**Question:**

*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.

*Related*