# Tag Info

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The standard perturbation inequality when perturbing only the matrix of the linear system is (in your notation) $$\frac{\|x-x_2\|}{\|x\|}\leq \frac{\epsilon}{1-\epsilon}, \qquad \epsilon=\mathrm{cond}(U)\frac{\|U-U_2\|}{\|U\|},$$ assuming that $\epsilon<1$ (I guess there's a mistake or a typo in your definition of $y$, it should be ...

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If the space $X$ is banach it is an easy consequences of the open mappig theorem. Anyway with the norm induced topology over $X$ you in fact are resizing a ball so it is a ball again and it is open by definition of the topology. So the map sends open ball in open ball therefore it is open. This reasoning heavily rely on the "absolute omogeneity" of the ...

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We have $|F(x)|\leqslant \sum_{j=1}^n|\lambda_j|\lVert x\rVert_\infty$, hence $\lVert F\rVert\leqslant \sum_{j=1}^n|\lambda_j|$. There is actually equality. To see that, we can construct a continuous function $x$ such that $x(t) \in [-1,1]$ for each $t\in [a,b]$ and if $i\in\{1,\dots,n\}$, then $x(t_i)=\begin{cases} 1&\mbox{ if }\lambda_i\gt 0;\\ ... 0 Your jacobian matrix is correct, but you did something wrong. You're taking$f=(f_1, \ldots ,f_n)$, where$f_k\colon \mathbb R^{\color{red}n}\to \mathbb R, x\mapsto \dfrac{x_k}{|x|}$, for each$k\in \{1, \ldots ,n\}$. So the entry$(i,j)$at each point$x\in \mathbb R^{\color{red} n}$is given by$\dfrac{\partial f_i}{\partial x_j}(x)$. Your computations ... 1 As I said in my comment, in a convex optimization setting, one would normally not use the derivative/subgradient of the nuclear norm function. It is, after all, nondifferentiable, and as such cannot be used in standard descent approaches (though I suspect some people have probably applied semismooth methods to it). Here are two alternate approaches for ... 1 Your work for (1) is correct. To complete it, just note that using AM-GM inequality you can conclude that the diagonal entries of$\Sigma +\Sigma ^{-1}$are all greater than or equal to two. Actually, (2) follows from the same observation as well. 2 Start with the SVD decomposition of$x$: $$x=U\Sigma V^T$$ Then $$\|x\|_*=tr(\sqrt{x^Tx})=tr(\sqrt{(U\Sigma V^T)^T(U\Sigma V^T)})$$ $$\Rightarrow \|x\|_*=tr(\sqrt{V\Sigma U^T U\Sigma V^T})=tr(\sqrt{V\Sigma^2V^T})$$ By circularity of trace: $$\Rightarrow \|x\|_*=tr(\sqrt{V^TV\Sigma^2})=tr(\sqrt{V^TV\Sigma^2})=tr(\sqrt{\Sigma^2})=tr(|\Sigma|)$$ where ... 3 The map$A \mapsto |\det A|^{1/n}$satisfies all the desired property as a counter-example except for the nondegeneracy. So we perturb this map in the following way: Let$\| \cdot \|$be any submultiplicative matrix norm on$M_{n\times n}$, and define $$\| A \|' = \| A \| + t |\det A|^{1/n},$$ where$t \geq 1$is a constant to be chosen later. Then$\| ...

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Without some assumptions about $A$, the trivial bound $$|Av|\,|A^{-1}v|\ge \|A\|^{-1}\|A^{-1}\|^{-1} \tag{1}$$ cannot be improved. Indeed, let $A$ be the transformation $(x,y)\mapsto (My,x/M)$. Then for the vector $v=(1,0)$ we have $Av=(0,1/M)=A^{-1}v$, since $A$ is its own inverse. Hence, equality holds in (1). Note that the singular vectors of $A$ and ...

