 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. Tags eigenvalue, eigenvector, find, polynomials, tff, tp2r>p2r Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mophiejoe Linear Algebra 1 April 1st, 2015 06:34 PM lakayii Linear Algebra 2 September 24th, 2012 07:48 AM mbradar2 Linear Algebra 5 March 18th, 2011 01:32 AM gudivada213 Linear Algebra 0 June 22nd, 2010 12:01 AM lakayii Algebra 1 December 31st, 1969 04:00 PM

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