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14

Let $A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ with the $l_1$ norm, then $\|A\| = 1$. Since $A^2 = 0$, we see that $\lim_n \|A^n\| = 0$, but we have $\lim_n \|A\|^n = 1$.


6

Rademacher's theorem: a locally Lipschitz function on a nonempty open subset of $\mathbb R^n$ is (Frechet) differentiable almost everywhere.


3

We want $|ab| = |a|\,|b|$. This fact comes up all of the time. (Especially in estimating geometric series.) Furthermore, we want $|\overline{a}| = |a|$. Just think about the geometry and this will be obvious. Now, let's talk about the complex numbers as of the form $x+iy$ for real $x$ and $y$. We can already calculate that $$ ...


3

In general, if you want to approximate $F$ to the first order around some point $u^k$, Taylor's formula says $$F(u) = F(u^k) + \Bbb d F (u^k) (u - u^k) + \frac 1 2 \Bbb d ^2 F (v) (u - u^k, u - u^k) ,$$ with $v$ some point on the line segment of endpoints $u$ and $u^k$. As you see, there are three terms showing up; let us analyze them one by one. The ...


2

Q1) This is false, take $A=Id$, then $\langle x, x\rangle =\|x\|^2$ for all $x$. So $m=\inf\{\langle Ax,x\rangle\mid x\in X, \|x\|=1\}$ can be positive. Q2) If $\|x\|\leq 1$, then we can write $\langle Ax, x\rangle =\|x\| ^2\langle \frac{Ax}{\|x\|}, \frac{x}{\|x \|}\rangle \leq \sup\{|<Ay,y>|\mid y\in X, \|y\|=1 \}$ since $\|x\|^2\leq 1$. The ...


2

If $\|\cdot\|_B$ is a cross norm, then yes, your argument shows that $D_1 \otimes_a D_2$ is dense in $B_1 \otimes_a B_2$ and therefore also dense in $B$. Without this assumption or something similar, the answer is no, as then there is nothing at all relating the $B$ norm to the $B_1, B_2$ norms. This really has nothing to do with tensor products, so take ...


2

The "geometric" sequence $x_1=0, x_2=1$, $x_{n+1}=qx_n, n\ge2$, is in $\ell^2$ iff $|q|<1$. For this choice we get $(y_n)=T(x_n)$ where for all $n\ge2$ we have $$ y_n=-x_n+\alpha x_{n+1}=x_n(-1+q\alpha). $$ In other words $(x_n)$ is an eigensequence of $T$ belonging to the eigenvalue $\lambda=\lambda(q,\alpha):=-1+q\alpha$. This implies that we can make ...


2

Write $\alpha = re^{i\varphi}$ with $r \geqslant 0$ and $\varphi \in [0,2\pi)$. Let $\lambda = -e^{-i\varphi}$. Then consider $$x_n = \sum_{k=2}^{n+1} \lambda^k e_k.$$ We have $\lVert x_n\rVert = \sqrt{n}$, and $$T(x_n) = \sum_{k=2}^n (\alpha \lambda^{k+1} - \lambda^k )e_k - \lambda^{n+1}e_{n+1} = - \sum_{k=2}^n (1+r)\lambda^k e_k - ...


2

Let $X$ be the set of $f:[0,1]\to[0,1]$ such that $f(1)=1$. Then $X$ is complete in the uniform norm, the identity function is in $X$, and it's easy to show that $\Phi$ is a strict contraction on $X$.


2

The spectral radius is not a norm; one reason is because there are nonzero matrices all of whose eigenvalues are zero. (Such matrices can't be diagonalizable, but nondiagonalizable matrices exist!) copper.hat gave one example. Your property does hold if you replace $\| \cdot \|$ with the spectral radius. If $A^n \to 0$, then there exists a norm, which can ...


2

If use the norm for matrix as $$\|A\|=\sqrt{\sum\limits_{i,j=1}^{n}|a_{ij}|^2}$$ We can prove $$ \|A\|^n \to 0\implies \|A^n\| \to 0 $$ Converse is not true as a counter example is given by copper.hat. First we prove the following: Lemma: $\hspace{2 mm}\|AB\|\leqslant\|A\|\|B\|$ Prove: \begin{align} ...


2

Lemma 1: Let $A$ and $B$ be two points having norm $1$. Then theere is no point on the line segment joining $A$ and $B$ which has norm greater than $1$. Proof: The norm is a convex function because $\|tx+(1-t)y\|\leq t\|x\|+(1-t)\|y\|$ for all points $x$ and $y$ and $t\in [0, 1]$. Lemma 2: If there exist three points $A, B$ and $C$ having norm $1$, then ...


2

The answer is yes, assuming that the dimension is $\geq2$. Define$$f:V\to\mathbb{R},\quad y\mapsto\|x+y\|-\|x-y\|.$$The function $f$ is clearly continuous. Pick any $0\neq y_0\in V$. If $f(y_0)=0$, we're good. Otherwise, take some path $\gamma:[0,1]\to V$ connecting $y_0$ with $-y_0$, without passing through $0$. Since $f(\gamma(0))$ and $f(\gamma(1))$ have ...


2

Let $n$ denote the norm in question, since we are in finite dimensions, it is equivalent to the usual Euclidean norm. (Presumably by parallel, you meant parallel in the Euclidean sense.) Let $p \in S \subset \mathbb{R}^n$, that is, $n(p)=1$. Let $\pi$ be the orthogonal projection onto the subspace $\{x | \langle p, x \rangle = 0 \}$. Since $n \circ \pi$ is ...


