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Q & A style.

Just wanted to share the following question which came in a competitive exam and so college level maths students may find it useful:

A non-zero matrix $A\in M_n(\mathbb{R})$ is said to be nilpotent if $A^k = 0$ for some positive integer $k\geq 2$. If A is nilpotent, which of the following statements are true?

  1. $k\leq n$ for the smallest such $k$.
  2. The matrix $I + A$ is invertible.
  3. All the eigenvalues of A are zero.

Here $M_n(\mathbb{R})$ is the real vector space of all $n\times n$ matrices with real entries.

I have given quite a clear explanation of my way of approach in solving it in the Answer section.

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  • $\begingroup$ What is the question? $\endgroup$ Jun 11, 2014 at 12:08
  • $\begingroup$ @ Algebraic Pavel .. the question is that which of the given statements are true. $\endgroup$
    – Debashish
    Jun 11, 2014 at 12:09
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    $\begingroup$ may be ... but we are always welcomed to share our knowledge in this site. This may be know to some and may not be known to others. $\endgroup$
    – Debashish
    Jun 11, 2014 at 12:22
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    $\begingroup$ @ Brian and others ... This is a Q&A style question which this site encourages to share. Both the question and answer are clear and so I thought I would share this as students may find it useful. I think it can be reworded to fit within the scope. Please do leave your comments. $\endgroup$
    – Debashish
    Jun 12, 2014 at 5:56
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    $\begingroup$ But I think the question is quite clear and meets the standard for introductory matrix problems. I have posted this in Q & A style and given quite a clear explanation of my way of approach in solving it. $\endgroup$
    – Debashish
    Jun 12, 2014 at 7:35

1 Answer 1

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Here is the solution. Any suggestion will be appreciated.

Clearly, $A^k$ has all its eigenvalues equal to zero. But eigenvalues of $A^k$ are just the $k$th power of the eigenvalues of $A$. Thus, all the eigenvalues of $A$ must be zero. Hence (3) is true. The eigenvalues of $I+A$ are $1+0,1+0,1+0$ i.e. $1,1,1$ and so it is invertible. In fact $(I+A)^{-1}=I-A+A^2-A^3+\ldots+(-1)^{k-1}A^k$. Hence, (2) is true. Since all the eigenvalues of $A$ are zero, so the characteristic equation of $A$ is $x^n=0$ and so by Cayley Hamilton theorem, $A^n=0$. Thus, the smallest such $k$ for which $A^k=0$ can at most be equal to $n$. Thus, $k\leq n$. Hence, (1) is also true.

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