The following stumps me:     Let $\mathbb K$ be a field. Let $A, B \in \mathbb K^{n \times n}$ where $B$ is nilpotent and commutes with $A$, i.e., $A B = B A$. Show that $$ \det(A+B)=\det(A) $$   I have no idea how to approach this I thought perhaps raise both sides to a power but nothing works. Thanks for all help.
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Assume that the matrices $A,\: B\in \mathbb{R}^{n\times n}$ satisfy  $$ A^k=0,\,\, \text{for some $\,k\in \mathbb{Z^+}$}\quad\text{and}\quad AB=BA. $$  Prove that $$\det(A+B)=\det B.$$
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Let $A \in {M_n}$ and $trcA=0$.why $A=M+N$  where $M$ and $N$  are nilpotent matrices?
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.
61531
How can you show that $$ \det(A + B^T) = \det(A^T + B)$$ for any $n\times n$ matrices $A$ and $B.$
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How to prove that an element $a$ is a nilpotent element of $\mathbb Z_n$ if and only if each prime divisor of $n$ is a divisor of $a$?
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Possible Duplicate:          Given a commutative ring $R$ and two nilpotent elements $r$, $s$ there exists an $n \in \mathbb{N}$ such that  $$ (r+s)^n = 0.$$  I want to prove that in order to show that the nilpotent elements of a commutative ring are closed under addition (to show that the nilpotent elements form an ideal within a commutative ring). Is that a good way to do it? I'm a little stuck.
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Consider the transpose inverse automorphism on $GL_n(\mathbb F)$ where $n\geq2$ and $|\mathbb F|&gt;2$. (i.e. $\mathbb F$ is a field, possibly infinite, with three or more elements). I want to show this automorphism is not inner. I was told to consider $\det(BAB^{-1}) = \det(A)$ and $\det(\,^TA^{-1})=\det(A)^{-1}$ and derive a contradiction.  However, in some fields (where the result holds) I fail to see the contradiction. What about $\mathbb F_3$? An element of $GL_2(\mathbb F_3)$ either has determinant $1$ or $2$, both of which are their own multiplicative inverses in $\mathbb F_3$. I'm not sure if my knowledge of fields and determinants is flawed or the hint is.
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On , I saw the following formula  $$\det\begin{bmatrix}A &amp; B\\ C &amp; D\end{bmatrix} = \det(AD-BC)$$  if $C$ and $D$ commute. Is this always true?   Or is there a good counter example for each $2 \times 2$ block matrices?
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(Question 9 in chapter 6.1 of Dummit and Foote). Prove that a finite group G is nilpotent if and only if whenever $a, b \in G$ with $(|a|, |b|) = 1$ then $ab = ba$. It says to use the following theorem:  Let G be a finite group, $p_1$, $p_2$, ... $p_s$ be the distinct primes dividing its order, and $P_i \in Syl_{p_i}(G)$. Then G is nilpotent iff $G \cong P_1 \times P_2 \times ... P_s$.  I believe I know the if direction: an element $a \in G$ corresponds to an element $(g_1, g_2, ... g_s) \in P_1 \times P_2 \times ... P_s$ and $|a| = lcm(|g_1|, |g_2|, ... |g_s|)$. If $b$ corresponds to $(h_1, h_2, ... h_s)$ then $(|a|, |b|) = 1$ implies each $(|g_i|, |h_i|) = 1$. Since the order of the elements divides $|P_i|$ a prime power, $|g_i|$ or $|h_i|$ has to be 1 or their gcd would not be 1. So one of every pair $g_i$ and $h_i$ has to be 1, so they commute, so $a$ and $b$ commute.  I'm not sure how to do the only if direction. Any pointers? Thank you
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I am trying to show that $[A,B^n] = nB^{n-1}[A,B]$ where A and B are two Hermitian operators that commute with their commutator. However, I am running into a little problem and would like a hint of how to proceed.  If A and B commute then $[A,B] = ABA^{-1}B^{-1} = e$ where e is the identity element of the group.  $$\therefore AB=BA$$  $$n=1; [A,B^1] = (1)B^0[A,B] = e$$ This statement is certainly true. however moving on to $n = 2$ I find...  $$[A,B^2] = AB^2A^{-1}B^{-2} = ABBA^{-1}B^{-1}B^{-1} = BBAA^{-1}B^{-1}B^{-1}$$  Where in the last step I have used the fact that A and B commute to rearange the terms. However, it is plain to see that this last term simply reduces to the identity as well and for the n = 2 case we have:  $$[A,B^2] = e \ne 2B[A,B] = 2Be = 2B$$  Clearly I have assumed something I shouldn't have. The fact that there is a multiplicative factor of n implies I should be adding things, but I thought if I kept it as general as possible, the answer should just fall out naturally. I don't want an answer please, only guidance.
