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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.