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0

Partial answer to $i)$: Using a generalization of the Perron-Frobenius theorem you can compute some of these norms. This is all explained in this paper (mainly Theorem 1 and 2). Actually they treat the more general case of such norms but for tensors instead of matrices. Note that this answer is also related to your problem. I summarize here a few results: ...


4

Actually your question is the combination of two questions. About your first question: You certainly know the Euclidean norm, $$\left\|v\right\|_2 = \sqrt{\sum_{k=1}^n \left|v_k\right|^2} = \left(\sum_{k=1}^n \left|v_k\right|^2\right)^{\frac{1}{2}}$$ derived from the Euclidean scalar product. Note that I used absolute value bars here, which are superfluous ...


2

Let's just think about norms in the plane. The ordinary Euclidean norm is $(x^2 + y^2)^{1/2}$. The unit circle for that norm is the ordinary unit circle. There are other useful norms. One is $|x| + |y|$. Its "unit circle" is the square with vertices $(\pm 1, 0)$ and $(0, \pm 1)$. These norms are part of a family of norms defined by $\|(x,y)\|_p = (|x|^p + ...


1

In $\mathbb R^n$, the norm $\|\cdot\|_\infty$ is the one whose unit ball is the $n$-dimensional cube $[-1,1]^n$. Thus $\|x\|_\infty$ is how much you have to scale that cube so that $x$ will lie on its surface. For $\mathbb C^n$, the definition is formally the same, though we might not be happy calling the unit ball a "cube". For $m\times n$ matrices, I ...


1

I think this calculation comes from two parts. First, the eigenvalue of $aa^T$, where $a$ is $n\times 1$ vector, is just $\|a\|_2^2=a_1^2+\ldots+a_n^2$ and (n-1) $0$s(This is because when $b$ is perpendicular to $a$, $a a^T b=0$. Thus the eigenvector is just $a$, thus $aa^Ta=a\|a\|_2^2$.) Secondly, if we want to compute eigenvector of ...


1

I assume here the spaces are finite dimensional (as in the question), and the vector spaces are real (for convenience, one can generalize to other fields too). Recall first that a norm is induced from an inner product if and only if it satisfies the polarization identity: $$ 2\|u\|^2 + 2\|v\|^2 = \|u+v\|^2 + \|u-v\|^2 $$ for every $u,v\in V$. Let $V, W$ ...


1

Let $a = (a_n)_{n\in \Bbb N}$ and $b = (b_n)_{n\in \Bbb N}$ be in $\ell^p$. If $\|a + b\|_p = 0$, the inequality is clear, so suppose $\|a + b\|_p > 0$. Since $$|a_n + b_n|^p = |a_n + b_n| |a_n + b_n|^{p-1} \le (|a_n| + |b_n|)|a_n + b_n|^{p-1}$$ for all $n\in \Bbb N$, then $$\|a + b\|_p^p = \sum_{n = 1}^\infty |a_n + b_n|^p \le \sum_{n = 1}^\infty ...


0

How about this: let $\mu$ be the counting measure on $\mathbb N$. Then $$\int_{\mathbb N} |a| \, d\mu = \sum_{n=0}^\infty |a_n|$$ for any sequence $a = (a_n)_{n \in \mathbb N}$. If $a,b \in L^p(\mathbb N,\mu)$, the Minkowski inequality gives you $$ \|a+b\|_{L^p(\mathbb N,\mu)} \le \|a\|_{L^p(\mathbb N,\mu)} + \|b\|_{L^p(\mathbb N,\mu)}$$ which says that $$ ...


1

Let $Y=1^T\Phi$, then the problem is to find the derivative of the function $\,L=\|Y\|_F^2$ Better yet, using the Frobenius product, the function can be written as $\,L=Y:Y$ Start by taking the differential $$\eqalign{ dL &= 2\,Y:dY \cr &= 2\,1^T\Phi:1^Td\Phi \cr &= 2\,11^T\Phi:d\Phi \cr &= 2\,11^T\Phi:\Phi'\circ d(XV) \cr &= ...


