# Tag Info

1

First, let's obtain some closed forms for $x$ and $y$. For $x$ it is quite trivial: $x = A^{-1} b$. For the $y$ we have $$y = (A + \Delta A)^{-1} (b + \delta b) = \left(A (I + A^{-1}\Delta A)\right)^{-1} (b + \Delta b).$$ Under the given conditions the matrix $I + A^{-1} \Delta A$ is invertible: $$\left(I + A^{-1} \Delta A\right)^{-1} = I + \sum_{k=1}^{\... 1 The condition number \kappa(A) = \| A \| \| A^{-1}\|  can take any norm. So I'll assume you're also using \|\cdot \|_{\infty}. First we have  \frac{\| A_{n} - B \|_{\infty} }{\| A_{n} \|_{\infty} } - \frac{1}{\kappa(A_{n})} = \frac{\| A_{n} - B \|_{\infty} }{\| A_{n} \|_{\infty} } - \frac{1}{\|A_{n}\|_{\infty} \| A_{n}^{-1} \|_{\infty}} Let's find \|... 0 Because y^* is optimal for the unconstrained optimization problem, we must have \nabla f(y^*)=0, i.e.$$A^\top (Ay^*-b)=0.$$You expanded the norms incorrectly: \|v-w\|_2^2 = \|v\|_2^2 - 2 v^\top w + \|w\|_2^2 (the middle term is not \|v\|_2 \|w\|_2). Note that$$\|Ay-b\|_2^2 - \|Ay^* - b\|_2^2 = \|Ay\|_2^2 - 2 b^\top Ay - \|Ay^*\|_2^2 + 2 b^\top A ...

1

Based on the comments above, it's easy to see that $$||\phi||_{op} \le \int^1_0 |a(t)|dt$$ and for equality you can consider the sign function $f = sgn(a)$, then for $a(t) \neq 0$ $$||\phi||_{op} \ge |\phi(f)|= \left|\int^1_0 sign(a) \cdot a(t)dt\right|$$ but the product, $sign(a) \cdot a(t) = |a(t)|$. Finally $$||\phi||_{op} \ge \int^1_0 |a(t)|dt$$ ...

3

It might help to separate $T$ into the composition of two simpler maps. Note that the map $s:C[0,1] \to \mathbb{R}^n$ defined by $s(f) = (f(t_1),...,f(t_n))$ has norm one (using the $l_\infty$ norm on $\mathbb{R}^n$) and is surjective. Furthermore, for any $y$ with $\|y\|_\infty =1$ it is easy to construct (by interpolation & extrapolation) some $f \in C[... 1 This looks like a force, and typically the force is not just a real number, it also usually has a direction. That's what the$x_i-x_j$is contributing. All the other quantities appearing combined to form a coefficient in front of the vector$x_i-x_j$. I can't see anything to suggest that$\|\cdot\|$means anything other than the Euclidean norm on real ... 4 WLOG, one can assume that$a_k\ne0$for all$k$(if not, just drop all the zero-values, and in the case that all are zero, the problem is trivial). Take$f(x)$a continuous function with these properties:$f(t_k)=\frac{|a_k|}{a_k}$.$|f(x)|\le 1$for all$x\in[0,1]$. For example, you can take the polygonal through the points$\{(t_k,f(t_k)),\,1\le k\le n\}\...

1

$\Vert f\Vert_K=0$ does not imply that $f=0$. First, it is possible that $K=\{z_0\}$, a single point. For example, if $K=\{0\}$ and $f(z)=z^2$, then $\sup_{z\in K}|f(z)|=0$, so $\Vert f\Vert_K=0$, but $f\neq 0$. This invalidates options A and B. If $K$ has non-empty interior, then it is still possible that a continuous function $f$ is zero on all of $K$, but ...

0

Hint. When $Q$ is an orthogonal matrix, how is the $j$-th diagonal element of $Q^TQ$ related to the entries of the $j$-th column of $Q$?

0

$\displaystyle || A ||_2 = \sqrt{\sum_{i,j=1}^{n,m} | a_{ij}|^2 } \ge 0\; \forall A \in \mathbb{R}^{n \times m}$ Because $|a_{ij}|$ and $|a_{ij}|^2$ are not negative, so their sum and root won't be negative. Let $\displaystyle ||A||_2 = \sqrt{\sum_{i,j=1}^{n,m} | a_{ij}|^2 } = 0$ $\displaystyle \Rightarrow \sum_{i,j=1}^{n,m} | a_{ij}|^2 = 0$ ...