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(1) Typically, to prove $\|A\|_1 \le C$, you prove $$\|Ax\|_1 \le C\|x\|_1$$ (and then take the sup over $x$ such that $\|x\|_1=1$, or over $x$ such that $\|x\|\ne 0$, whichever you find more convenient). So we want to show $$\|Ax\|_1 \le n\|A\|_\infty \|x\|_1 \tag{a}$$ A typical method of proving an inequality is by a chain of inequalities, like $$... 1 The Hilbert-Schmidt norm induces the standard topology on \mathbb R^N \cong \mathbb R^{n\times n} for N=n^2. So all you need to check is that polynomials are continuous on \mathbb R^N: The projections \iota_k\colon\mathbb R^N\to\mathbb R, x\mapsto x_k are continuous. Products f\cdot g of continuous functions f,g\colon\mathbb R^N\to\mathbb R are ... 0 1.) is not quite correct. For fixed x, we have x>1/n for sufficiently large n; and then, f_n(x)=1. So the pointwise limit, f, is the function which is identically 1. (I think this is what you really meant to say.) 2.) is not correct. Note you need to compute \int_0^1 | f_n(x)-1|^2\,dx. You'll have convergence in the 2-norm if and only ... 2 Yes you're right. In fact if A=(a_{ij}) then A^t=(b_{ij}) where b_{ij}=a_{ji} so if C=(c_{ij})=A^tA then$$c_{ij}=\sum_{k=1}^n b_{ik}a_{kj}=\sum_{k=1}^na_{ki}a_{kj}$$hence$$\operatorname{tr}(A^tA)=\sum_{i=1}^n c_{ii}=\sum_{i=1}^n\sum_{k=1}^na_{ki}^2$$Now to prove the given expression defines a norm you can for example prove that the map ... 1 This is indeed the Frobenius norm. So, if you've established that the Frobenius norm is indeed a matrix norm, it is sufficient to show that this is that, which is a norm. In order to show that they are the same, note the following: The ith diagonal entry of A^TA is given by$$ b_{ii}= \sum_{j=1}^n a_{ji}^2 $$So that$$ \operatorname{trace}(A^TA) = ...

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The Banach function spaces may be a better choice for your question. You can see here for the definition of a Banach function space or this book Function Spaces, Volume 1 (chapter 6). Answer to your question: The first thing is to introduce the definition of the Riesz-Fischer property. We say that a normed linear space $(X,\left\|\cdot\right\|)$ has ...

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I'll trade your question for an easier one, perhaps my thoughts on this two-dimensional problem may give you some ideas. Consider $\mathbb{R}^2$ with time-space coordinates $(t,x)$ and the metric $\eta = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]$. Note the interval $I((t_1,x_1),(t_2,x_2)) = (t_2-t_1)^2-(x_2-x_1)^2$ can be compactly ...

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Let $S$ be the set of all tuples of scalars $(\alpha_1,\dots,\alpha_n)\in K^n$ such that $|\alpha_1|+\cdots+|\alpha_n|=1$. Then the map $f:S\rightarrow [0,\infty)$ given as $$f(\alpha_1,\dots,\alpha_n)=\|\alpha_1 u_1+\cdots+\alpha_n u_n\|$$ is continuous. Because $S$ is compact, the continuous function $f$ attains its infimum on $S$ at some $x_0\in S$, ...

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Your example works. Assume the norms are equivalent. Then there exists a constant $C> 0$ s.t. $$\|P\|_1 \ge C \|P\|_2$$ for all $P$. Let $n$ be large enough such that $1/(n+1)< C$. Set $P (t)=t^n$. Then $$\frac{1}{n+1} = \|P\|_1 \ge C \|P\|_2 = C$$ Contradiction.

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For each $k$, $$\left\lVert\sum_{n=1}^{k}f_n\right\rVert \leqslant \sum_{n=1}^{k}\lVert f_n \rVert$$ Then take $k\to\infty$.

