Some questions on Nilpotent matrix [closed]

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.

• What is the question? Jun 11, 2014 at 12:08
• @ Algebraic Pavel .. the question is that which of the given statements are true. Jun 11, 2014 at 12:09
• 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. Jun 11, 2014 at 12:22
• @ 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. Jun 12, 2014 at 5:56
• 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. Jun 12, 2014 at 7:35

1 Answer

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.