1

The sup/max-norm trivially fulfills this: $$\|p\| := \max_{|x|\le1} |p(x)|.$$ This is a norm on these polynomials. Because for polynomials $\|p\|=0$ implies $p=0$. Then clearly for $x\in \mathbb C^2$ with $|x|\le 1$ $$|p(x)q(x)|\le |p(x)| \cdot |q(x)| \le \|p\|\cdot \|q\|.$$

2

If you take $\phi(f)=f(1)$, this is unbounded. Define $$f_n(t)=\begin{cases} 0,&\ 0\leq t\leq 1-\tfrac1{n^2}\\[0.3cm] nt+\tfrac1n-n,&\ t>1-\tfrac1{n^2} \end{cases}$$ Then $\phi(f_n)=n$, while $$\|f_n\|_2^2=\int_{1-\tfrac1n}^1(nt+\tfrac1n-n)\,dt=\tfrac1{3n^4}.$$

2

You indeed can do it using trigonometric functions. Following the suggestion of @Berci, define $$f_k(x) = -1+2|\sin(n\pi x)|^{1/k}$$ so that $u_n(f_k) = -1$ for all $k \in \Bbb{N}$. On the other hand, $f_k \to 1$ almost everywhere so by the Lebesgue dominated convergence theorem we have $$\lim_{k\to\infty} u(f_k) = \lim_{k\to\infty}\int_0^1 f_k(x)\,dx = \... 4 Assume \|A^{-1}\| < \frac1{\|A-B\|}. Then using the submultiplicativity of the norm we have$$\|I-A^{-1}B\| = \|A^{-1}(A-B)\| \le \|A^{-1}\|\|A-B\| < 1$$so A^{-1}B is invertible with inverse \sum_{n=0}^\infty (I-A^{-1}B)^n which converges since it converges absolutely. Hence B is invertible as well, which is a contradiction. 4 It is not true for any square matrix norm. For example, if \|\cdot\|_1 is a norm, so is \|\cdot\|_a=a\|\cdot\|_1, for every a>0, and observe that for sufficiently small a>0, we have \|A^{-1}\|_a\|A-B\|_a=a^2\|A^{-1}\|_1\|A-B\|_1<1. However, it is true for every induced norm, i.e., norm of the form$$ \|E\|=\sup_{\|x\|=1}\|Ex\|, $$where \|... 0 If u,v\in H^2(\Omega)\cap H_0^1(\Omega), then$$ \int_{\Omega}u\Delta v\,dx=-\int_{\Omega}\nabla u\cdot\nabla v\,dx. $$Hence$$ \int_{\Omega}u\Delta u\,dx=-\int_{\Omega}|\nabla u|^2\,dx $$and thus$$ \|u\|_{L^2}\|u\|_{H^2} \ge \int_{\Omega}u(u-\Delta u)\,dx=\int_{\Omega}(u^2+|\nabla u|^2)\,dx=\|u\|_{H^1}^2 $$which implies tha 0 You say that the case N=1 is "trivially convex". I'm sorry but I don't agree at all. I have been computing by hand at first, working at the elimination of the different absolute values, octant by octant, obtaining, as you say, one linear relationships, but not relationships with inequalities. Therefore the 3D set is a surface (S), not a volume, ... 1 First of all, for all x\in\ell^{\infty} we have: |f(x)|=|\sum_{n=1}^\infty x_n2^{-n}|\leq\sum_{n=1}^\infty |x_n|2^{-n}\leq\sum_{n=1}^\infty ||x||_{\infty}2^{-n}=||x||_{\infty}\sum_{n=1}^\infty 2^{-n}=||x||_{\infty} And so ||f||\leq 1. Also, let y=(1,1,1,1,...). Then ||y||_{\infty}=1, and so: ||f||\geq\frac{|f(y)|}{||y||_{\infty}}=|f(y)|=\sum_{n=1}^... 0 Although an answer is already provided and what I have in mind seems the same, I am still going to write one in order to break the problem down a bit more. The norms of operators over non-trivial (ie. not \{\vec{0}\}) normed linear spaces T:(V,\|\cdot\|_v)\rightarrow(W,\|\cdot\|_w) is defined as$$ \|T\|:=\sup\{\|T(v)\|_w:\|v\|_v=1\} $$And that applies ... 1 The error is where you have w^Tw = ||w||. This should be w^Tw = ||w||^2 because ||w|| = \sqrt{w^Tw}. Using the correction we have$$w^T\bigg(x^{(i)} - \gamma^{(i)}\frac{w}{||w||}\bigg) + b = 0w^Tx^{(i)} - \gamma^{(i)}\frac{||w||^2}{||w||} + b = 0w^Tx^{(i)} - \gamma^{(i)}||w|| + b = 0$$This leads to$$\gamma^{(i)} = \frac{w^Tx^{(i)} + b}{||w||...

