4 votes

Exercise 10, Section 3.4 of Hoffman’s Linear Algebra

If you're familiar with the idea of diagonalization, then since $S$ satisfies $S^2=S$ and is an operator on $\mathbb{R}^2$, the characteristic polynomial of $S$ is then $p(z)=z^2-z$. Since $S\neq I$ ...
user avatar
  • 3,226
3 votes
Accepted

Solution Verification: An equivalence relation on finite subsets of $\mathbb{N}$, defined by having equal sums of the elements in the sets

Your solution to $(1)$ neglects something: that $B$ is just the set $\{2,2,2\}=\{2\}$; this is why you should be more careful and write things explicitly as sets. (We delete duplicate entries in sets.)...
user avatar
  • 32.1k
3 votes
Accepted

Convex $n$-gons that can be decomposed into $2n$ right triangles

I assume by "rectangle triangle" you mean "right triangle". Take a convex $n$-gon $A$ with consecutive vertices $V_i$ for $1\leq i\leq n$. Fix $P=V_n$. Form $n-2$ (in general non-...
user avatar
  • 1,566
2 votes
Accepted

Correctness of the proof that $\lim_{n \to \infty}(x_{n}-x_{n-2})=0$ implies $\lim_{n \to \infty}\frac{x_{n}}{n}=0$.

If $n$ is odd numbers,we have \begin{align} \lim_{k \to \infty} \frac{x_{2k+1}-x_{2k-1}}{2k+1-(2k-1)}=\lim_{k \to \infty} \frac{x_{2k+1}-x_{2k-1}}2&=0\\ \\ \Rightarrow \lim_{k \to \infty} \...
user avatar
  • 6,182
2 votes

Applying a given formula to induction step.

I assume that you meant to write: $$x_{n+1} = \sqrt{2x_n-1}$$ The square root looks a bit weird in your post but I'm guessing that this is what you wanted. So, we have that: $$x_{n+1}-x_n = \sqrt{2x_n-...
user avatar
1 vote
Accepted

Is this proof that one can pull an existential quantification out of a universal quantification using a cartesian product of the power set correct?

The proof does look correct. And it doesn't require the Axiom of Choice. In fact, it constructs exactly the sort of set one might then apply the Axiom of Choice on as a next step.
user avatar
  • 5,444
1 vote
Accepted

Exercise 10, Section 3.4 of Hoffman’s Linear Algebra

Expanding on a way from my hint: You noted if $S \neq 0$ and $S \neq I$, we must have vectors $\alpha_1, \alpha_2 \in \mathbb{R}^2$ with $S \alpha_1 \neq 0$ and $S \alpha_2 \neq \alpha_2$. To satisfy $...
user avatar
  • 5,444
1 vote
Accepted

Matchstick dice game winning strategy

We can describe the current game state as $(n,d)$ where $n \in \mathbb{N}$ (including zero) is the number of nuts remaining and $d \in \{1,2,3,4,5,6\}$ is the number on the top face of the die, ...
user avatar
  • 5,444
1 vote
Accepted

Exercise 11, Section 3.5 of Hoffman’s Linear Algebra

Since $W_1\cap W_2$ is a subspace of $W_2$, there is a complementary subspace $U_2$ such that $W_2=(W_1\cap W_2)\oplus U_2$. Let $U$ be a complement of $W_1+W_2$ in $V$, So $U\oplus U$ is a complement ...
user avatar
  • 306
1 vote

Prove that two sets A and B with $A \cap B=\emptyset$, $\sup A = \sup B$, $\sup A \notin A$ and $\sup B \notin B$ cannot exist.

Let $A:=\left\{1-\frac{1}{2n-1}\mid n\in\left\{1,2,\dots\right\}\right\}.$ Let $B:=\left\{1-\frac{1}{2n}\mid n\in\left\{1,2,\dots\right\}\right\}.$
user avatar
  • 6,540

Only top scored, non community-wiki answers of a minimum length are eligible