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 \(D\).


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


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.