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1

Let $\lambda_a \colon b \mapsto a\cdot b$. Then it's easy to see that the operator norm of the matrix is at most $$\max_i \sum_j \lVert \lambda_{a_{ij}}\rVert_{\operatorname{op}}.$$ For a general Banach algebra, it is possible that $\lVert \lambda_a \rVert_{\operatorname{op}} < \lVert a\rVert$ for some $a$. Still, even if we take the operatornorm of the ...


0

I that think the key to answering this question is to recognize that Im(A) is a subspace of ℝ^m. It means that for a given b ∈ ℝ^m, there exists a point of Im(A) closest to b; namely, the orthogonal projection of b onto Im(A). We can write this projection as Ax, for some x ∈ ℝ^n, by the definition of Im(A). Thus, we have the result that for any other point ...


0

Yes, you could solve this using some proximal algorithm such as the Douglas-Rachford method. Let $C = \{ (A,x) \mid D = A + Mx \}$, and let $I_C$ be the indicator function of $C$: \begin{equation} I_C(A,x) = \begin{cases} 0 & \text{if } D = A + Mx, \\ \infty & \text{otherwise.} \end{cases} \end{equation} Your problem can be restated as ...


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} ...


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 ...


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 $\lim_n \|A\|^n = 1$.


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 $X$.


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, $\|T\|$ 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

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


1

If $\xi\in \ell^p$ with $\|\xi\|_p = 1$, then $$\|T\xi\|_p^p = \sum_{i=0}^\infty |\xi_{2i+1}|^p\leqslant\sum_{i=0}^\infty |\xi_i|^p=1, $$ so $T$ is bounded and $\|T\|\leqslant 1$. Let $\xi_0=(1,0,0,\ldots)$, then $\|\xi_0\|_p =1$ and $T\xi_0=\xi_0$, so $\|T\xi_0\|_p=1$ and $\|T\|\geqslant 1$.


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 $3$.


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

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 ...


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 ...


1

Hint: Write $u=(u_1,\ldots,u_n)$ and $v=(v_1,\ldots,v_n)$. By Young's inequality, $$|u_i \cdot v_i| \leq \frac{|u_i|^p}{p} + \frac{|v_i|^q}{q}. \tag{1}$$ Rewrite this inequality using the definition of $u$ and $v$. Sum $(1)$ over $i=1,\ldots,n$. Deduce that $$\sum_{i=1}^n |u_i v_i| \leq 1.$$ Conclude.


1

Your parallelogram equality does not seem correct. If a norm $\| v \|$ is given by an inner product, then $$ \|v+w\|^2 + \|v-w\|^2 = 2 (\|v\|^2 + \|w\|^2) $$ Now just pick $v = (1,0)$, $w=(0,1)$, and calculate both sides with arbitrary $p$, then you will get an equation which holds iff $p=2$.


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 ...


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| ...


1

If you have a "shear-projection" in 2-d which is not orthogonal, e.g. has matrix representation $A = (0,0;\alpha,1)$ for some $\alpha \neq 0$, then with the vector $v = (0,1)^T$, you have $Av = v$ so your inequality can't hold. Are you leaving out some conditions to prohibit this case?


1

Suppose that $z$ is such that $\|Az\| = \|A\|$. Then $$ \sup_{\|x\|=1,\|y\|=1} |\langle x,Ay \rangle| \geq \left|\langle Az/\|A\|,Az \rangle \right| = \frac 1{\|A\|} \left|\langle Az,Az \rangle \right| = \|A\| $$ This works in the finite dimensional case where the sup is really a max. In the infinite dimensional case, it still suffices to choose $z$ such ...


0

I think you're forgetting the fact that completeness requires Cauchy sequences. Specifically you need $\|f_n-f_m\|_\infty<\epsilon$ for all $n,m\geq N$. However, $\lim_{m\rightarrow\infty}\|f_n-f_m\|_\infty\rightarrow 1$ for any fixed $n$. So your example is not a Cauchy sequence in $L^\infty$. Double check that it is Cauchy in $L^p$.


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 $$ ...


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 - ...


6

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


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 ...


0

Hint: note that $$ \|A^n(x)\| = \left\| \pmatrix{ 3\cdot 2^n-2\cdot 3^n \\ 2 \cdot 2^n-2\cdot 3^n } \right\| = 3^{n}\left\| \pmatrix{ 3\cdot (2/3)^n-2 \\ 2\cdot (2/3)^n-2 } \right\| $$ and that as $n \to \infty$, we have $$ \left\| \pmatrix{ 3\cdot (2/3)^n-2 \\ 2\cdot (2/3)^n-2 } \right\| \to \left\| \pmatrix{ -2 \\ -2 } \right\| $$


0

The answer is "no" in general. I give an example, where the dual norm of the operator norm norm with respect to the primal norm does not coincide with the operator norm with respect to the dual norm. Take $V$ to be $\mathbb{R}^n$ equipped with the Euclidean norm $\|.\|_2$. By a suitable identification of dual space and primal space (this works well in finite ...


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} ...


0

You probably mean $||A||$ to be the norm of the operator $A$. Also, you probably mean $\ge $ rather then $>$. The answer is still No. Consider the operators $A = \left( \begin{array}{cc} 5 & 0 \\ 0 & 1\end{array} \right )$ and $B = \left( \begin{array}{cc} 2 & 0 \\ 0 & 2\end{array} \right )$.


0

This system produces a counterexample. \begin{bmatrix} \ 1 & 0 \\0 & 1 \end{bmatrix} Determinant of this is 1 \begin{bmatrix} \ 10 & 0 \\0 & 0 \end{bmatrix} Determinant of this is zero, less than the first matrix \begin{bmatrix} \ 1 & 0 \end{bmatrix} Each of the multiplications of the matrices by this vector are ...


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

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

Take $x\in M$, $x\ne 0$. Then $\|x-Qx\|<\|x\|$. Since $\|Qx\|\le \|x\|$, this shows that $Qx\ne 0$. Hence $Q|_M$ is injective and $\dim im(Q|_M)=\dim M$. Since $Q|_M$ is a restriction of $Q$, it follows $\dim im(Q)\ge \dim im(Q|_M)$. Now $M$ was the image of $P$, and hence $\dim im(Q)\ge \dim im (P)$. Analogously you obtain the reverse direction. This ...


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

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 $$ A = 4\pi^2 ab $$


0

Hint: Use the equality case of the Cauchy-Schwartz inequality and the fact that in your case the scalar product of $x$ and $y$ is positive.



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