1

Assume that $A$ is orthogonal, i.e. that $A^T A = I$ (observe that this is equivalent to $A^{-1} = A^T$). We consider $$|| A x ||^2 = (Ax)^T (Ax) = x^T A^T A x = x^T x = || x || ^2,$$ which shows that $|| A x || = || x ||$. Comment: On advise from another user, I posted a more thorough version of my earlier comment as this answer.

1

If $u=(u_1,u_2,\ldots,u_n)$, then\begin{align}u^\top u&=\begin{pmatrix}u_1\\u_2\\\vdots\\u_n\end{pmatrix}(u_1,u_2,\ldots,u_n)\\&=u_1^{\,2}+u_2^{\,2}+\cdots+u_n^{\,2}\\&=\|u\|^2.\end{align}

0

You want to use the actual definition of the operator norm, which is $$\|[AB]\| = \sup\{\|[AB]v\| : v\in \mathbb{R}^4 \text{ and } \|v\|=1\}$$ $$\|A\| = \sup\{\|Av\| : v\in\mathbb{R}^2\text{ and } \|v\|=1\}.$$ Given any vector $(x,y)\in \mathbb{R}^2$, you have the vector $(x,y,0,0)\in \mathbb{R}^4$ with $[AB](x,y,0,0)$ = $A(x,y)$. From this, the set that $\... 1 Notice that your condition implies that$(x_n)_n$is a Cauchy sequence. Namely, for any$\varepsilon > 0$by definition of$\lim_{j\to\infty}$there exists$j_0 \in \Bbb{N}$such that $$j \ge j_0 \implies \lim_{k\to\infty} \|x_j-x_k\| \le \frac\varepsilon2.$$ Now for any$j \ge j_0$by definition of$\lim_{k\to\infty}$there exists$k_0 \in \Bbb{N}$such ... 3 I presume that by$\|\cdot\|_\infty$for matrices you mean the matrix norm induced by the$\|\cdot \|_\infty$norm on vectors. Then you are right that $$\|A^{-1}\|_\infty = \frac{1}{\min \{ \|Ax\|_\infty \; : \; \|x\|_\infty = 1\}}$$ But there is no reason to think that the minimum is obtained for an integer point, and it is not true that$\|A^{-1}\|_\...

3

It is not true. Note that inf-norm on matrices is the max absolute sum of per row. Thus $$A=\begin{bmatrix}1 & -1\\-2 & 1\end{bmatrix}, A^{-1}=-\begin{bmatrix}1 & 1\\2 & 1\end{bmatrix}$$ is a counter example.

0

There is no completeness involved in this result so there is no point in looking at Cauchy sequences. A linear map $T$ from one normed linear space to another is continuous if and only if it is continuous at $0$ iff $\|Tx\| \leq C\|x||$ for some $C \in (0,\infty)$ (and all $x$). By hypothesis, the identity map from $(X,\|.\|_1)$ to $(X,\|.\|_2)$ is ...

2

\begin{align}A^n-B^n&=A^n-A^{n-1}B+A^{n-1}B-A^{n-2}B^2+\cdots-B^n\\ &=A^{n-1}(A-B)+A^{n-2}(A-B)B+\cdots+(A-B)B^{n-1} \end{align} $$\therefore\ \|A^n-B^n\|\le(\|A^{n-1}\|+\|A^{n-2}\|\|B\|+\cdots+\|B^{n-1}\|)\|A-B\|\le\frac{\|A\|^n-\|B\|^n}{\|A\|-\|B\|}\|A-B\|$$ If the intent of the question is whether $\|A^n-B^n\|$ is small compared to $\|A-B\|$ then ...

1

Answer. $B$: Only if $k\ge n$. First, we study the case $k=n$. If $p(x)=c_0+c_1x+\cdots+c_k x^k$, and $$p(a_j)=b_j, \quad j=0,1,\ldots,k,$$ then the above is a $(k+1)\times (k+1)$ linear system $$c_0+c_1 a_j+\cdots+c_n a_j^k=b_j, \quad j=0,1,\ldots,k,$$ with unknowns the $c_j$'s. The determinant of the matrix $M$ the system is the Vandermonde one, ...