2

Since the argument on the norm of the identity does not work here, let's try the unitary invariance. The Frobenius norm is unitarily invariant ($\|A\|_F=\|PAQ^*\|_F$ for any unitary $P$ and $Q$ of suitable dimensions), so if there were such vector norms inducing the Frobenius norm, they would need to be unitarily invariant as well. The only unitarily ...


1

It seems there is no such formula. Vaguely speaking even for simple cases this problem is NP-hard. For details see this paper


1

Since $$ d:X\times X\to [0,\infty) $$ is a metric, it satisfies the triangle inequality: $$ d(a_1,a_2)\le d(a_1,a_3)+d(a_3,a_2) \quad \forall a_1,a_2,a_3\in X. $$ Given $a=(a_1,a_2)\in X\times X$, we have for every $x=(x_1,x_2)\in X\times X$: $$ d(x_1,x_2)\le d(x_1,a_1)+d(a_1,a_2)+d(a_2,x_2), $$ and therefore $$\tag{1} d(x_1,x_2)-d(a_1,a_2)\le ...


1

You could prove that the metric $d$ is a continuous function from $A\times A$ to $\mathbb{R}$ by considering the product topology on $A\times A$. Let $M = \left({A, d}\right)$ be a metric space. Let $\tau$ be the topology on $A$ induced by $d$. Let $\left({A \times A, T}\right)$ be the topological product of $(A, \tau)$ and $(A, \tau)$. ...


1

Reference : 355p in the book ${\it mathematical\ analysis}$- Apostol Since $U$ is convex then there exists a line between $u$ and $v$ when $u,\ v\in U$. Fix $u,\ v$: MVT : Let ${\bf f} : \mathbb{R}^n\rightarrow \mathbb{R}^m$. For every ${\bf a}\in \mathbb{R}^m$, there exists ${\bf z}\in \overline{{\bf uv}} $ : $$ \langle {\bf a}, {\bf f}({\bf ...


0

Certainly $||x|| = \lambda||x|| + (1-\lambda)||x||$ so the paper is asserting that $$ ||x-\lambda\alpha_1-(1-\lambda)\alpha_2|| < \lambda||x-\alpha_1|| + (1-\lambda)||x-\alpha_2||\;.\tag1$$ This is equivalent to showing $$ ||\lambda(x-\alpha_1)+(1-\lambda)(x-\alpha_2)|| < ||\lambda(x-\alpha_1)|| + ||(1-\lambda)(x-\alpha_2)||\;.\tag2$$ The triangle ...


1

Here's a second-derivative proof for $\mathbb{R}^n$. The gradient and Hessian do not exist at the origin, but everywhere else they are given by $$\nabla f(x) = \|x\|^{-1} x, \quad \nabla^2 f(x) = \|x\|^{-1} I - \|x\|^{-3} xx^T$$ Strict convexity requires that the Hessian be positive definite; that is, $v^T(\nabla^2 f(x))v>0$ for all $v\neq 0$. But suppose ...


1

Not every normed space will provide an inner product, because there are normed spaces which are not prehilbert. The additional point to be satisfied is parallelogram law. The result you need is just the Jordan-Von Neumann Theorem.


4

The function $$ \lVert \mbox{ }\rVert:\mathbb{R}^n\to \mathbb{R},\, f(x)=\lVert x\rVert $$ is certainly convex. In fact, given $t\in [0,1]$ and $x,y\in \mathbb{R}^n$ we have: $$ \lVert tx+(1-t)y\rVert=\lVert tx+(1-t)y\rVert\le \lVert tx\rVert+\lVert (1-t)y\rVert=t\lVert x\rVert+(1-t)\lVert y\rVert. $$ But is it STRICTLY CONVEX? i.e. if $t\in (0,1)$ and $x,y ...


0

It is easier to work with the definition that a function f is convex if for any $\lambda \in [0,1]$ and for all $x,y \in (a,b)$: $$ f((1-\lambda)x + \lambda y) \leq (1-\lambda)f(x) + \lambda f(y)$$ Then: $$ ||(1-\lambda x) + \lambda y ||^2 = \sum_{i} |(1-\lambda x_i) + \lambda y_i|^2 \leq \sum_{i} (1-\lambda) |x_i|^2 + \lambda |y_i|^2 = (1-\lambda)||x||^2 ...


