Suppose that $N \in M_{n,n}(\mathbb{C})$ is nilpotent (that is, $N^k = 0$ for some integer $k &gt; 0$). 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.
32265
Possible Duplicate:          Given $A^{2012}=0$ prove that $A+I$ is invertible and find an expression for $(A+I)^{-1}$ in terms of $A$. ($I$ is the identity matrix).
76155
If $A\in M_{n}$ has $n$ distinct eigenvalues and if $A$ commutes with a given matrix $B\in M_{n}$, how can I show that $B$ is a polynomial in $A$ of degree at most $n-1$? I think first I need to show that $A$ and $B$ are simultaneously diagonalizable, but I don't know how to begin. Any advice is much appreciated.
11622
Given a ring $R$ (with unity), prove that if $a \in R$ is nilpotent, then $a+1,a-1$ are both invertible.   Suppose $a^n=0$. Then $1=1-a^n=(1-a)(1+a+\ldots+a^{n-1})$, so $1-a$ is invertible.  If $n$ is odd we can use $1+a^n=(1+a)(1-a+\ldots+a^{n-1})$.  But what if $n$ is even?
62821
I would like to prove the following proposition, which is given as an exercise in Hoffman and Kunze:     If $A$ is a $2\times 2$ matrix with coefficients in $\mathbb{C}$, then $A$ is similar either to a matrix of the form $\begin{pmatrix} a &amp; 0 \\ 0 &amp; b \end{pmatrix}$ or to a matrix of the form $\begin{pmatrix} a &amp; 0 \\ 1 &amp; a \end{pmatrix}$.   A hint directs the reader to prove that if $N$ is an nilpotent matrix (also in $M_{2}(\mathbb{C})$), then either $N=0$ or $N$ is similar to $\begin{pmatrix} 0 &amp; 0 \\ 1 &amp; 0 \end{pmatrix}$. I have proven this claim (by supposing that $N \neq 0$ and showing that the transformation induced by multiplication by $N$ has matrix representation $\begin{pmatrix} 0 &amp; 0 \\ 1 &amp; 0 \end{pmatrix}$ under a particular basis).  However, I'm not really sure how to use this knowledge to prove the proposition in question. Any steps in the right direction would be appreciated. Also, I'd love to see any other proofs of the proposition.  Thanks in advance!
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If $A\in M_{n\times n}\left(\mathbb{R}\right)$ is such that $ (A+I)^3=0$, show that $A$ is an invertible matrix and find the inverse of $A$.  My idea was: \begin{eqnarray*} 0&amp;=&amp;\left(A+I\right)^3\\ &amp;=&amp;A^3+3A^2I+3AI^2+I^3\\ &amp;=&amp;A^3+3A^2+3A+I, \end{eqnarray*} then $$I=-A^3-3A^2-3A,$$ so $$I=A\left(-A^2-3A-3I\right).$$ It follows that $$A^{-1}=-A^2-3A-3I.$$ Now I found the inverse matrix, but how does this show that an inverse actually exists?
46430
Possible Duplicate:          An element $a$ of a ring $R$ is nilpotent if $a^n = 0$ for some positive integer $n$. Let $R$ be a commutative ring, and let $N$ be the set of all nilpotent elements of $R$.  (a) I'm trying to show that $N$ is an ideal, and that the only nilpotent element of $R/N$ is the zero element.  (c) What are the nilpotent elements of $R = \mathbb{Z}_{24}$? And what is the quotient ring $R/N$ in that case? (what known ring is it isomorphic to?)
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Consider the matrix $$ M=\left(\begin{matrix} 0&amp;0&amp;0&amp;1\\ 0&amp;0&amp;0&amp;0\\ 0&amp;0&amp;0&amp;0\\ 0&amp;0&amp;0&amp;0\\ \end{matrix}\right). $$ Is there a matrix $P\in{\Bbb C}^{4\times 4}$ such that $P^n=M$ for some $n&gt;1$?    One obvious fact is that if such $P$ exists, then $P$ must be nilpotent. However, I have no idea how to deal with this problem. Furthermore, what if $M$ is an arbitrary nilpotent matrix with index $k$?
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I have to show that all $n \times n$ nilpotent matrices of order n are similar. My initial approach was to show that for all nilpotent matrices their minimun characteristic polynomial is of the form: $$\lambda^n$$  Is this sufficient?  Can someone show me a formal approach to this problem?  Thanks!
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Prove that a matrix with only zero eigenvalues must be nilpotent.   How will I be able to prove this?
88261
Can anyone show me how to find invertible elements in $\mathbb{Z}/m\mathbb{Z}$, $m$ is say $28$?  Also I'm not very clear about what it mean by 'invertible element in $\mathbb{Z}/m\mathbb{Z}$'
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Suppose $A$ is invertible $n\times n$ matrix. Show that $A^{-1}$ can be written as a polynomial of degree at most $n-1$, i.e.  $$A^{-1} = c_{n-1}A^{n-1} + \dots + \ c_{1}A + c_{0}I$$  Are there any tips on how may this proof be approached?
5337
I wish to prove the following:  if $T:X\to X$ is a nilpotent (and the     linear space $X$ is finite-dimensional), then there exists a basis     of $X$ such that the matrix representation of $T$ is upper     triangular with zero diagonal elements.  I was trying to prove $\{x_1,...,x_{k+1},v, Tv, ..., T^{n-k-2}v\}$ is a basis for $X$ where $x_i$ are basis for $N(T)$ and $v,..., T^{n-k-2}v$ are basis for $R(T)$ where $T^{n-k}=0$ and $n-k$ is minimal. However the matrix representation w.r.t this basis is diagonal. Is there an easy way to prove this ?  I know that for all $T:X\rightarrow X$, there exists a basis that the matrix representation is  upper triangular. How is this useful?
87769
Given the following identity: $$\sum_{i=0}^k (-1)^i \binom ni = (-1)^k \binom {n-1}{k}.$$  This is provable by induction. However, I wonder if there is a way to prove this in a combinatorial fashion (something like subsets, groups, choosing $n$ balls from a box with $k$ balls, et cetera.)  Thank you.
53375
if $a,b$ are nilpotent elements of a commutative ring $R$, show $a+b$ is also nilpotent  So then $a^n=0, b^m = 0, n,m \in \mathbb{Z}^+$ I know this is solvable using the binomial theorem but I would much rather solve it another way if possible. The answer using the binomial theorem isn't making the most sense at the present moment.
49333
Let $A$ be an $n\times n$ complex matrix. If $A^m=I_n$ for some positive integer $n$. How to show that $A$ is diagonalizable?
99031
Let $A$ and $B$ be $n$ $\times$ $n$ real matrices such that $AB$=$BA$=$0$ and $A+B$ is invertible. Which of the following are always true?  $1$. rank $(A)$=rank $(B)$.  $2$. rank $(A)$$+$rank $(B)$ =$n$  $3$. nullity $(A)$ $+$ nullity $(B)$ = $n$  $4$. $A-B$ is invertible.  I am not getting how to do. Atleast it is clear that Rank$(A+B)$= $n$. help me!
34125
Establish if the matrix $A\in M_3 (\Bbb{Z}_9)$  $$     \begin{bmatrix}     1 &amp; 2 &amp; 3 \\     2 &amp; 3 &amp; 0 \\     0 &amp; 3 &amp; 6     \end{bmatrix} $$  is invertible, and if so, find the inverse matrix.