Finding eigenvalues

May 2010
20
1
Let \(\displaystyle T: P_2 \mapsto P_2\) be defined by T(p)=xp'. Find eigenvalues of T.

I'm not sure what the notation of T(p)=xp' means and how to go about finding the eigenvalues.

Thanks.
 
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dwsmith

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Mar 2010
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Florida
Let \(\displaystyle T: P_2 \mapsto P_2\) be defined by T(p)=xp'. Find eigenvalues of T.

I'm not sure what the notation of T(p)=xp' means and how to go about finding the eigenvalues.

Thanks.
That is the linear transformation which is x times the derivative of the polynomial.
 
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May 2010
20
1
That is the linear transformation which is x times the derivative of the polynomial.
Ok, so p is a polynomial? How would I go about finding eigenvalues for said polynomial?
 

dwsmith

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Mar 2010
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Ok, so p is a polynomial? How would I go about finding eigenvalues for said polynomial?
I am assuming P is a polynomial since we are mapping \(\displaystyle P_2\) on \(\displaystyle P_2\). Unless your book notates this differently, I am almost positive these are polynomials; however, different books notate \(\displaystyle P_2\) differently. For instance, mine says \(\displaystyle P_2\) are polynomial of degree 1 or less. Thus, the 2 in \(\displaystyle P_2\) is the amount of elements {1, x} but some books refer to \(\displaystyle P_2\) as degree two or less, {\(\displaystyle 1\), \(\displaystyle x\), \(\displaystyle x^2\)}. What does your book say?
 
May 2010
20
1
I am assuming P is a polynomial since we are mapping \(\displaystyle P_2\) on \(\displaystyle P_2\). Unless your book notates this differently, I am almost positive these are polynomials; however, different books notate \(\displaystyle P_2\) differently. For instance, mine says \(\displaystyle P_2\) are polynomial of degree 1 or less. Thus, the 2 in \(\displaystyle P_2\) is the amount of elements {1, x} but some books refer to \(\displaystyle P_2\) as degree two or less, {\(\displaystyle 1\), \(\displaystyle x\), \(\displaystyle x^2\)}. What does your book say?
I believe in my case it's the 2nd degree polynomial.
 

dwsmith

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Mar 2010
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I believe in my case it's the 2nd degree polynomial.
Ok so your linear transformation goes from \(\displaystyle ax^2+bx+c\rightarrow xp'(x)=x(2ax+b)=2ax^2+bx\).

Now you need to find the eigenvalues of the linear transformation.
 
May 2009
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I would do the following:

A basis for \(\displaystyle P_{2} \) is \(\displaystyle \{1,x,x^{2} \} \)

\(\displaystyle T(1) = 0 \)

\(\displaystyle T(x) = x \)

and \(\displaystyle T(x^{2}) = 2x^{2} \)

so the transformation matrix is \(\displaystyle \begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0\\
0 & 0 & 2
\end{bmatrix} \)

which has eigenvalues 0,1, and 2
 

HallsofIvy

MHF Helper
Apr 2005
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I believe in my case it's the 2nd degree polynomial.
The "set of all 2nd degree polynomials" does not form a vector space. As dwsmith said, \(\displaystyle P_2\) is the set of all polynomials of degree 2 or less.

Random Variable showed how to represent this transformation as a matrix- worth knowing for itself.

But for this problem, you don't have to do that. As dwsmith said, if \(\displaystyle p= ax^2+ bx+ c\), then \(\displaystyle T(p)= xp'= x(2ax+ b)= 2ax^2+ bx\). If \(\displaystyle \lambda\) is an eigenvalue for T, we must have some a, b, c, not all 0, such that \(\displaystyle T(p)= 2ax^2+ bx= \lambda(ax^2+ bx+ c)\). Since that must be true for all x, we can set corresponding coefficients equal: \(\displaystyle 2a= \lambda a\), \(\displaystyle b= \lambda b\), and \(\displaystyle 0= \lambda c\).

If a is not 0, then \(\displaystyle \lambda= 2\), if b is not 0, then \(\displaystyle \lambda= 1\), and if c is not 0, then \(\displaystyle \lambda= 0\).

That is, 0 is an eigenvalue with eigenvectors any multiple of 1, 1 is an eigenvalue with eigenvectors any multiple of x, and 2 is an eigenvalue with eigenvectors any multiple of \(\displaystyle x^2\), exactly what Random Variable got.
 
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