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Let $A=(a_{i,j})$. The $1\times 1$ minors show that the diagonal entries $a_{i,i}$ have to be zero. The $2\times 2$ minors then show that for $i\ne j$, either $a_{i,j}=0$ or $a_{j,i}=0$. So we can for …

Take the group of matrices of the form $\left(\begin{smallmatrix}A&B\\0&A^{-1}\end{smallmatrix}\right)$ with $A\in U(2)$. Inside this, choose two generators for a free group as the values of $A$, in s …

Since $k'$ is finite dimensional over $k$, you can move it from inside to outside a Hom in the last isomorphism below. $$\operatorname{\rm Hom}_{k'G}(k'\otimes_k C,k'\otimes_k D)\cong \operatorname{\r …

I presume you mean invertible in $M_{n\times n}(\mathbb{Q})$. Here's an example. $$\left(\begin{matrix} 1&2&3&4&5&6\\ 2&1&3&4&5&6\\ 1&3&2&4&5&6\\ 3&1&2&4&5&6\\ 2&3&1&4&5&6\\ 3&2&1&4&5&6 \end{matrix}\r …

Here's a counterexample: For $U$ take $\left(\begin{smallmatrix} 0& 0 &0& 0& 1& 1& 1& 1\\ 0& 0 &0 &1& 0& 1& 1& 1\\ 0& 0 &0& 0& 1& 0& 0& 1\\ 0& 0 &0& 0& 0& 0& 0& 1\\ 0& 0 &0& 0& 0& 0& 1& 0\\ 0& 0 &0& 0 …

No. An easy three dimensional example has basis $u$, $v$, $w$ with $A(u)=w$, $A(v)=0$, $A(w)=0$, $B(u)=0$, $B(v)=w$, $B(w)=0$. Since $AB=BA=0$, $A$ and $B$ commute. This is indecomposable but not gene …

Let $\mathbb{F}$ be a field of characteristic two, and let $K=\mathbb{F}(t)$, the field of rational functions over $\mathbb{F}$. Let $p(x)=x^2-t$. If $A$ is a symmetric matrix with entries in $K$ then …

Let $A=\left(\begin{smallmatrix}1&0&0\\1&1&0\\0&0&1\end{smallmatrix}\right)$ and $B=\left(\begin{smallmatrix}1&0&0\\0&1&0\\1&0&1\end{smallmatrix}\right)$. Then your hypotheses are satisfied, with $v=\ …

Since the order of $GL(n,2)$ is less than $2^{n^2}$, $a(n)$ is at least $2^{2^n−1−n^2}$. I think it's closer to that than to $2^{2^n}$. In particular, this shows that $$\lim_{n\to\infty}\frac{\log_2\l …