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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 ...


5

The inner product does commute, but scalar vector multiplication doesn't work with it. $(v\cdot u)u$ is in the direction of (parallel to) $u$ and $(u\cdot u)v$ is in the direction of $v$, thus if $u\nparallel v$, the vectors point in entirely different directions (let alone being equal). In fact, the equality holds if and only if there is some ...


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 ...


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.


3

This is true. Assume that $f$ is not zero at a point $x_0$. By continuity you can find a small ball $B_r(x_0)$ such that $|f(x)| \geq \delta > 0$ for all $x \in B_r(x_0) \cap [0,1]$. But this means that the norm must be greater than $\delta \cdot | B_r(x_0) \cap [0,1]|$ (where $| B_r(x_0) \cap [0,1]|$ is the size of the interval $B_r(x_0) \cap [0,1]$), ...


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)})} = ...


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 ...


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


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 ...


2

As pointed out by felipeh, if $0\leqslant X_{n+ 1}(\omega)\leqslant X_n(\omega)$ for each $n$ and each $\omega$, and $\mathbb E[X_n]\to 0$, then $X_n\to 0$ a.s. (the almost convergence for the subsequence actually holds for the whole sequence). An other case of interest is when the series $\sum_n\mathbb E|X_n-X|$ is convergent. In this case, an ...


2

If $x = \sum_i a_i v_i$ where $\{v_i\}$ form an orthonormal basis, $||x|| = \sqrt{\sum_i |a_i|^2}$. This is just Pythagorean theorem.


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 $[0,1]$).


2

If $u = (1,0,0)$ and $v = (1,1,1)$, then $v\cdot u = 1$ and $u\cdot u = 1$. So $(v\cdot u)u = u$, but $(u\cdot u)v = v \neq u$. If either $u$ or $v$ is the zero vector, equality holds. So assume neither $u$ nor $v$ is the zero vector. If equality holds, then $v = Au$, where $A = (v\cdot u)/(u\cdot u) \in \Bbb R$. Conversely, if $v = tu$ for some $t\in \Bbb ...


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

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

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 ...


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 $$ \lvert x \rvert_p^p = |x_1|^p + \ldots + |x_k|^p.$$


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

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 ...


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 $$ \|A\|=\frac nm \max_j \sum_{i=1}^m |a_{ij}| $$


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

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 $x_n$, $y_n$, $a$, ...


1

It suffices to show that at least one $x$ with $\|x\|_{\mathbb{R}^n}=1$ achieves the specific bound. Let $$i^*=arg\max_{i=1,2,\ldots,m}\sum_{j=1}^n{|a_{ij}|}$$ i.e. $i^*$ is the index of the row of $A$ with the largest sum of absolute values of its elements. Then, if we select $\xi_j=sign(a_{i^*j}$) we have that $\|x\|_{\mathbb{R}^n}=1$ and ...


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

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

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

Your definition is missing an absolute value: it should be $\|f\|_p = (E(|f|^p))^{1/p}$. Of course, you can drop the absolute values if $f$ is real and nonnegative. The intuition for the $p$ norm is that it is a generalization of the familiar norms $\|f\|_1 = E(|f|)$ and $\|f\|_2 = (E(|f|^2))^{1/2}$. Larger values of $p$ assign more weight to points where ...



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