# Tag Info

1

Simple example of two non-equivalent norms on infinietly-dimensional space: Consider space of all contnuously differentiable functions $X = C^1 [0,1]$. Then equipping it with the norm: $$\|f \|_{C^1} = \sup \limits_{x \in [0,1]} |f| + \sup _{x \in [0,1]} |f'|$$ gives us a complete space (Banach space), but if we consider norm: $$\|f \|_\infty = \sup ... 4 If A is nonsingular, then AA^{-1} = I, so$$ 1 = ||I|| = ||AA^{-1}|| \leqslant ||A||\cdot||A^{-1}||. $$In general, then 1 \leqslant ||A||\cdot||A^{-1}|| \implies ||A||^{-1} \leqslant ||A^{-1}||. Equality is thus not necessarily guaranteed for arbitrary nonsingular A; however, the inequality above implies that equality may occur. Consider an ... 0 for any two polynomials P(x),Q(x) :$$L\left(P(x^2)+ x Q(x^2)\right) = P(x^2)$$now think to :$$\min_{Q} \quad||P(x^2) + x Q(x^2)|| = \int_{-1}^1 \left|P(x^2)+x Q(x^2)\right| dx$$because P(x^2) is an even function while x Q(x^2) is an odd function, the minimum is attained for Q(x^2) = 0. thus :$$||L(P(x^2)+x Q(x^2))||=||P(x^2)||\le ...

1

[Answer under repair; see comments below] Take $p = p_{n}(x) = x^{2n} - x^{2n+1}$, so that $L(p) = x^{2n}$. We note that $$\|p\| = \int_{-1}^1 |p(t)|\,dt = 2\int_0^1 [t^{2n} - t^{2n+1}]\,dt = 2 \left[ \frac {1}{2n} - \frac{1}{2n+1}\right] = \frac{1}{n(2n+1)}$$ On the other hand, $$\|L(p)\| = \int_{-1}^1 t^{2n}\,dt = \frac{2}{2n+1}$$ From there, it ...

