equation, either $\displaystyle \lambda= 0$ or $\displaystyle a_2= 0$. If $\displaystyle a_2= 0$, then, from the second equation either $\displaystyle \lambda= 0$ or $\displaystyle a_1= 0$. If $\displaystyle a_1= 0$ then, from the third equation, either $\displaystyle \lambda= 0$ or $\displaystyle a_0= 0$. But $\displaystyle a_2= a_1= a_0= 0$ gives the "trivial" vector, 0.

So at some point in that, we must have $\displaystyle \lambda= 0$. In that case, the three equations become 0= 0, $\displaystyle 2a_2= 0$ so $\displaystyle a_2= 0$, and $\displaystyle a_1+ 2a_2= a_1= 0$. But that says nothing about $\displaystyle a_0$! The eigenvalue is 0 and the set of eigenvectors corresponding to eigenvalue 0 is the set of all quadratic functions of the form $\displaystyle 0x^2+ 0x+ a_0= a_0$ where $\displaystyle a_0$ can be any number- in other words, the set of all constant functions.
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Last edited by Country Boy; July 10th, 2016 at 05:06 AM.
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