0

Any solution to $Ax=b$ (if it exists) can be written as $x=x_0+z$, where $z\in N(A)$ and $x_0\in N(A)^{\perp}$ and $x_0$ is that "solution without nullspace components". Since $x_0\perp z$, the Pythagorean theorem gives that $\|x\|^2=\|x_0\|^2+\|z\|^2$. The minimum-norm solution is hence attained for $z=0$.


1

For your second question the above answer is good.... For the first one let $x,y \in \overline{D(T)}$, i.e., there exist sequences $(x_n),(y_n)$ in $D(T)$ such that $x_n \longrightarrow x$ and $y_n \longrightarrow y$, then $(x_n+y_n)$ is a sequence in $D(T)$ (as $D(T)$ is a vector space) such that $x_n+y_n \longrightarrow x+y$ and if $\alpha$ is any scalar ...


1

If the system is unstable, the $H_\infty$ norm is infinite. You have a discrete time system with a pole greater than 1 in absolute value hence the answer inf. First one doesn't check for stability and converts it to state space system. The discrepancy is due to the fact that norm() command only checks for $L_\infty$ not necessarily $H_\infty$ (for stable ...


1

You can define $G$ as follows: if $x_n \in D(T)$ and $w \in W$ with $x_n \to w$, then $G(y) = \lim_{n \to \infty} T(x_n)$ (use boundedness of $T$ to show that this is well-defined). Then if $x_n \to w$ and $y_n \to v$, and $a,b$ are scalars, you want to show that $G(aw + bv) = a G(w) + b G(v)$. Well, what sequence (defined in terms of $x_n$, $y_n$, $a$, ...


2

Yes, the closure of a vector space is a vector space. Yes, $D(T)$ might not be a closed set. For example, polynomials form a non-closed subspace of $C[0,1]$ (the continuous functions on $[0,1]$).


1

First, suppose $T$ is bounded. Let $\{x_n\}$ be a sequence in $X$ converging to zero. Then, $\{\lVert x_n\rVert\}$ also converges to zero and hence is bounded (by some $M$). Thus, $\lVert Tx_n \rVert \leq \lVert T \rVert \lVert x_n \rVert \leq \lVert T \rVert M < \infty$. Conversely, suppose $T$ maps zero-convergent sequences to bounded sequences. If $T$ ...


1

The $2$-norm of a vector is the length of the vector (or perhaps the square of the length of the vector. This notation isn't completely standardized). More generally, a $p$-norm of a vector $x = (x_1, ..., x_k)$ usually refers to either $$ \lvert x \rvert_p = \sqrt[p]{|x_1|^p + \ldots + |x_k|^p}$$ or $$ \lvert x \rvert_p^p = |x_1|^p + \ldots + |x_k|^p.$$


0

No, this is not correct. Note that $x_n \to x$ implies always $$\|x_n-x\| \leq \epsilon$$ as $n \to \infty$ for any $\epsilon>0$; that's simply the definition of convergence of a sequence. If you want to show that a subspace $X \subseteq Y$ is closed, you have to show that for any convergent sequence $x_n \to x \in Y$ it holds that $x \in X$. For ...


0

Fix $w\in W$. We have for each n, $\|v^*-w_n\|\le \|v^*-w\|$ Now, take the limit as $n\to \infty$, and use the fact that the norm is continuous to justify exchanging the limit with the norm, to conclude that $\|v^*-w_0\|\le \|v^*-w\|$


0

Yes, you can write Holder's inequality for random vectors (for $\mathbb{R}^n$-valued functions more generally). $$\mathbb{E}\left[\sum_{i=1}^n\int_{\Omega} |X^i(\omega)Y^i(\omega)|\mathrm{d}\mathbb{P}\right]\leq \mathbb{E}\left[\sum_{i=1}^n\int_{\Omega} |X^i(\omega)Y^i(\omega)|^p\mathrm{d}\mathbb{P}\right]^{1/p}\mathbb{E}\left[\sum_{i=1}^n\int_{\Omega} ...