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The function $$f(x) = \begin{cases} 1, &\text{if x is rational}, \\ 0, &\text{if x is irrational} \end{cases}$$ is measurable and bounded, hence integrable on any bounded interval. But $\sup_{x\in[0,1]} |f(x)-P(x)| \ge \frac12$ for any polynomial $P$ (indeed, any continuous function $P$).

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$\left \| f-g \right \|_{p}^{p}=$ $\int_{\{f>g\}}(f-g)^{p}dx+\int_{\{g>f\}}(g-f)^{p}dx$. Let's look at $\int_{\{f>g\}}(f-g)^{p}dx$. $=p\int_{\{f>g\}}\int_{g(x)}^{f(x)}(t-g(x))^{p-1}dtdx$ $\stackrel{h=\int \frac{dh}{ds}ds}{=}(-1)p(p-1)\int_{\{f>g\}}\int_{g(x)}^{f(x)}\int_{t}^{g(x)}(t-s)^{p-2}dsdtdx=$. ...

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I am assuming that $||P_1||$ should be $||P||_1$ and similarly for the other norm. Convergence in $||~\cdot~||_2$ is uniform convergence and that is a very strong convergence whereas convergence in $||~\cdot~||_1$ is convergence in $L^1$ and that is a relatively weak convergence. The sequence of standard tent functions all of height 1 do not converge in ...

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It should be $$\|\lambda L \| = \sup_{a \ne 0} \dfrac{\|\lambda L a\|}{\|a\|} = \ldots$$

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Choose $g={f \over \|f\|}$. Then $\|g\| =1$ and $\langle g,f \rangle = \|f\|$.

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Hint: The Cauchy Schwarz inequality states that $$|\langle f,g \rangle|\leq\|f\|\cdot \|g\|$$ For the other direction, it suffices to show that there exists a vector $g$ with $\|g\| = 1$ such that $|\langle f,g \rangle| = \|f\|$. In particular, we may take $$g = \frac{f}{\|f\|}$$ Note that for this $g$, we have $$\left\langle g,f \right\rangle = ... 3 Yes. You need to show that x^t A y defines an inner product. Then the result follows. The simplest way to show that this is true is in two steps Since A is positive definite, we can write A=B^T B. Next use the change of basis \hat x = B x and transfer the standard inner product from the new basis to original as shown next$$ \hat x = B x, \hat y = ...

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Sketch of Proof: \begin{align} \|A\|_W&=\sup_{\|x\|_W\leq1}\|Ax\|_W\\ &=\sup_{\|Wx\|\leq1} \|WAx\|\\ &=\sup_{\|y\|\leq1} \|WAW^{-1}y\|\\ \\&=\|WAW^{-1}\| \end{align}

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By definition $$\|A\|_W=\sup_{x\ne 0}\frac{\|WAx\|}{\|Wx\|}=\sup_{x\ne 0}\frac{\|WAW^{-1}(Wx)\|}{\|Wx\|}.$$ But as $W$ is non-singular $$\big\{Wx:x\in\mathbb R^n\smallsetminus\{0\}\big\}=\big\{y:y\in\mathbb R^n\smallsetminus\{0\}\big\},$$ and hence $$\sup_{x\ne 0}\frac{\|WAW^{-1}(Wx)\|}{\|Wx\|}=\sup_{y\ne 0}\frac{\|WAW^{-1}y\|}{\|y\|}=\|WAW^{-1}\|.$$

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Complementing the very well detailed answer of Zev Chonoles, I think you've got it in the opposite direction. $T$ as is defined in the title is the mapping $x\mapsto T(x)=x^T$, such that $T(x)(y)=x^Ty$ is a linear transformation of the vector $y$. This means that the domain of $T$ is $(\mathbb R^n, \Vert \cdot \Vert_\infty)$ and its codomain is the dual ...

