# Finding eigenvalues

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• May 17th 2010, 08:11 AM
wattkow
Finding eigenvalues
Let $T: P_2 \mapsto P_2$ 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, 11:58 AM
dwsmith
Quote:

Originally Posted by wattkow
Let $T: P_2 \mapsto P_2$ 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.
• May 17th 2010, 04:24 PM
wattkow
Quote:

Originally Posted by dwsmith
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?
• May 17th 2010, 04:29 PM
dwsmith
Quote:

Originally Posted by wattkow
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 $P_2$ on $P_2$. Unless your book notates this differently, I am almost positive these are polynomials; however, different books notate $P_2$ differently. For instance, mine says $P_2$ are polynomial of degree 1 or less. Thus, the 2 in $P_2$ is the amount of elements {1, x} but some books refer to $P_2$ as degree two or less, { $1$, $x$, $x^2$}. What does your book say?
• May 17th 2010, 04:35 PM
wattkow
Quote:

Originally Posted by dwsmith
I am assuming P is a polynomial since we are mapping $P_2$ on $P_2$. Unless your book notates this differently, I am almost positive these are polynomials; however, different books notate $P_2$ differently. For instance, mine says $P_2$ are polynomial of degree 1 or less. Thus, the 2 in $P_2$ is the amount of elements {1, x} but some books refer to $P_2$ as degree two or less, { $1$, $x$, $x^2$}. What does your book say?

I believe in my case it's the 2nd degree polynomial.
• May 17th 2010, 04:38 PM
dwsmith
Quote:

Originally Posted by wattkow
I believe in my case it's the 2nd degree polynomial.

Ok so your linear transformation goes from $ax^2+bx+c\rightarrow xp'(x)=x(2ax+b)=2ax^2+bx$.

Now you need to find the eigenvalues of the linear transformation.
• May 17th 2010, 05:38 PM
Random Variable
I would do the following:

A basis for $P_{2}$ is $\{1,x,x^{2} \}$

$T(1) = 0$

$T(x) = x$

and $T(x^{2}) = 2x^{2}$

so the transformation matrix is $\begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0\\
0 & 0 & 2
\end{bmatrix}$

which has eigenvalues 0,1, and 2
• May 18th 2010, 06:22 AM
HallsofIvy
Quote:

Originally Posted by wattkow
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, $P_2$ 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 $p= ax^2+ bx+ c$, then $T(p)= xp'= x(2ax+ b)= 2ax^2+ bx$. If $\lambda$ is an eigenvalue for T, we must have some a, b, c, not all 0, such that $T(p)= 2ax^2+ bx= \lambda(ax^2+ bx+ c)$. Since that must be true for all x, we can set corresponding coefficients equal: $2a= \lambda a$, $b= \lambda b$, and $0= \lambda c$.

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

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 $x^2$, exactly what Random Variable got.
• May 18th 2010, 11:09 PM
wattkow
Thanks fellas.