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Problem Type:True/False?

Let (V) be the vector space of polynomials with real coefficients in variable (t) of degree ( \le 9). Let (D:V\to V) be the linear operator defined by (D(f):=\frac{df}{dt}). Then (0) is an eigenvalue of (D).


Hint:

If 0 were an eigenvalue, what would be its eigenvector?


Discussion:


There are several ways to do this. One possible way is to find out the matrix representation of (D) with respect to standard basis ( {1,t,t^2,…,t^n})( or any other basis) and observe that it is a (strictly) upper triangular matrix with all diagonal entries 0 and therefore the determinant of (D) is 0. This implies that D is not injective, so there is some nonzero vector to which when (D) is applied gives the zero vector. Therefore, (D) has 0 eigenvalue.

Another way to do this is by observing that (D(1)=0(1)), therefore 0 is an eigenvalue of (D) with 1 as an eigenvector.


HELPDESK

  • What is this topic: Linear Algebra
  • What are some of the associated concept: eigenvector,eigenvalue
  • Book Suggestions: Linear Algebra done Right by Sheldon Axler