1

Let $n=2$. If $k=1$, let $\|.\|$ be the corresponding defined norm. If $a_0=1, a_1=-1$, then we have $\|x^2-1\|=0$, even thought $x^2-1 \ne 0$. In fact, if $n=2$ and $k=1$, then we have $\|(x-a_0)(x-a_1)\|=0$. In general, if $k+1 \le n$, then we have $\|\prod_{i=0}^k (x-a_i) \|=0$ and $\prod_{i=0}^k (x-a_i)$ is of degree $k+1$ and non-zero. Hence we need $k \... 0 We can say$\lvert \int_0^t f(s)\,ds \rvert^2\leq \|1\|_{L^2([0,t])}^2\,\|f\|_{L^2([0,t])}^2=t \|f\|_{L^2([0,t])}^2 \leq t \|f\|_{L^2([0,1])}^2$. Hence,$\|Tf\|^2=\int_0^1 \lvert \int_0^t f(s)\,ds \rvert^2\,dt \leq \|f\|_{L^2([0,1])}^2 \int_0^1 t\,dt =\frac 12 \|f\|_{L^2([0,1])}^2$. Taking square roots shows $$\|Tf\|\leq \frac{\sqrt{2}}{2}\|f|.$$ What am I ... 1 Hint: Let$g(t)=1$for$t \geq 0$and$g(t)=0$for$t <1$. Then$Tf=f*g$. Show that$T^{n}f=f*g^{(n)}$where$g^{(n)}$stands for$g*g*...*g$($n-$fold convolution). Show by direct computation that$g^{(n)}(t)=\frac {t^{n}} {n!}$for$t >0$and$0$for$t <1$. This gives$\|T^{n}\| \leq \frac 1 {n!}$. So$\|T^{n}\|<\|T\|^{n}$for$n >1$. 2 In general$\|B^{-1}\|\ge\frac{1}{\|B\|}$since$1=\|BB^{-1}\|\le\|B\|\|B^{-1}\|$. Also,$\|I-A\|\le1+\|A\|. Hence combining the two, $$\|(I-A)^{-1}\|\ge\frac{1}{\|I-A\|}\ge\frac{1}{1+\|A\|}$$ 1 \begin{align}||(1,0)|| & = \sqrt{\langle (1,0),(1,0)\rangle}\\ & = \sqrt{2 \cdot 1 \cdot 1 + 0 \cdot 0}\\ & = \sqrt{2}\end{align} \begin{align}||(0,1)|| & = \sqrt{\langle(0,1),(0,1) \rangle}\\ & = \sqrt{2 \cdot 0 \cdot 0 + 1 \cdot 1} \\ & = \sqrt{1} \\ & = 1\end{align} 1 The norm ofx=(1,0)$is$\sqrt{\langle x,x\rangle}=\sqrt{2x_1x_1+x_2x_2}=\sqrt{2\cdot1\cdot1+0\cdot0}$. Now can you compute the norm of$(0,1)$? 1 Show that$f(x,y)=\|x+y\|^2$and$g(x,y)=\|x-y\|^2$are continuous from$(\Bbb R^n)^2$to$\Bbb R.$The inner product$<x,y>=\frac {f(x,y)-g(x,y)}{4}.$3 Suppose that we have finitely many matrix$A_1,\dots,A_n$to consider. Let$\tilde A$denote the matrix $$\tilde A = \pmatrix{A_1\\ \vdots\\ A_n}.$$ We see that$\sum_{i=1}^n A_i^\dagger A_i = \tilde A ^\dagger \tilde A, so that $$\left\| \sum_{i=1}^n A_iA_i^\dagger\right\|_{\infty} = \left\|\tilde A^\dagger \tilde A \right\|_\infty = \left\| \tilde A\... 1 The Schatten. p-norm of a matrix A\in M_n(\mathbb C) is the p-norm of the vector containing the singular values of A. That is, \|A\|_p=\|\left(\sigma_1(A),\ldots,\sigma_n(A)\right)\|_p. When A is a square matrix whose size is at least 2\times2, \|A\|_p is not an induced norm unless p=\infty, because \|I_n\|_p=n^{1/p}>1 when p is ... 0 \def\p{\partial} For ease of typing, define the vector$$w=Zx-z$$Write the function in terms of this new vector, then calculate its gradient.$$\eqalign{ \lambda &= \|w\|^2 = w^Tw \\ d\lambda &= 2w^Tdw = 2w^TZ\,dx = (2Z^Tw)^Tdx \\ \frac{\p\lambda}{\p x} &= 2Z^Tw = 2(Z^TZx - Z^Tz) \\ }$$Exactly as claimed by the solution set. Yay! 1 Normalization ensures \int f(x)\ dx = 1, not that the integral of the square of the functions equals 1. 