1

A minor remark: from the statement of the problem, one deduces that $A$ is invertible. Since $\| I-AB_0 \| < 1$, $AB_0$ is also invertible, whence it follows that $B_0$ is invertible too (but this will not be used). Note that $$B_k = 2 B_{k-1} - B_{k-1} A B_{k-1} ,$$ therefore $$I - A B_k = I - 2 A B_{k-1} + A B_{k-1} A B_{k-1} = (I - A B_{k-1})^2$$ ...


1

For every positive element $a$ one has $a\leq\Vert a\Vert I$


1

First, $\|T\|$ is a single number describing the size of the $3\times3$ matrix. It cannot depend on any coordinate $x_1,x_2,x_3$. You can solve the problem by writing $|T(x)|=((2x_2)^2+(x_1)^2+(3x_3)^2))^{1/2}$ and playing with that, but let me give a more elaborate answer. There are several possible norms you could use on the space of $3\times3$ matrices, ...


1

No. By definition $$\|T\|_{2\to2} = \max_{\|x\|_2 = 1} \|Tx\|_2$$ Now this special operator norm can be found by taking the square root of the largest eigenvalue of $T^H T$. In this case, since $T$ is a permutation of a diagonal matrix, you can simply read it off as $3$.


1

For $A$ matrix denote by $\rho(A)$ the spectral radius of $A$, $= \max \{ |\lambda_i|\}$, with $\lambda_i$ are the eigenvalues of $A$. We have $A^n \to 0$ if and only $\rho(A)^n \to 0$ if and only if $\rho(A)<1$. For proof one can use the Jordan canonical form. Now if $||\cdot ||$ is any algebra norm on $M_n(\mathbb{R})$ ( for example coming from a ...


1

It seems the following. Surprisingly, the answers turned out very similar. $\ell^1$-norm. $$\|(x,y)\|=|x|+|y|.$$ $$\|T(x,y)\|= \|(ax+cy,bx+dy)\|=| ax+cy |+|bx+dy|\le$$ $$|ax|+|cy|+|bx|+|dy|=|a||x|+|c||y|+|b||x|+|d||y|= (|a|+|b|)|x|+(|c|+|d|)|y|\le$$ $$ \max\{|a|+|b|,|c|+|d|\}(|x|+|y|)=\max\{|a|+|b|,|c|+|d|\}\|(x,y)\|.$$ From the other side, if $|a|+|b|\le ...


1

Another way to proceed would be by writing out the surface integral using differential forms. To this end we need to set up a chart and a coordinate system on the Torus, fortunately we need only one chart and the coordinate system can be made global (using the coordinates already introduced in the question $u,v$), the surface integral will be $$ S=\int_{M} ...


1

Use Pappus theorem. If the radius of the transversal section of the torus is $r$ then its perimeter is $2\pi r$ and Pappus theorem states that the surface of the torus (it is a revolution surface) equals $A=2\pi r \cdot 2 \pi R$ where $R$ is the radius of rotation that generates the torus. In your case this is $$ A = 4\pi^2 ab $$


1

Note that $e+v$ satisfies the definition given in 2.2 since $e \in T$ and the norm of $v$ is less than or equal to 1, so $e + v$ must be in the subgradient. $\langle e+v,h\rangle = \langle e+v-y,h\rangle$ because $\langle y,h\rangle = 0$ (since $y$ is in the image of the adjoint while $h$ is an element of the nullspace). This follows from the fact that $$Ah ...


1

First of all, for any real square matrix of size $n$, symmetric or not, positive definite or not, the equation $\vec v'(t) = M \vec v(t), \; t \in [0, T], \tag{1}$ with the initial condition $\vec v(0)$ at $t = 0$, always admits the unique solution $\vec v(t) = e^{Mt}\vec v(0); \tag{2}$ that (2) solves (1) is easily seen by direct differentiation; (2) ...


1

Strictly speaking, the answer to the question in your question title is "yes" if you replace the strict inequality $>$ by weak inequality $\ge$. There will always be at least a one-dimensional subspace of the whole vector space in which the inequality is true. But the answer for the question about whether it's true for the whole space is of course "no" if ...


1

Direct calculation gives $$\|A^n x\|^2 = x' A^{\prime n} A^n x = x' S^{\prime -1}D^nS'SD^nS^{-1} x.$$ If the eigenvalues are $\lambda_1,\lambda_2$, then $D^n = \begin{pmatrix} \lambda_1^n & 0 \\ 0 & \lambda_2^n \end{pmatrix}$, and so, letting $T = S'S$, $$ D^n T D^n = \begin{pmatrix} \lambda_1^{2n} T_{11} & \sqrt{\lambda_1 \lambda_2}^{2n} T_{12} ...


1

Let $\sigma_1, \sigma_2, \cdots, \sigma_m$ denote all non-zero singular values of $A$, the identity below holds: $$ \sum_{i=1}^m \sigma_i^2 = \|A\|_F^2 $$ Also note that $\sigma_1^2, \sigma_2^2, \cdots, \sigma_m^2$ are all non-zero singular values of $A^TA$, thus $$ \sum_{i=1}^m \sigma_i^4 = \|A^TA\|_F^2 $$ By Cauchy-Schwarz inequality, we have $$ ...


1

Since $||A||_P = (\Sigma_j\Sigma_i|a_{ij}|^p)^{1/p}$, if for any pair $|a_{ij}| \ge |b_{ij}|$ holds, I think $||A||_p \ge ||B||_p$


1

An example of a matrix norm where the submultiplicative inequality is not tight (except when $A$ or $B$ is zero): Taking $|\cdot|$ to be any norm (for example, take the spectral norm), we note that the norm $\|\cdot\|$ defined by $\|A\| = c|A|$ is another matrix norm for any $c > 1$. We note that if $|AB| = |A| \cdot |B|$, then $$ \|AB\| = c\cdot|A| ...



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