1

Your notation is a little idiosyncratic. More precisely, $\|x\| = \inf \{ r \ge 0 | x \in r B \}$. Suppose $x = 0$, then $x \in rB$ for all $r >0$, hence $\|x\| = 0$. If $\|x\| = 0$, then there are $r_k \ge 0$ such that $r_k \to 0$ such that $x \in r_k B$. Since $\cap_k r_k B = \{0\}$ (from 5.), we see $x = 0$. Suppose $\lambda = 0$, then $0 = \| ... 1 Indeed, the only requirement to have the inequality that you wrote for the Frobenius norm and for arbitrary matrices is that the product$AB$is defined. If you are looking for a reference see for example the book Numerical Linear Algebra, by Trefethen and Bau, page 23. 3 This follows from Ptolemy's inequality, which states that for any quadrilateral$ABCD$, we have $$\overline{AB}\cdot \overline{CD}+\overline{BC}\cdot \overline{DA} \ge \overline{AC}\cdot \overline{BD}.$$ Identifying$A,B,C,D$with$0,z,x,y$, this gives $$\left\|z-0\right\|\left\|y-x\right\|+\left\|x-z\right\|\left\|0-y\right\|\ge ... 1 The "functions" in L^p spaces, including L^\infty, are actually equivalence classes, where f and g are equivalent if they are equal almost everywhere. So any statement you make about f(x) can only be interpreted in the "almost everywhere" sense. However, for f \in L^\infty you can choose a representative of the equivalence class that is bounded ... 1 Only for almost all x. This follows from the definition of the "norm". One possible (equivalent) definition is that \|f\|_\infty is the smallest number M>0 such that |f(x)|\le M for almost all x. 0 Hint: draw a picture of x,y,z. What does hypotheses mean? Then using triangle inequalities 1 Let us take \|B\|_{2}=sup \dfrac{\|BX\|_{2}}{\|X\|_{2}} \ \ (\forall X \neq 0, X \in \mathbb{R}^n) or equivalently: \|B\|_{2}=sup \|BX\|_{2}\ \ \ (\forall X \neq 0 \ s.t. \|X\|_{2}=1) (induced norm). Remark: \|B\|_{2} can have a different meaning (Frobenius norm). Let us consider the rank-one matrix B=vu^T (its image space is generated by v). ... 0 l_p metric on \mathbb R^m:$$ d_p(x-y) = \left(\sum_{i=1}^m \big|x_i-y_i\big|^p\right)^{1/p} $$1\le p < \infty. I will let you do the l_\infty metric. 1 (some changes in order to improve precision) About your last interrogation. Any normed vector space defines naturally a metric space by the relationship d(x,y)=\|x-y\|_p. Thus, indeed, any \|\cdot\|_p norm induces naturally an \ell_p "metrics" (synonym : "\ell_p distance"). A point of vocabulary about the words "metrics" vs. "distance". "Metrics" ... 0 The answer can be found in "Calculus without derivatives" of Prof. Jean-Paul Penot. I refer to Lemma 3.94 p.251. The answer is the following : an equivalent norm on the dual X^* is the dual norm of an equivalent norm on X if and only if it is weak-* lower semicontinuous. The major idea is the following. Let us consider the notations introduced in the ... 1 The best intuition I can get for your result comes from one dimension. Think of u as being 1 on an interval (a,b) and then diminishing down to 0 outside a bigger interval (c,d)\supset (a,b). As you exponentiate u to higher and higher powers, it becomes dominated by what happens on (a,b) and goes to zero outside (a,b). Also, no matter how ... 0 Let S=\mathrm{span}\{x\} and P the orthogonal projection onto S. Then Py=cx for some c\in F. Thus \begin{multline}\|Py\|^2=|\langle Py,Py\rangle|=|\langle Py,y\rangle|\\=|\langle cx,y\rangle|=|c\langle x,y\rangle|=|c|\|x\|\|y\|=\|cx\|\|y\|=\|Py\|\|y\|...(*)\end{multline} If Py=0, done. if Py\neq 0 , then by (*) we have \|Py\|=\|y\|. ... 2 Define z:=x-\frac{\left\langle y,\,x \right\rangle}{\left\langle x,\,x \right\rangle}y. Rearrange \left\langle z,\,z \right\rangle \geq 0, with equality iff z = 0, to \left|\left\langle x,\,y \right\rangle\right|^2 \leq \left\langle x,\,x \right\rangle \left\langle y,\,y \right\rangle. This inherits the above equality condition, which implies x,\,y ... 1 Note: the correct statement you should prove is$$|\left \langle x,y \right \rangle|=\left \| x \right \|\cdot \left \| y \right \|\Leftrightarrow x,y\text{ are not linearly independent}$$(there is a distinction of cases to be made for x=0, for instance) A classic approach: Consider the polynomial (in \lambda)$$ P(\lambda) = \lVert x+\lambda ... 0 I’m guessing from your comment that you don’t know what$Q_8$is. It has eight elements,$\{\pm1,\pm i,\pm j, \pm k\}$with the property that$\{\pm1\}$commute with everything, and such that$i$,$j$, and$k$satisfy the combination rules that you know:$ij=k=-ji$,$jk=i=-kj$, and$ki=j=-ik$. Now you see it? 4 As Travis said in the comments, $$\| a b\| = \| a\| \| b\|$$, in this case we have$ab=1$so we can write, $$1 = \|a\| \|b\|$$, and if we square both sides we can write, $$1 = (a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2).$$ All the$a_n$and$b_n$are integers. We then have that the product of two integers is equal to$1$. Under what ... 1 No. Don't satisfies Triangle inequality. For example d([0 0 0],[2 2 2])=12 > d([0 0 0], [1 1 1]) + d( [1 1 1], [2 2 2])=6. 2 Note that $$\dfrac{1}{\inf_{\|{\bf x}\|=1} \|A{\bf x}\|} = \sup_{\|{\bf x}\|=1} \dfrac{1}{\|A{\bf x}\|} = \sup_{\|{\bf x}\|=1} \dfrac{\|A^{-1}(A{\bf x})\|}{\|A{\bf x}\|} = \sup_{{\bf x} \in \mathbb{R}^n \setminus \{{\bf 0}\}} \dfrac{\|A^{-1}(A{\bf x})\|}{\|A{\bf x}\|}$$ Since$A$is nonsingular, we have$A(\mathbb{R}^n \setminus \{{\bf 0}\}) = \mathbb{R}^n ...