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Let $A,B,C \in \ M(3,\mathbb{R})$ be such that $A$ commutes with $B $and $B$ commutes with $C$. $B$ is not scalar matrix. Then, $A$ commutes with $C$.   Is this true or false? I think it is not true, but cannot think of a counterexample. Any suggestions?
38731
If $A$ and $B$ are two $4\times3$ matrices such that $A\mathbb{v}=B\mathbb{v}$ for all $\mathbb{v}\in\mathbb{R}^3$, then $A=B$?  If it's not true, can you give me an example of it?  Thank you.
34722
It is known that if two matrices $A,B \in M_n(\mathbb{C})$ commute, then $e^A$ and $e^B$ commute. Is the converse true?     If $e^A$ and $e^B$ commute, do $A$ and $B$ commute?   Edit: Addionally, what happens in $M_n(\mathbb{R})$?  Nota Bene: As a corollary of the counterexamples below, we deduce that if $A$ is not diagonal then $e^A$ may be diagonal.
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I am trying to prove a result, for which I have got one part, but I am not able to get the converse part.  Theorem. Let $R$ be a commutative ring with $1$. Then $f(X)=a_{0}+a_{1}X+a_{2}X^{2} + \cdots + a_{n}X^{n}$ is a unit in $R[X]$ if and only if $a_{0}$ is a unit in $R$ and $a_{1},a_{2},\dots,a_{n}$ are all nilpotent in $R$.  Proof. Suppose $f(X)=a_{0}+a_{1}X+\cdots +a_{n}X^{n}$ is such that $a_{0}$ is a unit in $R$ and $a_{1},a_{2}, \dots,a_{r}$ are all nilpotent in $R$. Since $R$ is commutative, we get that $a_{1}X,a_{2}X^{2},\cdots,a_{n}X^{n}$ are all nilpotent and hence also their sum is nilpotent. Let $z = \sum a_{i}X^{i}$ then $a_{0}^{-1}z$ is nilpotent and so $1+a_{0}^{-1}z$ is a unit. Thus $f(X)=a_{0}+z=a_{0} \cdot (1+a_{0}^{-1}z)$ is a unit since product of two units in $R[X]$ is a unit.   I have not been able to get the converse part and would like to see the proof for the converse part.
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Let A be a commutative ring with 1.   1) Prove that a sum of a nilpotent element and an invertible element is invertible.   2) Prove that if $f=a_0+a_1x+\dots+a_nx^n \in A[x]$  a) $\exists f^{-1}\in A[x] \Leftrightarrow a_0$ is invertible and the other coefficients are nilpotent.  b) f is nilpotent $\Leftrightarrow $ all its coefficients are nilpotent.  p.s. Those are the first two in a series of problems. The rest easily follow from each other. I'm only struggling with the first two.
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Let $B$ be a given nilpotent $n\times n$ matrix with complex entries. Let $A = B-I$   find out $\det(A)$. What if B is orthogonal or skew symmetric matrix? Then can we say anything about its trace and determinant?
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Let $A$ be a $  2 \times  2 $ matrix and  a positive integer $k \geq 2$. Prove that $A^k = 0 $ iff $A^2 = 0$.  I can make it to do this exercise if I have $ \det (A^k) = (\det A)^k $. But this question comes before this.  Thank you very much for your help!