Select Page TIFR 2013 Problem 38 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Linear Algebra Done Right by Sheldon Axler. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## 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