1

A good first step is to perform some experiments with small matrices! Here tools such as MATLAB can be valuable when disproving identities. You will find that the statement under investigation is false in general, but it is certainly true if the matrices $B^{-1}A$ and $A^{-1}B$ are symmetric! There is condition which occurs frequently in pure mathematics, ...

0

Another counterexample on any Banach space $X$: choose $$d(x,y) = f(\|x - y\|_{X})$$ where $f : [0,\infty) \to [0,\infty)$ is monotonically increasing $f(x) = 0$ iff $x = 0$ is subadditive This is not induced by a norm unless $f$ is linear.

2

As Carl Christen said you just need to find the eigenvalues of your $A^{-1}_n$ and get the max. The characteristic polynomial of $A^{-1}_n$ is: $$p_0(\lambda)=1$$ $$p_1(\lambda)=\lambda - 4$$ $$...$$ $$p_n(\lambda)=(\lambda - 4) p_{n-1} + p_{n-2}$$ Being the resolvant a tridiagonal matrix of the form $$A^{-1}_n- \lambda I = \left[\begin{array}{cccc} ... 1 Not necessarily. First of all note that if d is a metric induced by a norm then it must satisfy the following two properties (1) d(x,y)=d(x+z,y+z); (2) d(ax,ay)= |a|d(x,y) for all scalars a and vectors x,y,z. So, the discrete metric will be a trivial counter example to your question. 2 I think there's no general inequality which holds: The sequences$$a_0 = 2, a_k = 0, k > 0$$and$$b_0 = \frac{1}{2}, b_k = 0,k > 0$$should show that (take i.e. p = 1, q = 2). To prove l^p \subseteq l^q, you have to take a sequence (a_k) \in l^p and show that it is in l^q. First, observe that we can assume without loss of generality that ... 2 Your large matrix is also symmetric, so computing the 2-norm is a question of computing the eigenvalues. I recommend that you write your large matrix as a Kronecker product E \otimes A^{-1} for a suitable symmetric matrix E. Known properties of the Kronecker product will then solve your problem. 0 Call your big matrix B. \|B\|_2^2 is just the sum of squares of all entries. So, if you find \left\|A_n^{-1}\right\|_2^2 = x, and notice your large matrix has exactly 2(n-1) copies of A_n^{-1}, so$$ \|B\|_2^2 = 2(n-1)x... $$3 \newcommand{\norm}[1]{\left\|{#1}\right\|} First of all you have to know or demonstrate that$$\norm{S \otimes T} \leq \norm{S}\norm{T}$$Then notice that the matrix M you address is$$M=J \otimes I$$where I is the identity matrix and J the generalization of this matrix$$J = \left[\begin{array}{cccc} 0 & 1 & 0 & 0\\ 1 & 0 & 1 ...

0

The $x$ is an element that you can take norm of, and $0$ is an neutral element under addition (ie such that $0+f = f$). In your example it would translate to: $||f||_1\ge 0$ $||f||_1 = 0$ iff $f(t)=0$ for all $t\in[a,b]$ $||\alpha f||_1 = |\alpha|\cdot||f||_1$ $||f + g||_1 \le ||f||_1+||g||_1$ So for example for the first we note that since $|f(t)|\ge0$ ...

