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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- May 17th 2010, 07:11 AMwattkowFinding eigenvalues
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. - May 17th 2010, 10:58 AMdwsmith
- May 17th 2010, 03:24 PMwattkow
- May 17th 2010, 03:29 PMdwsmith
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 17th 2010, 03:35 PMwattkow
- May 17th 2010, 03:38 PMdwsmith
- May 17th 2010, 04:38 PMRandom Variable
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 - May 18th 2010, 05:22 AMHallsofIvy
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. - May 18th 2010, 10:09 PMwattkow
Thanks fellas.