My Math Forum Find eigenvector and eigenvalue of T:P2(R)->P2(R) Tf=f''+f'

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 July 9th, 2016, 07:46 PM #1 Newbie   Joined: Jul 2016 From: Mexico Posts: 3 Thanks: 1 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 $=0$ ? Thanks from manus
 July 10th, 2016, 04:31 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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. Thanks from vChav Last edited by Country Boy; July 10th, 2016 at 05:06 AM.

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