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A pair $(V,\|\cdot\|)$ denotes a vector space $V$ over $\mathbb{R}$, together with a norm function $\|\cdot\|:V\to\mathbb{R}$. Thus, $(\mathbb{R}^n,\|\cdot\|_1)$ and $(\mathbb{R}^n,\|\cdot\|_\infty)$ mean "the vector space $\mathbb{R}^n$ equipped with the $L^1$ norm", and "the vector space $\mathbb{R}^n$ equipped with the $L^\infty$ norm", respectively. ...

1

By "Euclidean norm", I take it to mean a norm that arises from an inner product. It is known that a norm on a vector space arises from an inner product if and only if it satisfies the Parallelogram Law: $$2\|x\|^2 + 2\|y\|^2 = \|x - y\|^2 + \|x + y\|^2$$ You should be able to find matrices $A,B$ such that the operator norm does not obey this equation.

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It can be done in a trivial sort of way, let $\rho \in (m,M)$, then let $\|x\|_* = \rho \|x\|$. Then $m\|x\| \le \rho \|x\| = \|x\|_* \le M \|x\|$.

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First, note that $||A||_{op} \leq {\lambda_1}^\frac{1}{2}$ where ${\lambda}_1$ is $A^TA$'s max eigenvalue. The proof is as follows: $||Av||^2 = <Av, Av> = <A^TAv,v>$ where $<,>$ denotes a dot product. Note that $A^TA$ is a non-negative matrix, so we can apply the Spectral theorem. $<A^TAv,v>$ $=$ ...

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The finite difference method computes a point-wise approximation of $u_{\mathrm{true}}$. You have to do a sort of post-processing of the FDM approximation $u_h$ for which you can compute/approximate its derivative. E.g., in 1D, it is reasonable to reconstruct a $\tilde{u}_h$ which is linear on each interval such that $\tilde{u}_h(x_i)=u_h(x_i)$ in the point ...

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The matrix $A^TA$ is symmetric and thus it is orthogonal diagonalizable: $$A^TA=U^TDU,$$ where $U$ is an orthogonal matrix, i.e., $U^TU=I$, $\|Ux\|=\|x\|$, for all $x\in\mathbb R^n$, and $D$ diagonal: $$D=\mathrm{diag}(\lambda_1,\ldots,\lambda_n),$$ where the $\lambda_j$'s are non-negative, as $A^TA$ is clearly positive definite. We can assume that ...

