Tag Info

Hot answers tagged

10

Here's the way I like to think about it. I'll start with the finite dimensional space $\Bbb{R}^n$ because it looks like that's where you are, but I'll give an analogy for infinite dimensional spaces as well. The quantity $z^Tx$ represents a linear functional on $\Bbb{R}^n$, that is a linear function which eats a vector and spits out a real number: $$ ...


4

The dual space is a space of linear functionals. If we want to define a norm on the dual space, we do what we always do to measure the "size" of a linear transformation: we use an operator norm. Alternatively, the dual norm of $z$ is the matrix norm of the matrix $z^T$.


3

0) $\|Ax\|\leq \|A\|\|x\|$ because $\|A\|$ is essentially defined to be the smallest number $C$ for which $\|Ax\|\leq C\|x\|$ holds. For $x\neq0$ this directly follows from the usual definition and the definition of supremum. Just divide by $\|x\|$ and take the supremums on both sides. 1) The induced norm is always a matrix norm, the axioms follow from the ...


2

Any induced norm satisfies $\|A\|\ge \rho(A)$, where $\rho(A)$ is the spectral radius of $A$, i.e. the largest absolute value of its eigenvalues. This fact is stated without proof in Wikipedia; here is a proof. Any induced norm is of the form $$\|A\|=\max_x\frac{\|Ax\|}{\|x\|}.$$ Let $x^*$ be an eigenvector corresponding to the eigenvalue $\lambda^*$ ...


2

Dual norm is a particular case of the support function, specifically it is the support function of the unit ball of the original norm. When the unit ball is smooth enough $\|z\|_*$ is the Euclidean distance from the origin to the hyperplane with the normal vector $z$ (of unit Euclidean length) tangent to the ball. The equation of this hyperplane is ...


2

Let's be precise about what "proving $\geq$" entails. What you need to show is that for any $\epsilon > 0$, there exists a matrix norm $\|\cdot\|$ such that $$ \|A\| < \rho(A) + \epsilon $$ The trick then, is to tailor a suitable example $\|\cdot \|$ to your matrix $A$ and choice of $\epsilon$. I suggest the following: given the vector norm $|\cdot|$ ...


2

A common use is in topologizing certain function spaces. For example, let $\Omega \subset \mathbb{R}$ be open. We want a topology on $C^\infty(\Omega)$ the collection of smooth functions. Intuitively we'd like convergence $f_n \to f$ to imply the convergence locally of all derivatives. So what we do is we use a family of semi-norms to do it. $\|f\|_{K,n}$ is ...


2

$P=I - \frac{v v^T}{v^T v}$ is the orthogonal projection onto $v^\perp$. Proof: Clearly, $P$ is symmetric. $P^2 = (I - \frac{v v^T}{v^T v}) (I - \frac{v v^T}{v^T v}) = I - 2 \frac{v v^T}{v^T v} + \frac{v v^T}{v^T v} \frac{v v^T}{v^T v} = I - 2 \frac{v v^T}{v^T v} + \frac{v (v^T v) v^T}{(v^T v)^2} = I - 2 \frac{v v^T}{v^T v} + (v^T v) \frac{v v^T}{(v^T ...


2

Thanks to Daniel Fischer's comment, this lemma is actually really trivial to prove. Let's generalize! Given any metric space $(M, d)$, we define the closed ball of radius $r > 0$ centred at $c \in M$ to be: $$ B_r[c] = \{x \in M \mid d(x, c) \leq r\} $$ Using this new notation, we now want prove the following lemma: Generalized Lemma: If $x,y \in ...


2

Hint:$|f(x)g(x)|=|f(x)|\,|g(x)|\leq\|f\|_{\infty} |g(x)|,\forall x\in X\,\,\,$


2

Let $(e_i)_{i\in I}$ be an an orthonormal basis, and $x =\sum_{i\in I} y_i e_i$ with norm $1$, i.e. $\sum_{i\in I} |y_i|^2 = 1$, then $$\|Tx\| \leq \sum_{i\in I}|y_i|\|Te_i\| \leq \sqrt{\sum_{i\in I} |y_i|^2}\sqrt{\sum_{i\in I}\|Te_i\|^2} = \|T\|_{HS}$$


2

Hint: If you are given a proof for those estimates, look at them. The proofs contain some other inequalities that you may be more familiar with. When does equality hold in them? A different kind of hint: What special kind of vectors can you come up with in $\mathbb R^n$? What are some simple nonzero vectors? The first inequality is not optimal and thus ...


1

This is not true. The $H^r$ norm controls the supremum of $f$ if and only if $2r>n$ where $n$ is the dimension of the space ($n=1$ in your case), by the Sobolev-Morrey embedding. In the borderline case $r=1/(2n)$ the embedding fails. This is typical of embeddings. An example is easier to give in the periodic setting, on $\mathbb R/(2\pi \mathbb Z)$. Let ...


1

Suppose the circle is centered at $(0,0)$ $$x_1^2+y_1^2\leq r^2$$ $$x_2^2+y_2^2\leq r^2$$ $$(x_1-x_2)^2+(y_1-y_2)^2 \leq x_1^2+x_2^2+y_1^2+y_2^2+2|x_1x_2+y_1y_2|$$ $$=x_1^2+x_2^2+y_1^2+y_2^2+2|(x_1,y_1)\cdot(x_2,y_2)|\leq 2r^2+2||(x_1,y_1)||||x_2,y_2||=4r^2$$


1

No, you apply the norm (in our case $\| \cdot\|_\infty$) to the vector difference to get the distance: $$\|a - b\|_\infty = \|(1-4,-2-2)\|_\infty = \max \{|-3|,|-4| \} = 4$$


1

I'm following up this slightly on Michaels comment that you perhaps should look at the numbers before you rule out anything or start spending days on learning and coding in some other language. To begin with, on my machine, Gurobi takes roughly 0.3-0.4 seconds to solve a problem of this size. Hence, we are looking at some 30 hours of core computations. That ...


1

You can use this nice result for the differential of the trace $$ \eqalign { d\,\mathrm{tr}(f(A)) &= f'(A^T):dA \cr } $$ to write $$ \eqalign { d\,\mathrm{tr}((x^Tx)^{\frac {1} {2}}) &= \frac {1} {2} (x^Tx)^{-\frac {1} {2}}:d(x^Tx) \cr &= \frac {1} {2} (x^Tx)^{-\frac {1} {2}}:(dx^T x + x^T dx) \cr &= x(x^Tx)^{-\frac {1} {2}}: dx ...


1

Perhaps an example will help? Consider the "taxicab metric" on $\mathbb{R}^2$ which is defined by $$||x||_1 = |x_1|+|x_2| \,\, \text{where $x = (x_1,x_2)$} $$ The self-isomorphisms of $\mathbb{R}^2$ with respect to this metric are exactly the translations, and compositions of translations with one of the eight norm-preserving isomorphisms, each of which is ...



Only top voted, non community-wiki answers of a minimum length are eligible