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New answers tagged duality-theorems

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I would use the theorem regarding Dimension of Sum and Intersection of Vector Spaces: for $M,N$ subspaces of $V$, you have $$\dim (M+N) +\dim (M \cap N ) = \dim (M) + \dim(N).$$ And apply it with $M= \ker \psi, N =\ker N$. You have $\dim(M+N)= n$ due to the hypothesis on $\psi,\phi$.

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Note that $\ker \phi \cup \ker \psi\subseteq V$ thus $\dim (\ker\phi \cup \ker \psi)\leq n$ Thus \begin{align} \dim (\ker\phi \cup \ker \psi)&=\dim (\ker\phi) +\dim (\ker \psi)-\dim(\ker \phi \cap \ker\psi )\\ & = n-1+n-1-\dim(\ker \phi \cap \ker\psi )\leq n \end{align} so $\color{red}{\dim(\ker \phi \cap \ker\psi )\geq n-2}$ And since $\phi\neq ... 0$d^2=x^2+y^2$is the square of distance from$(0,0)$to$(x,y)$. $$\\$$ 1. When$(x,y)$is within$\space (x−1)^2+(y−1)^2≤1$, then$d^2$is minimum when$(x,y)=(1-\frac{\sqrt2}{2},1-\frac{\sqrt2}{2})$. (Imagine which point on the circle is closest from the origin.)$\therefore d^2 \geq \left(1-\frac{\sqrt2}{2}\right)^2=\frac{3}{2}-\sqrt2$$$\\$$ 2. When ... 0 I'm too lazy to use KKT, so I'm providing a non-KKT solution. Maybe, someone else will help with the KKT requirement. With only the first constraint, from$(x-1)^2+(y-1)^2\leq 1$, we have by AM-GM that$x^2+y^2+1\leq 2x+2y\leq 2\sqrt{2}\sqrt{x^2+y^2}$. Thus, if$r:=\sqrt{x^2+y^2}$, then$r^2-2\sqrt{2}r+1\leq 0$, which means$\sqrt{2}-1 \leq r\leq ...

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It is true that "duality" is a quite broad concept in many fields and the field of mathematics is not the exception. For example the Wikipedia website shows many forms of duality in Mathematics. As mathematicians we want to abstract the meaning of a word into a unique concept that ecompasses all situations. For duality we need to define two spaces of ...

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By the first isomorphism theorem the map $T$ factors over $V/\ker T$ as $$V\ \stackrel{\pi}{\longrightarrow}\ V/\ker T\ \stackrel{\overline{T}}{\longrightarrow}\ \operatorname{im}T,$$ where $\pi$ is the canonical quotient map and $\overline{T}$ is an isomorphism. Note that $\pi=\overline{T}^{-1}\circ T$. If $\phi\in(\ker T)^{\circ}$ then $\ker ... 0 My teacher released the answers, hopefully this helps someone else: Primal :$Max\ c'x\ subject\ to\\ Ax = 0\\ x \ge 0$Dual:$Min\ 0'p\ subject\ to\\ A'p \ge c$Assuming$c > 0$From now on youl have to excuse me if I write A'p vs p'A. Just know that they mean the same thing. (a) If some x* satisfies$Ax\* =0$then we can say that the primal is ... 1 For any$V$, we have$|V^*|=|k|^{\dim V}\geq 2^{\dim V}>\dim V$. The point of the assumption$\dim V\geq |k|$is that it implies (via$|V|=\max(|k|,\dim V)$) that$\dim V=|V|$, so we get$|V^*|>|V|$, which then clearly implies$\dim V^*>\dim V$. The case of arbitrary commutative rings actually follows more or less immediately from the field case. ... 4 If$R$is noncommutative and$P$is a right$R$-module then$\text{Hom}_R(P, R)$naturally has the structure of a left$R$-module (and vice versa if$P$is a left$R$-module), so you can't even ask for this isomorphism because the two objects belong to different categories. But this is still false if$R$is commutative. Take$R = \mathcal{O}_K$to be the ... 1 One way to do things is by counting dimensions. In particular, let$r$be the dimension of the image of$T$, and let$n$be the dimension of$V$. By the rank-nullity theorem,$\ker T$has dimension$n - r$. Then, we note that$(\ker T)^0$must have dimension$r$(you may have to prove this yourself). On the other hand, note that$T'$has the same rank ... 1 You are right to be skeptical; your limit$a$does not work. Hint: Knowing that $$\sup_k |a_{n,k} - a_{m,k}| → 0$$ means that for each$k$,$a_{n,k}∈ℝ $is a ... Proof that the map$\phi: \ell^1 → c_0^*$given by$T(x)(a) = \sum a_i \overline{x_i}$is an isometric isomorphism. 1 Most of your question is answered by Theorem 3.2 in Simon's Trace Ideal and Their Applications. It is not an easy read because of the awkward notation, but it is worth it. Yes,$L^q(H)$is the dual of$L^p(H)$, basically because Hölder holds,$|\text{Tr}(TS)|\leq\|T\|_p\|S\|_q$. I cannot say that the dualities in the sequence spaces follow from the ... 2 To show that$j_Yy = y^{**}$, let$y^* \in Y'$and$x^*\in X'$a Hahn-Banach extension of$y^*. We have \begin{align*} (j_Yy)(y^*) &= y^*(y)\\ &= x^*(y)\\ &= j_Xx(x^*)\\ &= \tilde y(x^*)\\ &= y^{**}(x^*|_Y)\\ &= y^{**}(y^*) \end{align*} Hence,j_Yy = y^{**}$. 3 Broadly, two mathematical set-ups are dual when they can be transformed into each other by a simple exchange of symbols and terminology. For example, a general topological space may be defined in terms of its open sets, or dually in terms of its closed sets: corresponding to infinite unions and finite intersections of open sets, in the first formulation, are ... 8 A duality is a pair of related concepts that display a one-to-one translation symmetry, usually (not always) as the result of some form of involution operator. In classical logic, the operators,$\vee$and$\wedge$form a dual, and negation is their involution operator. This is expressed through deMorgan's Laws:$\$\neg(A\vee B) = \neg A\wedge \neg ...

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