3

Let the eigenvectors be $v_1$ and $v_2$ as you described. Define an application $\Bbb S^1\to \Bbb R$ by $$f(a,b)=\frac{\|A(av_1+bv_2)\|}{\|av_1+bv_2\|}.$$ This is a continuous function defined on a compact, hence it takes all possible values between its $\inf$ and $\sup$. We know that $f(1,0) = \frac 23$ and $f(0,1)=\frac 95$, therefore $\sup f\ge \frac ...


2

Hint Consider the function $$f: v \mapsto ||Av|| - ||v||$$ on $\mathbb{R}^2$ and a path $\gamma$ in $\mathbb{R}^2$ from $v_1$ to $v_2$. Note that, as stated, you need not actually construct a vector $v$ for which $||Av|| = ||v||$, only show that such a vector exists. Remark More generally, using the same idea one can show that given any linear ...


1

I believe the error is in the last inequality/step: first, why do you have $0 < \frac{\mu}{M}\leq 1$? (you need it for your argument to go through, as $\alpha$ was specifically assumed to be in $(0,1]$.) Note also, crucially, that this $\alpha$ should not depend on $f$... and yours does.


1

There's an implicit hypothesis that $T$ is self-adjoint (otherwise the desired formula might be false - take antysymetric $T$). Let $$a=\sup_{\|x\|=1}|(Tx,x)|.$$ You can write that $$(Tx,y) = \frac14(T(x+y),(x+y))-\frac14(T(x-y),(x-y)),$$ hence $$|(Tx,y)|\le \frac a4 (\|x+y\|^2+\|x-y\|^2) =\frac a2 (\|x\|^2+\|y\|^2) $$ We also know that ...


1

Using the block inversion formula, we have $$ A^{-1}=\pmatrix{*&*\\*&S^{-1}}, \quad S:=A_{22}-A_{12}^TA_{11}^{-1}A_{12}. $$ The Cauchy interlacing theorem implies that $$ \lambda_{\min}(A^{-1})\leq\lambda_{\min}(S^{-1})\leq\lambda_{\max}(S^{-1})\leq\lambda_{\max}(A^{-1}) $$ and hence, since for SPD $X$, $\lambda_{\max}(X^{-1})=1/\lambda_{\min}(X)$ ...


3

(1) Define $F \colon c_0 \to \def\K{\mathbf K}\K$ by $$ F(x) = \sum_{n=0}^\infty x_n^n $$ Then $F$ is continuous, as the series converges locally uniform, but $F$ is unbounded, as the elements $x^{(n)} = (1, \ldots, 1,0, \ldots) \in \bar B_{c_0}$ have $$ \def\norm#1{\left\|#1\right\|} \norm{x^{(n)}} = 1,\qquad \def\abs#1{\left|#1\right|}\abs{F(x^{(n)})} = ...


1

This is a general inequality concerning bounded linear operators between normed spaces. The Sturm-Liouville setting is actually irrelevant here. Let $X,Y$, be normed spaces, and let $T:X\to Y$ be linear and bounded. That is, the set$$\{Tx|\|x\|=1\}\subset Y$$is bounded. The operator norm of $T$ is defined by$$\|T\|=\sup_{\|x=1\|}\|T(x)\|,$$and the ...


3

Hint: See the Schur Complement.


2

Look at functions which are $-1$ on $x<- \epsilon$, $1$ on $x > \epsilon$ and go linearly from $-1$ to $1$ on the interval $[-\epsilon,\epsilon]$. Making $\epsilon$ arbitrarily small gives a function which is still in $C[-1,1]$ with norm $1$ and the value of $f$ for these is $2-\epsilon^2$. So, $||f|| \geq 2 - \epsilon^2$ for any $\epsilon >0$. ...


