# Finding eigenvalues

#### wattkow

Let $$\displaystyle T: P_2 \mapsto P_2$$ 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.

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#### dwsmith

MHF Hall of Honor
Let $$\displaystyle T: P_2 \mapsto P_2$$ 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.

• HallsofIvy

#### wattkow

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

MHF Hall of Honor
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 $$\displaystyle P_2$$ on $$\displaystyle P_2$$. Unless your book notates this differently, I am almost positive these are polynomials; however, different books notate $$\displaystyle P_2$$ differently. For instance, mine says $$\displaystyle P_2$$ are polynomial of degree 1 or less. Thus, the 2 in $$\displaystyle P_2$$ is the amount of elements {1, x} but some books refer to $$\displaystyle P_2$$ as degree two or less, {$$\displaystyle 1$$, $$\displaystyle x$$, $$\displaystyle x^2$$}. What does your book say?

#### wattkow

I am assuming P is a polynomial since we are mapping $$\displaystyle P_2$$ on $$\displaystyle P_2$$. Unless your book notates this differently, I am almost positive these are polynomials; however, different books notate $$\displaystyle P_2$$ differently. For instance, mine says $$\displaystyle P_2$$ are polynomial of degree 1 or less. Thus, the 2 in $$\displaystyle P_2$$ is the amount of elements {1, x} but some books refer to $$\displaystyle P_2$$ as degree two or less, {$$\displaystyle 1$$, $$\displaystyle x$$, $$\displaystyle x^2$$}. What does your book say?
I believe in my case it's the 2nd degree polynomial.

#### dwsmith

MHF Hall of Honor
I believe in my case it's the 2nd degree polynomial.
Ok so your linear transformation goes from $$\displaystyle ax^2+bx+c\rightarrow xp'(x)=x(2ax+b)=2ax^2+bx$$.

Now you need to find the eigenvalues of the linear transformation.

#### Random Variable

I would do the following:

A basis for $$\displaystyle P_{2}$$ is $$\displaystyle \{1,x,x^{2} \}$$

$$\displaystyle T(1) = 0$$

$$\displaystyle T(x) = x$$

and $$\displaystyle T(x^{2}) = 2x^{2}$$

so the transformation matrix is $$\displaystyle \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 2 \end{bmatrix}$$

which has eigenvalues 0,1, and 2

• wattkow and HallsofIvy

#### HallsofIvy

MHF Helper
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, $$\displaystyle P_2$$ 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 $$\displaystyle p= ax^2+ bx+ c$$, then $$\displaystyle T(p)= xp'= x(2ax+ b)= 2ax^2+ bx$$. If $$\displaystyle \lambda$$ is an eigenvalue for T, we must have some a, b, c, not all 0, such that $$\displaystyle T(p)= 2ax^2+ bx= \lambda(ax^2+ bx+ c)$$. Since that must be true for all x, we can set corresponding coefficients equal: $$\displaystyle 2a= \lambda a$$, $$\displaystyle b= \lambda b$$, and $$\displaystyle 0= \lambda c$$.

If a is not 0, then $$\displaystyle \lambda= 2$$, if b is not 0, then $$\displaystyle \lambda= 1$$, and if c is not 0, then $$\displaystyle \lambda= 0$$.

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 $$\displaystyle x^2$$, exactly what Random Variable got.

• wattkow

Thanks fellas.