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Eigenvector and eigenvalue for differential operator

kalish

Member
Oct 7, 2013
99
My friends and I have been struggling with the following problem, and don't understand how to do it. We have gotten several different answers, but none of them make sense. Can you help us?

**Problem statement:** Let $V$ be the vector space of real-coefficient polynomials of degree at most $3$. Let $D:V \rightarrow V$ be the differential operator; $D(p(x))=\frac{d}{dx}p(x)$. Give an example of an eigenvector for $D$. What is the corresponding eigenvalue?

We ended up getting that $\frac {d}{dx}p(x)=\lambda p(x)$, so that $p(x)=\frac{x^2}{2}$. Is this correct?

Thanks.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
My friends and I have been struggling with the following problem, and don't understand how to do it. We have gotten several different answers, but none of them make sense. Can you help us?

**Problem statement:** Let $V$ be the vector space of real-coefficient polynomials of degree at most $3$. Let $D:V \rightarrow V$ be the differential operator; $D(p(x))=\frac{d}{dx}p(x)$. Give an example of an eigenvector for $D$. What is the corresponding eigenvalue?

We ended up getting that $\frac {d}{dx}p(x)=\lambda p(x)$, so that $p(x)=\frac{x^2}{2}$. Is this correct?

Thanks.
I don't think so, since:
$$D\Big(\frac{x^2}{2}\Big) = x \ne \lambda \cdot \frac{x^2}{2}$$

Let's pick a generic polynomial in V:
$$p(x) = ax^3 + bx^2 + cx + d$$
where $a, b, c, d \in \mathbb R$.

Then we have:
$$D(p(x)) = \lambda p(x)$$
$$3ax^2 + 2bx + c = \lambda ax^3 + \lambda bx^2 + \lambda cx + \lambda d$$
Which solutions can you find for $\lambda$? And for $a, b, c, d$?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
My friends and I have been struggling with the following problem, and don't understand how to do it. We have gotten several different answers, but none of them make sense. Can you help us?

**Problem statement:** Let $V$ be the vector space of real-coefficient polynomials of degree at most $3$. Let $D:V \rightarrow V$ be the differential operator; $D(p(x))=\frac{d}{dx}p(x)$. Give an example of an eigenvector for $D$. What is the corresponding eigenvalue?

We ended up getting that $\frac {d}{dx}p(x)=\lambda p(x)$, so that $p(x)=\frac{x^2}{2}$. Is this correct?
$p(x) = \frac{x^2}{2}$ is not an eigenvector, because $D(p(x)) = x$, which is not a scalar multiple of $p(x)$.

You might find it helpful to represent the operator $D$ by a matrix. Take the set $\{x^3,x^2,x,1\}$ to be a basis for $V$. Then the matrix of $D$ with respect to that basis is $\begin{bmatrix}0&0&0&0 \\ 3&0&0&0 \\ 0&2&0&0 \\ 0&0&1&0 \end{bmatrix}.$ Now use the usual algebraic method to calculate eigenvectors and eigenvalues for that matrix. Finally, you have to figure out what that tells you in terms of polynomials.

Edit. *sigh* As usual, I like Serena beat me to it. But he and I have given two different approaches to the problem, so I hope that one of them works for you.
 

kalish

Member
Oct 7, 2013
99
I don't think so, since:
$$D\Big(\frac{x^2}{2}\Big) = x \ne \lambda \cdot \frac{x^2}{2}$$

Let's pick a generic polynomial in V:
$$p(x) = ax^3 + bx^2 + cx + d$$
where $a, b, c, d \in \mathbb R$.

Then we have:
$$D(p(x)) = \lambda p(x)$$
$$3ax^2 + 2bx + c = \lambda ax^3 + \lambda bx^2 + \lambda cx + \lambda d$$
Which solutions can you find for $\lambda$? And for $a, b, c, d$?
Hi,
This is where we got stuck, because we couldn't solve for the values without knowing if lambda, a, b, c, d are nonzero...
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
Hi,
This is where we got stuck, because we couldn't solve for the values without knowing if lambda, a, b, c, d are nonzero...
Consider two cases:
  1. $\lambda = 0$
  2. $\lambda \ne 0$
What can you say about a, b, c, and d in each of those cases?
Note that the coefficient of each power of x must match left and right.