Finding eigenvalues

**wattkow** · May 17th 2010, 07:11 AM

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.

**dwsmith** · May 17th 2010, 10:58 AM

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.

**wattkow** · May 17th 2010, 03:24 PM

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?

**dwsmith** · May 17th 2010, 03:29 PM

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?

**wattkow** · May 17th 2010, 03:35 PM

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.

**dwsmith** · May 17th 2010, 03:38 PM

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.

**Random Variable** · May 17th 2010, 04:38 PM

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 $\begin{bmatrix}<br /> 0 & 0 & 0 \\<br /> 0 & 1 & 0\\<br /> 0 & 0 & 2<br /> \end{bmatrix}$

which has eigenvalues 0,1, and 2

**HallsofIvy** · May 18th 2010, 05:22 AM

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.

**wattkow** · May 18th 2010, 10:09 PM

Thanks fellas.

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