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I have a practice problem on linear transformations that I'd like help on. I have to find the eigenvalues and eigenvectors of

$$T(ax^2 + bx + c) = bx^2 + cx .$$

So I know the general idea is that you solve $T(v) = \lambda v$, so from there I can get the expression:

$$bx^2 + cx = \lambda ax^2 + \lambda bx + \lambda c ,$$ which implies that

$$ \begin{align*} b &= \lambda a \\ c &= \lambda b \\ 0 &= \lambda c \end{align*} $$

but I'm stuck there. I know you need one of $a, b, c$ to be non-zero, but that only works if $\lambda = 0$. Even after I find the eigenvalues, how exactly do I find eigenvectors, since there's no matrix given?

Thanks in advance.

Edit: corrected my claim to state that at least one of $a, b, c$ should be non-zero.

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It is certainly not necessary that all of $a, b, c$ are nonzero. What is true is that the polynomial $ax^2 + bx + c$ is nonzero; i.e., at least one of $a, b, c$ should be nonzero. – Srivatsan Dec 13 '11 at 5:04
But in this case though, if any one of the variables is zero, I believe all three will be zero, given the relationships. – John Doe Dec 13 '11 at 5:06
If such a thing is true, then you should say so carefully. [Your wording is a bit misleading.] And the claim in your comment is false. For instance, I claim that $x^2 = x^2 + 0 x + 0$ is an eigenvector of $T$. Can you verify this? – Srivatsan Dec 13 '11 at 5:09
I see, that makes sense. I just jumped ahead and assumed 0 = $\lambda$a forces a to be zero, but that's not true if $\lambda$ is zero. – John Doe Dec 13 '11 at 5:11
up vote 6 down vote accepted

You only need $a$, $b$, and $c$ to not all be zero, but some of them can be zero.

Start with $\lambda c=0$, because it gives you the most information: either $\lambda=0$ or $c=0$.

If $\lambda=0$, then $c$ has to be zero (so that $c=\lambda b$ holds), and $b$ has to be zero (so that $b=\lambda a$ holds), but $a$ seems like it can be anything. That suggests trying polynomials of the form $ax^2$ with $a\neq 0$ and verifying that they are eigenvectors for $\lambda=0$.

If $\lambda\neq 0$, then $\lambda c = 0$ means that we must have $c=0$. Then $c=0=\lambda b$ tells you that $b$ must be zero. And then...

Once you have an eigenvalue $\lambda$, you find the eigenvectors by solving $T(v)=\lambda v$, $v\neq\mathbf{0}$.

Let me also add that if you are more comfortable with matrices, you can always compute a matrix for $T$: first pick a basis for your vector space; for example, $1$, $x$, and $x^2$. Then compute the value of $T$ at the basis, and express the answers in terms of the basis: $$\begin{align*} T(1) &= x\\ &= 0(1) + 1(x) + 0(x^2).\\ T(x) &= x^2\\ &= 0(1) + 0(x) + 1(x^2).\\ T(x^2) &= 0\\ &= 0(1) + 0(x) + 0(x^2). \end{align*}$$ These give you the columns of the matrix of $T$ relative to the basis $\beta=[1,x,x^2]$, so that $T$ is represented (with respect to this basis) by $$\left(\begin{array}{ccc} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{array}\right).$$ Finding the eigenvalues and eigenvectors is now straightforward; you will obtain your answers in terms of coordinate vectors relative to the basis $\beta$, which you can then translate back into polynomials written the usual way.

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@JohnDoe: If a choice of $\lambda$ forces $a=b=c=0$, that tells you that $\lambda$ is not an eigenvalue. Remember that an eigenvalue is a scalar $\lambda$ for which there is a vector $\mathbf{v}\neq\mathbf{0}$ such that $T(\mathbf{v})=\lambda\mathbf{v}$. If no nonzero vector solves the equation, $\lambda$ is not an eigenvalue. As to checking they are eigenvectors for $\lambda=0$, just verify that $T(\mathbf{v})=0\mathbf{v}=\mathbf{0}$ and that they are not the zero vector. That's the definition of being an eigenvector corresponding to $0$. – Arturo Magidin Dec 13 '11 at 5:16
Thanks for the clarification. That makes complete sense to me. – John Doe Dec 13 '11 at 5:23

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