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An equivalent definition for the Hilbert-Schmidt norm is $||A||_{HS} = (\sum_j \sigma_j(A)^2)^{1/2}$ where $\sigma_k(A)$ is the $k$-th singular value of the matrix $A$. These are just the eigenvalues of the matrix $\sqrt{A^*A}$. Then you can show that the largest singular value of $A$ equals the operator norm of the matrix $A$. Now clearly, $||A|| = ... 3 In your proof, you should start with a Cauchy sequence$\{(x_n,y_n)\}\subset X\times Y$and show that this sequence is convergent in$X\times Y$. First observe that if$\{(x_n,y_n)\}$is a Cauchy sequence in$X\times Y$, then both$\{x_n\}$and$\{y_n\}$are Cauchy sequences in$X$and$Y$respectively, since $$... 2 What kinds of operations are may we assume are permissible? I mean, to be perfectly honest, this seems evident, since$$\|\Sigma^{1/2} x\|_2=\sqrt{\|\Sigma^{1/2}x\|_2^2}=\sqrt{\langle\Sigma^{1/2}x,\Sigma^{1/2}x \rangle} =\sqrt{x^T\Sigma^{1/2}\Sigma^{1/2}x}=\sqrt{x^T\Sigma x}.$$But of course, this assumes you accept the definition of a symmetric matrix square ... 3 A condition is "T is dense in [0,1] for the usual topology". Indeed, in this case, x(t)=0 for each t\in T implies by continuity the same for t in the closure of T. If T is not dense in [0,1], then [0,1]\setminus \overline T contains a non-empty open set, hence a small interval. We then construct a continuous function which doesn't vanish ... 0 Clearly$$ \int_a^b f(x)\,dx=\lim_{n\to\infty}\frac{b-a}{n}\sum_{i=1}^n f(t_{n,i}), $$where t_{n,i}=a+i(b-a)/n, and$$ \Big\|\sum_{i=1}^n f(t_{n,i})\Big\|_2\le \sum_{i=1}^n \|f(t_{n,i})\|_2, $$due to the triangular inequality. But$$ \lim_{n\to\infty}\frac{b-a}{n}\sum_{i=1}^n \|f(t_{n,i})\|_2=\int_a^b\|f(x)\|_2\,dx, $$etc... 0 This is one solution I adopted for similar problems in order to avoid discretization and curve fit$$k^2=\frac{\int_a^b x^2 g(x) \, dx}{\int_a^b f(x) \, dx}$$I even wonder if this does not correspond exactly to the least square fit solution for an infinite number of data points for a<x<b. 0 This is a consequence of the Uniform Boundedness Principle, which hold for a Banach space, and not a normed one. However, you can think of S as a subset of X^{**} which is indeed a Banach space. 1 for item c) you must prove the following result: Theorem: Let E be a vector space of finite dimension. Then all norms on E are equivalent. Thus, the example of Emin solves the item c) in case R^n. For item d), you should seek norms in spaces of infinite dimension. Regards 2 This is absolutely right. Well done! You're not ignoring the relationship between z and t; it is naturally captured by the model. The fact that they are both present in the objective ensures that an increase in one must be accompanied by a decrease in the other. It is possible to prove that at the optimal point, it must be the case that z_i=|x_i|, ... 0 Let (V,\mid\mid\cdot\mid\mid) be a normed vector space over \mathbb{R}; then the closed unit ball \overline{B}(0 , 1) is convex. Proof: Clearly \overline{B}(0 , 1) is non empty, so let x,y\in\overline{B}(0 , 1) and let \lambda\in[0,1]. Consider z=\lambda x+ (1-\lambda)y Now;$$\mid\mid z\mid \mid= \|\lambda x+ (1-\lambda)y \| \leq ... 0 We simply use the triangular inequality repeatedly: $$\|u\|=\|(u-v)+v\|\le \|u-v\|+\|v\|,$$ and thus $$\|u\|-\|v\|\le \|u-v\|\le \|u\|+\|-v\|=\|u\|+\|v\|. \tag{1}$$ Similarly $$\|v\|=\|(v-u)+u\|\le \|v-u\|+\|u\|,$$ and thus $$\|v\|-\|u\|\le \|u-v\|. \tag{2}$$ Now$(1)$and$(2)$imply that $$\big|\|u\|-\|v\|\big|\le \|u-v\|\le \|u\|+\|v\|.$$ 5 The first $$||u-v||=||u+(-v)||\le ||u||+||-v||=||u||+||v||$$ and the second $$||u||=||u-v+v||\le ||u-v||+||v||\Rightarrow ||u||-||v||\le ||u-v||$$ and by symmetry we have the other inequality so we conclude. 0 Hint:$\left\|u\right\| = \left\|(u-v)+v\right\|\left\|v\right\| = \left\|(u-v)+u\right\|\left\|u-v\right\| = \left\|u+(-v)\right\|$0 If$1 \leq p < q \leq \infty$then we have$l^p \subsetneq l^q$. Proof: For the inclusion$l^p \subset l^q$: The case$q= \infty$is clear. Now let$q< \infty$and let$x=(x_n)_{n \in \mathbb{N}} \in l^p$. We want to show that $$||x||_{l^q} \leq ||x||_{l^p}.$$ Note that this inequality is independent under multiplying$x$with a scalar$\lambda ...

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You proof is correct but yes, it works only for operator norms. Note that the proof mentioned in nayrb's answer requires sub-multiplicativity ($\|AB\|\leq\|A\|\,\|B\|$), which does not hold for all norms. It does however hold for the operator norms.

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