0 The Frobenius norm is unitarily invariant, i.e. \|U A V\|_F = \|A\|_F, for any U and V unitary. So we may assume that A = \text{diag}(d_1, d_2, \cdots, d_n) is diagonal. (The left hand side can be rewritten as \|U D V^\dagger (V y)\|_F, where UDV^\dagger is the singular value decomposition of A and y = V^\dagger x). Letting y = (y_1, y_2, \... 1 The reason why Bolzano-Weierstrass holds in every \mathbb{R^n} and its independent of the norm that you have on \mathbb{R^n} is based on the following facts (which are consequences of the finite dimensionality of the space). Proposition 1 A normed linear space (X,||.||) (for example \mathbb{R^n}) is finite dimensional if and only if its closed unit ... 1 The answer to your question is in two steps. I assume you know the Bolzano-Weierstrass in (\mathbb{R},|\cdot|), stating that a closed interval [a,b] is compact. First step The metric space \left(\mathbb{R}^n,\|\cdot\|_{\infty}\right) is the product metric space \prod_{i=1}^n \left(\mathbb{R},|\cdot|\right) with the supremum distance. Moreover, a ... 2 All norms are equivalent on finite dimensional vector spaces. That is, there are positive c,C such that$$c\|x\|_1\leq \|x\|_2\leq C \|x\|_1$$for all vectors x. In other words, they generate the same topology. 0 The singular value decomposition (SVD) of the real matrix \mathbf W is a factorization of the form$$\mathbf W=\mathbf U\mathbf \Lambda \mathbf V^Twhere \mathbf U and \mathbf V are unitary matrices and the diagonal of \mathbf \Lambda\triangleq\operatorname{diag}\left\{w_1,w_2,\cdots,w_m\right\}, w_k\geq 0\forall k, contains all singular of \... 1 Let V:=[v_1,\ldots,v_k] written as column vectors. Then V^\top V=I means that v_i are orthonormal. Now, for any vector x,\begin{align} \|VJV^\top x\|_\infty&\le\|VJV^\top x\|_2\\ &=\|(v_1\cdot x)v_1+\cdots-(v_k\cdot x)v_k\|_2\\ &=\sqrt{\sum_{i=1}^k(v_i\cdot x)^2}\\ &=\|(v_1\cdot x)v_1+\cdots+(v_k\cdot x)v_k\|_2\\ &=\|VV^\top x\|_2\... 1 Although this question is for previous years, I wanted to give a more complete and descriptive answer that might be useful to others. \frac{d |a|^2}{da}=\frac{d (a a^{*})}{da}=\frac{1}{2}\left(\frac{d( a a^{*})}{d \Re{a}}-i \frac{d( a a^{*})}{\Im{a}}\right) \\ \frac{1}{2}((a+a^{*})-(a-a^{*}))=a^{*} $$a is complex number in \mathbf{C}. 2 This statement is not true. As an example, consider the matrix$$ A = \pmatrix{0 & 1\\0 & 0}, $$and let \|\cdot\|_k denote the matrix norm defined by$$ \|A\|_k = \|S_kAS_k^{-1}\|, \quad S = \pmatrix{k & 0\\ 0 & 1}, $$where \|A\| denotes the induced Euclidean norm (maximal singular value, AKA spectral norm) of A. We find that for ... 0 \def\p{\partial} To reduce the vistual clutter, define the following variables$$\eqalign{ A &= K_{ts}(K_{ss} + \lambda I)^{-1} \\ x &= y_s \\ b &= y_t \\ }$$Then the objective function can be written in the standard form$$\eqalign{ L &= \tfrac 12 \|Ax-b\|^2 \\ }$$whose gradient and optimum$$\eqalign{ \frac{\p L}{\p x} &= A^T(Ax-b) ... 0 After thinking a long time, I have an idea to prove it. Leta,b\in[0,1]$. Then, $$a=T(a,a)\leq T(a,b)\leq T(a,1)=a.$$ So, I can conclude$T(a,b)=a\$.

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