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# Nilpotent matrix eigenvalues (TIFR 2014 problem 11)

Question:

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 diagonalizable-nilpotent-matrix-tifr-2013-problem-8 ) So this disproves options A and C. ## By Ashani Dasgupta

Ph.D. in Mathematics from University of Wisconsin Milwaukee (USA)
Founder - Faculty at Cheenta