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Suppose that $N \in M_{n,n}(\mathbb{C})$ is nilpotent (that is, $N^k = 0$ for some integer $k > 0$). Show that $I+N$ is invertible, and find its inverse as a polynomial in $N$.

I think I got the first part down "intuitively". Noticing that $N$ is nilpotent, so $N$ will be a matrix with a diagonal (any diagonal) with just $1$'s as its entries. Then if I add the identity matrix to it, the diagonal will definitely have $1$ as a diagonal, and so $\det(N+I) = 1$ hence invertible. Is there a more standard way to prove this rather than just "talking" through it?
Also, I'm unsure how I could approach the second part.

similarityinvariance1
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    Try $I-N+N^2-N^3+N^4-N^5+...$ – OR. Jun 24 '17 at 03:02
  • $N$ will have $\color{Red}0$s in its diagonal entries - it will be strictly upper triangular - at least with respect to some basis. – anon Jun 24 '17 at 03:08
  • Note that $$\begin{pmatrix} 6 & 9 \ -4 & -6 \end{pmatrix}^2 = 0 $$ so your intuition is flawed. – Trevor Gunn Jun 24 '17 at 03:14
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    An appropriate change of basis will make any complex nilpotent matrix look like OP has described, so his intuition is not far off. – D_S Jun 24 '17 at 03:21

2 Answers2

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$(I+N)(I-N+N^2-N^3+...+(-1)^{k-1}N^{k-1})=I$

Chinnapparaj R
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  • I'll still give you an upvote, but for future reference: This is a bit short to be an answer. A response like this fits better into the comments section. – Ken Duna Jun 24 '17 at 03:11
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    Dear @Ken Duna: are you joking ? An answer is an answer, even with a single character. I find quite dismaying the number of people who revel in censoring others because they don't fancy this or that, at their discretion. The language is perfectly respectuous, and on top of that the answer is correct. If the policies of stackexchange were that some correct answers must be refused because of some totally irrelevant features then it would be deeply flawed (but i don't think the founders of stackexchange mean anything like that). Thank you Chinnapparaj for the correct answer. – plm May 15 '23 at 10:06
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Something that comes up over and over again in different parts of math is the geometric series. For motivation, consider the infinite sum

$$S = 1 + x + x^2 + \cdots$$

Here $x$ could be a complex number, a matrix, a linear operator, or something else. For the moment, just pretend the infinite sum makes sense. To find a closed expression for $S$, notice that $S$ appears on the right hand side of the equation again if you factor out $x$:

$$1 + x + x^2 + \cdots = 1 + x(1+ x + \cdots) = 1 + xS$$

Now you can solve for $S$ as

$$S = \frac{1}{1-x}$$

In particular, $S$ is the inverse of $1-x$. As I mentioned, the infinite sum $S$ may or may not make sense. If $x$ is an element of a metric space in which addition is defined, the usual definition of $S$ is the limit as $n$ goes to infinity of $1 + x + x^2 + \cdots + x^n$, provided this limit exists. If $x$ is a nilpotent matrix, then $S$ also makes sense, because it is just a finite sum. Reasoning as I've done above will produce for you an inverse of $1+x$.

D_S
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