1

The first condition says $||x||\geq0$ for all $x\in X$. In your case $X=C[a,b]$, such that elements of $X$ are continuous functions $f:[a,b]\to\mathbb{R}$. In other words, you have to prove $||f||\geq0$ for all $f\in X$. Since $|f(t)|\geq0$ for any $t\in[a,b]$, you have $\sup_{t\in[a,b]}|f(t)|\geq0$, proving $||f||\geq0$. I think you'll be able to prove ...

1

I'd say the main argument would be that this formulation is not really advantageous. The thing is that for $\dim X < \infty$, $B[0,1]$ and $S$ are compact and $\left\Vert Tx \right\Vert$ is continuous. Hence, there exists some $x_0 \in S$ such that $$\left\Vert T \right\Vert = \left\Vert Tx_0 \right\Vert$$ making the supremum a maximum. The supremum over ...

3

It is meant to be 1-Lipschitz with respect to $\| \cdot \|_0$. Otherwise the claim would be false, consider for example the norm $\| \cdot \|_0 := 2 \| \cdot \|$ for $y = 0$ and an arbitrary $x \ne 0$.

0

In physics vector-valued $L^{2}$-functions occur frequently. As an example let $f$ be a function from $\mathbb{R}^{n}$ into the vector space $V$ with $% |.|_{V}$ the norm on $V$. Then we denote the $L^{p}$-space as $L^{p}(\mathbb{% R}^{n},V,d^{n}x)$ or something similar. Now \begin{equation*} \parallel f\parallel _{p}^{p}=\int d^{n}x|f(x)|_{V}^{p} ...

1

I guess that by "$u$ is a vector", you mean "$u$ is a vector-valued function", right? In this case, indeed the definition would be to use the definition of $l^p$ norm inside the integral term.