2

For any complex vector $$\|c\|=\sqrt{\langle c,c\rangle}=\sqrt{\sum_{i=1}^nc_i\overline{c}_i}.$$


4

No: $Tr \left ( \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \right ) = 0$ yet this matrix is not zero. Putting in absolute values doesn't help either, since we have matrices like $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. That said, the Frobenius norm $\| A \|_F$ is $\sqrt{Tr(A^T A)}$ and is indeed a norm, albeit not an induced norm.


0

My thoughts on your question: The condition $\|\Phi^T(t)\Theta(t)\|\geq \|\Psi^T(t)\Theta(t)\|$ is equivalent to $\Theta^T(t)[\Phi(t)\Phi^T(t)-\Psi(t)\Psi^T(t)]\Theta(t)$. This holds true for every $\Theta(t)$ if the matrix $$\Phi(t)\Phi^T(t)-\Psi(t)\Psi^T(t)\geq 0$$ i.e. if it is positive semidefinite. Also it holds true that $$\Phi^T(s)=G(s)\Psi^T(s)$$ ...


0

Let's try it: \begin{align} \|\textbf{x}+\textbf{y}\|^2 & = \langle\textbf x+\textbf y,\textbf x+\textbf y \rangle = \langle \textbf x,\textbf x \rangle + \langle \textbf x,\textbf y \rangle + \langle \textbf y,\textbf x \rangle + \langle \textbf y,\textbf y \rangle \\[5pt] & = \|\textbf x\|^2 + \langle \textbf x, \textbf y\rangle + \langle \textbf ...


0

Let $y \in V \setminus W$ with $\def\norm#1{\left\|#1\right\|}\norm y = 1$. Then, as $W$ is closed $d := \def\dist{\mathop{\rm dist}}\dist(y, W) > 0$. Choose $\delta > 0$ such that $\frac{d}{d+ \delta} > 1-\epsilon$. There is some $w \in W$ with $\norm{y-w} < d + \delta$. We let $v := \frac{y-w}{\norm{y-w}}$. Then $\norm v = 1$ and $$ \dist(v,W) ...


1

if $x \in V$ is non-zero then $||x|| \gt 0$, so set: $$ x' = \frac{x}{||x||} $$ it is now evident that $||x'|| = 1$


1

What is the spectral radius of $A^TA$? it is its largest eigenvalue. The square root of the largest eigenvalue of $A^TA$ is the largest singular value. Since the trace is the sum of eigenvalues, the inequality follows. If $A$ is symmetric and real, all eigenvalues are real, and if it is positive definite, all eigenvalues are positive. Since the spectral norm ...


1

Suppose the maximum $\max_{k=1,\ldots,n}\sum_{i=1}^{m}|\alpha_{ik}| = M$(say) is attained for $ k = k_0$, i.e, $M = \sum_{i=1}^{m}|\alpha_{ik_0}|$. Then take the vector $x= (0,\ldots,1,\ldots,0)$ with $1$ at the $k_0$ position. Notice that $\|Tx\|_{\mathbb{R^m}} = M$. Since there is a vector such that this supremum is attained, it follows that $\|T\| = M$.


1

It will be the same as the operator norm you get for the usual $1$-norm up to a factor. In particular, if $A=(a_{ij})$ is $n\times m$, then $$ \|A\|=\frac nm \max_j \sum_{i=1}^m |a_{ij}| $$


8

Yes. For example, we can define the Frobenius norm by $$ \|A\|_F = \sqrt{\sum_{ij}|a_{ij}|^2} = \sqrt{\operatorname{trace}(A^*A)} $$ in general, we have $\|I\|_F = \sqrt{n}$. This is a matrix-norm (AKA a sub-multiplicative matrix-norm) in that it is a norm under the usual definition and satisfies $\|AB\|_F \leq \|A\|_F\|B\|_F$. (This norm tends to be ...


1

For linear operators, it's always true that boundedness and continuity are equivalent. There's no difference in that regard between finite/infinite dimensions. The weird/nice thing about finite dimensions is this: since there are only finitely many directions to stretch, a linear map between finite dimensional spaces is forced to be bounded (and therefore ...



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