Do you see the goldfish? It's a top view of a goldfish, because that's one direction in which a fish would look symmetric.
Anyway, you construct it by adding two edges to certain even numbers, namely those that aren't double a prime $2p$. If they are of that form, then connect an edge from t...
Question for myself: Let $A$ by a square matrix. If we may partition this matrix as $A=[A_1|A_2]$ where $A_1,A_2$ both have full rank, must $A$ itself have full rank?
I guess that's equivalent to: Suppose I have vectors $\{u_k\}_{k=1}^n,\{v_k\}_{k=1}^m \subseteq \mathbb{R}^{n+m}$. If both sets of vectors are linearly independent, must their union be linearly independent?
The first way I wrote that made me think $A$ would have to have full rank, but that rewriting of it makes me dubious...
ok, yeah: Consider the case where $n=m$ and $A_1=A_2$ are full rank. Then $A$ will have rank $n=m<n+m$.
In my case of interest I further have $A_1^\top A_2=0$. This would rule out the previous example since then $A_1^\top A_2=A_1^\top A_1=0$ and therefore $A_1=A_2=0$
in that case I'd have $A^\top A = \begin{bmatrix} A_1^\top A_1 & 0 \\ 0 & A_2^\top A_2\end{bmatrix} = A_1^\top A_1 \oplus A_2^\top A_2$
Then the rank of the direct sum is the sum of the ranks, so $\text{rank}(A^\top A)=\text{rank}(A_1^\top A_1)+\text{rank}(A_2^\top A_2)=n+m$. But $\text{rank}(A^\top A)=\text{rank}(A)$, so $A$ is indeed full rank.
for example rock beats scissors, so if a is the prob you play rock and b the prob the enemy plays scissors, you will get a term contributing + ab in your profit, you have a - ab term
Right, but since they can't play rock we have an advantage. However, essentially to minimize their loss the $b = \frac{1}{3}$ is saying that they should play paper one-third of the time and play scissors two-thirds of the time
From this though, I am not sure what we should play or what the value of $a$ should be
Actually, it's a little unclear as I look at it. The problem only specifies that -you- gain/lose a dollar if you win/lose, not that said dollar goes to your opponent
$\Bbb Q(\zeta_{p^2})/\Bbb Q$ has a subextension of degree $p$ over $\Bbb Q$ by Galois theory (since its Galois group is $\Bbb Z/p \times \Bbb Z/p-1$): why is it unique?
Yeah, that's another. I'm fond of the one I gave because the link to the name is clearer (the function $Q(v)=v^\top M v$ is nonnegative for all possible $v$, so it's definitely not negative)
@Tuki yeah.
but, again, this all depends on what exactly you mean by $A^\top A\geq 0$
fixed
For the importance, consider the matrix $A=\begin{pmatrix} 1 & -2 \\ 0 & 1\end{pmatrix}$
Problem
Let $V$ be finite-dimensional inner product space and $U$ it's subspace. Let's
assume that $\vec{v} \in V$. Show that $\vec{v} \in U \iff \text{proj}_U(\vec{v}) = \vec{v}$
Attempt to show
Let's assume subspace $U$ has orthogonal basis ($\vec{u}_1,\vec{u}_2,\vec{u}_3,\dots \vec{u}_n$) ...
@tuki btw, the proof that $A^\top A$ is positive semi-definite is trivial for the definition I gave: For any vector $v$, one has $v^\top (A^\top A)v=(Av)^\top Av = \|Av\|^2$ which is certainly nonnegative
so $A^\top A\geq 0$ is immediate from that definition of "matrix $\geq 0$"
@Emolga Let $\chi_r$ be the characteristc function on the ball of radius $≤r$, then you have $\langle \chi_r, f\rangle =r$. Check that $\chi_r\to\chi_1$ as $r\to 1$ in $L^2$ norm, then $r=\langle \chi_r, f\rangle \to \langle \chi_1, f\rangle$ and the right hand side is $1$
to link this to your problem, if $\{e_k\}$ is an orthonormal basis then the trace can be written as $\text{tr}(A^\top A)=\sum_{k} e_k^\top A^\top A e_k$
and since $v^\top A^\top A v\geq 0$ for all $v$, that trace is a sum of nonnegative quantities
@Emolga i think the statement $f, g\in L^2 \implies \overline f\cdot g\in L^1$ is dominated convergence (+ Cauchy-Schwarz for $\Bbb C$), which is necessary for the inner product to be defined so that is whats going on the background
I think the clever thing to do is form the partitioned matrices $A=[\alpha|\alpha_\bot]$ and $B=[\beta|\beta_\top]$, in which case $A,B$ are full-rank square matrices.
Can't quite see to the end of it tho, which is annoying
@s.harp think I've got a proof if I assume the matrices $(\alpha,\beta_\bot)$ and $(\beta,\alpha_\bot)$ are nonsingular
But I can't tell if that's stronger than the premises provided.
Suppose $(\alpha,\beta_\bot)$ is singular. Then $(\alpha,\beta_\bot)(\alpha,\beta_\bot)'=\alpha\alpha'+\beta_\bot\beta_\bot'$ would have to be singular. But $\alpha$ and $\beta_\top$ are both assumed to be full-rank.
Problem
Let $V$ be finite-dimensional inner product space and $U$ it's subspace. Let's
assume that $\vec{v} \in V$. Show that $\vec{v} \in U \iff \text{proj}_U(\vec{v}) = \vec{v}$
Attempt to show
Let's assume subspace $U$ has orthogonal basis ($\vec{u}_1,\vec{u}_2,\vec{u}_3,\dots \vec{u}_n$) ...
if your subspace is a line (that is, the 1-dimensional subspace of vectors parallel to some v) then the orthogonal complement is the plane which is perpendicular to that line (that is, the 2-dimensional subspace of vectors perpendicular to that v)
$V\oplus W$ has a more special meaning, but if $V$ and $W$ are both subspaces of the same vector space $X$ then $V\oplus W$ only makes sense if $V\cap W=\{0\}$, in such a case $V\oplus W = V+W$
well $u = c_1 u_1 + c_2 u_2 + c_3u_3 \dots c_n u_n$ where $n \in \mathbb{N}$ and $c \in \mathbb{R}$ and where $(u_1,u_2,u_3,\dots,u_n)$ is orthogonal basis for U
Do you see the goldfish? It's a top view of a goldfish, because that's one direction in which a fish would look symmetric.
Anyway, you construct it by adding two edges to certain even numbers, namely those that aren't double a prime $2p$. If they are of that form, then connect an edge from t...
Problem
Let $V$ be finite-dimensional inner product space and $U$ it's subspace. Let's
assume that $\vec{v} \in V$. Show that $\vec{v} \in U \iff \text{proj}_U(\vec{v}) = \vec{v}$
Attempt to show
Let's assume subspace $U$ has orthogonal basis ($\vec{u}_1,\vec{u}_2,\vec{u}_3,\dots \vec{u}_n$) ...
