# Linear map between polynomial functions

Ok so I don't understand how to find the matrix representing the linear map from P2 to P2.

I don't understand how I will write a polynomial in column vector form and by that I mean e.g, If we have:

I know that if I have a vector e.g (0,0,1) and apply the transformation then I'll get (1,1,0) but if I have my linear transformation between polynomial functions I am not able to understand how exactly I will write the transformation defined by Lf=f' in column vector form, even my input function which will be a polynomial of the form ax^2+bx+c I don't know how to write in column vector form.

Any help would be much appreciated.

First of all, note that it is not strictly necessary to find the matrix of the linear transformation in order to find its eigenvalues. In fact, I'm almost certain that this is not the intended approach here.

That being said, the key to finding the matrix for any linear map is selecting a basis, and noting how $T$ (I meant $L$, but I've typed too much to fix it now) acts upon that basis.

In this case, $\mathcal B = \{1,x,x^2\}$ is a basis of $\mathcal P_2$. We note that $$T \pmatrix{1\\0\\0}_{\mathcal B} = T(1(1)+0(x)+0(x^2)) = T(1) = 0 = 0(1) + 0(x) + 0(x^2) = \pmatrix{0\\0\\0}_{\mathcal B}\\ T\pmatrix{0\\1\\0}_{\mathcal B} = T(0(1) + 1(x) + 0(x^2)) = T(x) = 1 = 1(1) + 0(x) + 0(x^2) = \pmatrix{1\\0\\0}_{\mathcal B}\\ T \pmatrix{0\\0\\1}_{\mathcal B} = T(0(1) + 0(x) + 1(x^2)) = T(x^2) = 2x = 0(1) + 2(x) + 0(x^2) = \pmatrix{0\\2\\0}_{\mathcal B}$$ The matrix of the transformation relative to the basis $\mathcal B$ is therefore the matrix consisting of the columns on the right, that is $$[T]_{\mathcal B} = \pmatrix{0&1&0\\0&0&2\\0&0&0}$$

The intended solution is probably as follows:

Let $\lambda$ be an eigenvector of $L$. By definition, this means that there is some "vector" (i.e. polynomial) $f$ satisfying $$L(f) = \lambda f \implies f'(x) = \lambda f(x)$$ From here we need to show that there will only be such a polynomial when $\lambda = 0$. You could solve this as mfl did in his answer, or you could solve it as a differential equation, noting that the general solution is only a polynomial when $\lambda = 0$.

• I know it says here to avoid comments like 'thanks' but I'm still going to say thanks a ton, this was very helpful. Nov 29 '14 at 19:41
• Glad I could help :) Nov 29 '14 at 19:45

Assume that $p=ax^2+bx+c$ is an eigenvector of $L$ with eigenvalue $\lambda.$ Then

$$2ax+b=L(p)=a\lambda x^2+b\lambda x+\lambda c.$$

If $\lambda\ne 0$ then you get $a\lambda=0$ since $L(p)$ has no terms in $x^2.$ This imply $a=0.$ Thus, $p=bx+c.$ Repeating the argument,

$$b=L(p)=b\lambda x+\lambda c,$$ from where $b=0.$ Finally $p=c.$ But

$$0=L(p)=\lambda c$$ implies $c=0.$ That is, $p=0.$

Thus, $L$ doesn't admit any nonzero eigenvalue.

There is no need to obtain a matrix representation of $L$ for this problem.

If $Lp = \lambda p$, then $Lp$ and $\lambda p$ must have the same degree. Hence if $\lambda \neq 0$, we must have that $p$ is a constant function (the derivative of a non-constant polynomial has degree one less that the polynomial), and hence it must be zero. So there are no non-zero eigenvalues.

Since $L(x \mapsto 1) = 0$, we see that $0$ is an eigenvalue.