For reference, if $R$ denotes either $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$, a matrix $A = (a_{ij}) \in M_{n}(R)$ is called unitriangular if all diagonal entries $a_{ii}$ are $1$ and all entries $a_{ij}$ $(i>j)$ below the diagonal are $0$ (i.e., an upper triangular matrix with ...
Let $G_{1},G_{2},\cdots, G_{n}$ be any finite collection of finite groups. I need to prove that there exists a finite group $\mathbf{G}$ such that each $\mathbf{G_{i}}$ is a homomorphic image of $\mathbf{G}$.
Since this problem is in the section we're doing on direct products, I let $G = G_{1}...
I need to prove that any two nontrivial subgroups of $\mathbb{Q}$ have a nontrivial intersection as part of a larger proof that $\mathbb{Q}$ cannot be represented as a nontrivial direct product (Yes, I realize that there are other ways to prove that $\mathbb{Q}$ cannot be represented as a nontriv...
Today when joining the chat thus revising how $0=1$ implies triviality in rings. I then started to wonder whether the same thing hold in semirings (aka rig, which are generalisation of rings throwing away additive inverses) if the annihilator axiom
$$\forall a\hspace{1mm}0a=0 $$
is thrown away
...
Question: Let $V$ be an inner product space finitely generated over $\Bbb C$ and let $\alpha$ be an endomorphism of $V$ satisfying $\alpha \alpha^* = \alpha^2$. Show that $\alpha$ is selfadjoint.
**Edited after more thought:
I know that in order to be selfadjoint $\alpha^*=\alpha$. Becuase this...
For reference, if $R$ denotes either $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$, a matrix $A = (a_{ij}) \in M_{n}(R)$ is called unitriangular if all diagonal entries $a_{ii}$ are $1$ and all entries $a_{ij}$ $(i>j)$ below the diagonal are $0$ (i.e., an upper triangular matrix with ...
