5

Yes, we can conjugate $B_1$ and $B_2$ simultaneously by a matrix $S$, whose columns are a basis of generalized eigenvectors of $B_1$, such that $S^{-1}B_1S$ is in Jordan normal form, and both $S^{-1}B_1S$, $S^{-1}B_2S$ are upper-triangular. This is possible, because commuting operators leave invariant generalized eigenspaces. On the other hand, not both of $...


4

The result does not hold. Here is a counterexample: take $X=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$, $E=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, $F=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$, and consider the matrices $$A:=\begin{bmatrix} [0]_{2 \times 2} & [0]_{2 \times 2} & I_2 & [0]_{2 \times 1} \\ [0]_{2 \times 2} & ...


3

The claim is not true. Suppose that $n= 3$ and take $A=\lbrace E_{1,2}+E_{2,3}, E_{3,2}-E_{2,1}\rbrace$, and let $S$ be the span of $A$. Note that all matrices from $S$ are nilpotent. However, if we compute $E(X)$, $E^2(X)$ and $E^3(X)$, we see that the entries from $E(X)$ and $E^3(X)$ are equal up to a scalar $1$ or $2$. In fact, the coefficients $x_{21},x_{...


3

It seems that here is a example of $k$ such transforms of the $(k+2)$-dimensional space. Let $(e_0,\dots,e_{m+1})$ be some fixed basis of the space. The $i$th transform $\phi_i$ acts as follows: $\phi_i(e_0)=\phi_i(e_i)=0$, $\phi_i(e_j)=e_0$ for $1\leq j\leq m$, $j\neq i$, and $\phi_i(e_{m+1})=e_i$. Thus $\mathop{\rm Im} \phi_i=\langle e_0,e_i\rangle$, and ...


2

The more general fact is that, if $N \subset \mathrm{Mat}_{n \times n}(k)$ is a subrng (ring without identity) whose every element is nilpotent, then we can choose a basis to make every element of $N$ simultaneously strictly upper triangular. This is a useful lemma which, as far as I know, doesn't have a standard name. Note that, if $N_1$ and $N_2$ are ...


2

The answer is positive. Details and link to Bourbaki, Lie groups and Lie algebras, Chap VII, are given in the answer here.


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