2

I am also new to analysis like you if almost having $\epsilon$ level of experience in "real analysis". Let $f \in \mathcal{C}[a,b]$ and suppose further that $f \neq 0$, hence there is some $c \in [a,b]$ with $f(c) \neq 0$. If $c = a$, then $f(a) \neq 0$ and since $f$ is countinuous at $x = a$, using definition of continuity at $a$ for $f$ we have: $\epsilon ... 2 I'll leave some hints. For a constant$c$, $$\left\Vert cf\right\Vert ={\it \int_{a}^{b}\left|cf(x)\right|dx=\int_{a}^{b}\left|c\right|\left|f(x)\right|dx=\cdots}$$ The triangle inequality is an application of the triangle inequality on$|\cdot|$: $$\left\Vert f+g\right\Vert ... 1 We're looking for a vector norm |\cdot|_T on X such that$$\|A\|_T = \max_{x \in X} \frac{|Ax|_T}{|x|_T} \left(= \max_{x \in X, |x|_T = 1} |Ax|_T\right),$$i.e. \|\cdot\|_T is induced by |\cdot|_T. Let |x|_T = |T^{-1}x|. (Is this a norm?) Then we have:$$\max_{x\in X} \frac{|Ax|_T}{|x|_T} =\max_{y\in X} \frac{|ATy|_T}{|Ty|_T} =\max_{y\in X} ... 1 Write$A = B + C$where$B = (A+A^t)/2$is symmetric and$C= (A-A^t)/2$is antisymmetric. We have$I + \epsilon A = \exp (\epsilon A) + O(\epsilon^2)$. Now $$\exp \epsilon A = \exp \epsilon(B+C) = (\exp \epsilon B) (\exp \epsilon C) + O(\epsilon^2) = (I + \epsilon B) \exp \epsilon C + O(\epsilon^2)$$ where the second equality follows from the series for ... 0 We assume that$A\not= 0$. Let$spectrum(A)=(\lambda_i)$where$|\lambda_1|\geq |\lambda_2|\geq\cdots$and$\Sigma(A)=(\sigma_i)$where$\sigma_1\geq \sigma_2\geq\cdots$. Then$\rho(A)=[\lambda_1|\leq \sigma_1=||A||_F=\sqrt{\sum_i \sigma_i^2}$. Thus$|\lambda_1|=\sigma_1$and$\sigma_2=\cdots=\sigma_n=0$. Finally$rank(A)=1$and$A=uv^*$where$u,v$are ... 2 As comments already pointed out, you can't have a physical space with more than 3 spatial dimensions. So the concept of length as something which can be measured with a ruler breaks down. You can simply continue to use the term “length” for the norm, since it helps intuition, it is compatible with everyday experience in lower dimensions and there is no ... 1 Suppose$S=\{0\}$. Then$dim(S)=0$. Thus$\dim(S^\perp)=n$, so$S^\perp=\mathbb{R}^n$. But$\mathrm{span}(\{v_1,...,v_m\})=S^\perp$which implies$n=\mathrm{dim(S^\perp)}=\mathrm{dim}(\mathrm{span}(\{v_1,...,v_m\}))\le m<n$. This contradiction shows$S\neq\{0\}$1 Let$A$be the$m\times n$matrix with the vectors$\def\v#1{{\bf#1}}\v a_i$as its rows. Then$S=\ker A$and so $$\dim S={\rm nullity}(A)=n-{\rm rank}(A)\ge n-m>0\ .$$ 1 Consider$\mathbb{R}^2$and$x = (1,1), y = (1,-1)$. Then$x \cdot y = 0$, but$\|x\|_1 = \|y\|_1 = \|x+y\|_1 = 2$. 1 Since$\|(a,b)\|_1=|a|+|b|, we get that \begin{align} \|(a,b)+(c,d)\|_1 &=\|(a+c,b+d)\|_1\\ &=|a+c|+|b+d| \end{align} However, \begin{align} \|(a,b)\|_1+\|(c,d)\|_1 &=|a|+|b|+|c|+|d| \end{align} Thus,\|(-1,1)+(1,1)\|_1=2$, yet$\|(-1,1)\|_1+\|(1,1)\|_1=4$What is true is that if$x\cdot y=0then \begin{align} \|x+y\|_2^2 ... 3|x\cdot y| \le \sum_{i=1}^n |x_i\bar y_i|\le \sum_{i}\left((\sup_j|x_j||y_i|)\right)=\sup_j|x_j|\sum_i|y_i|=\|x\|_{\infty}\|y\|_12 You know that \lvert x_i\rvert \le \max_{j=1,\dotsc,n}\lvert x_j\rvert for each i=1,\dotsc, n. Hence \lvert x_i\rvert\le \lVert x\rVert_\infty for each i=1,\dotsc,n. Now, \begin{align*} \lvert x\cdot y\rvert &= \left\lvert\sum_{i=1}^nx_iy_i\right\rvert \le \sum_{i=1}^n \lvert x_i\rvert\cdot \lvert y_i\rvert\\ &\le \sum_{i=1}^n\lVert ... 0 I'm going to add this as a separate answer, but it is more or less the same answer, phrased differently. Suppose that \mathbb{Q} is dense in \mathbb{R}. Then for each r \in \mathbb{R} there is some sequence of \{q_n\} in the rationals such that \{q_{n}\}_{n=1}^{\infty}=r. Let \mathbb{B}=\{B(q,1/n) \mid q \in \mathbb{Q}, n \in \mathbb{N}\}. ... 0 Let U be open in \mathbb R^n. For every x\in U, there exists q_x\in \mathbb Q^n and r_x\in \mathbb Q such that x\in B(q_x,r_x)\subset U. ThereforeU = \cup_{x\in U} \{x\} \subset \cup_{x\in U} B(q_x,r_x) \subset U.$Therefore$U = \cup_{x\in U} B(q_x,r_x),$and there are at most countably many distinct$B(q_x,r_x).\$

Top 50 recent answers are included