Eigenvector and eigenvalue for differential operator

[kalish1]

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 like Serena]

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]

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.

 

[kalish1]

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...

 

[I like Serena]

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.