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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