TIFR 2014 Problem 11 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.

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## Problem:

Let \(A\) be an \(nxn\) matrix with real entries such that \(A^k=0\) (0-matrix) for some \(k\in\mathbb{N}\).

Then

A. A has to be the 0-matrix.

B. trace(A) could be non-zero.

C. A is diagonalizable.

D. 0 is the only eigenvalue of A

## Discussion:

Let \(v\) be an eigenvector of \(A\) with eigenvalue \(\lambda\).

Then \(v \neq 0\) and \(Av=\lambda v\).

Again, \(A^2 v=A(Av)=A(\lambda v)=\lambda Av= (\lambda)^2v\).

We continue to apply A, applying it k times gives: \(A^k v=(\lambda)^k v\).

By given information, the left hand side of the above equality is 0.

So \(\lambda^k v=0\) and remember \(v \neq 0\).

So \(\lambda =0\).

Therefore \(0\) is the only eigenvalue for \(A\).

So D is true.

We analyse the question a little bit further, to check it satisfies no other options above.

We know \(trace(A)=\) *sum of eigenvalues of A*= \(\sum 0 =0\)

So option B is false.

Take \(A=\begin{bmatrix} 0 & 1 // 0 & 0 \end{bmatrix} \).

Then \(A^2 =0\). But \(A\) is not the zero matrix.

Also, if \(A\) were diagonalizable then the corresponding diagonal matrix would be the zero matrix. Which would then imply that \(A\) is the zero matrix, which in this case it is not. (See TIFR 2013 Probmem 8 Solution-Diagonalizable Nilpotent Matrix ) So this disproves options A and C.

## Helpdesk

**What is this topic:**Linear Algebra**What are some of the associated concept:**Eigenvectors,Characteristic Polynomial**Book Suggestions:**Introduction to Linear Algebra by Gilbert Strang