# How to create these Effects in photoshop? [closed]

Edit > Transform > Skew

In photoshop one can simply draw the box (=single layer) and add the texts.Insert the model as a layer, make it a little gray by reducing the contrast:

Create a new layer.

Edit > Transform > Skew

In photoshop one can simply draw the box (=single layer) and add the texts.Insert the model as a layer, make it a little gray by reducing the contrast:

Create a new layer.

I am curious how to create this folded/bent icon in Inkscape?Fairly simple, like @joojaa said: Cut the image into 3 pieces and skew each.

This question already has an answer here:

Tapered/sloped text effect in Illustrator (NOT skew or perspective)

2 answers

Create path you want the text like.write text you want.

How can I create this sort of parallelogram in Illustrator?As others have sorta tried to say you can use the Shear tool (Skew is in Photoshop).

Is the set of real $2n \times 2n$ skew-symmetric matrices having positive Pfaffians path connected?By definition, the Pfaffian is a polynomial in the entries $a_{ij}$ ($i<j$) such that $Pf(A)^2=\det A$, and $Pf(J_n)=1$, where

$$J_n=\begin{pmatrix} 0_n & I_n \\ -I_n & 0n \end{pmatrix}.

The elements of such a ring are called skew polynomials or Ore polynomials.For Ore polynomials the usual polynomial addition holds.

Is there an implemented algorithm that will take a polynomial $f(x,y)\in\mathbb Z[x,y]$ and a prime $p$, and determine whether the equation $f(x,y)=0$ has a solution over $\mathbb Q_p$?My recollection is that Nils Bruin did this is in fair generality in Magma.

I am trying to find the original reference for a lemma attributed to Cohn (as in Schur-Cohn method):

Let $A(z)$ be a palindromic or skew-palindromic polynomial, and denote its derivative by $A'(z)$.Then $A(z)$ and $A'(z)$ have the same number of zeros outside the unit circle.

I already have the candidates for the generators (certain polynomials of $16$ variables) but I need to find the relations between them.My question is the following - is it possible to force Magma, Sage or other program to find these polynomial relations (hopefully the whole ideal of relations.

This is proved in Prop 1.7.

Hi everyone,

let $D$ be a skew field, which is finite dimensional over its center $k$.Assume that $k$ is a number field, and let $\mathcal{O}_D$ be the set of elements $z\in D$ which are roots of a monic polynomial with coefficients in $\mathcal{O}_k$.

Moreover, are there any standard way to compute $gl\dim(S)$ when $S$ is non-commutative?Then let $\sigma_1$ be the $k$-algebra automorphism of $R_1$ defined by $\sigma_1(x_1) = a_{21} x_1$, and let $R_2$ be the skew-polynomial ring

$$

R_2 = R_1[x_2; \sigma_1].

From Kaplansky's PI-theorem it then follows that $K$ cannot satisfy a polynomial identity (the theorem says that primitive PI-algebras have finite dimension over their center).There, a GPI (generalized polynomial identity) has coefficients from the center $F$.

The pfaffian can be defined as $\sqrt{{\rm det}(A) } $ when $A$ is skew symmetric, or explicitly $${\rm pf}(A) = \frac{1}{2^n n!Let $F \in k[x_0, \ldots, x_n ]$ be a homogeneous polynomial of degree $d$.

When I was a very little hare, a big grey wolf told me about the following skew polynomial algebra, which I never understood.My question is whether the following construction is a part of some bigger abstract construction and whether it is been written anywhere.

Let $k$ be a field and $q\in k -0$.Define $S_q := k\langle x, y\rangle/(xy - qyx)$, where $k\langle x,y\rangle$ is the non-commutative polynomial ring.

In using MAGMA, I always get type errors.For example, I have made an empty set $e := []$, how would I make MAGMA print out the type of $e$?

The Pfaffian $\text{pf}$ is defined for a skew-symmetric matrix which is also a polynomial of matrix coefficients.One property for Pfaffian is that $\operatorname {pf} (A)^{2}=\det(A)$ holds for every skew-symmetric matrix A.

Imagine I have two skew-symmetric square matrices $A$, $B$.) Now I am interested in the square root of the determinant of $AB+I$, where $I$ is the identity matrix,

$$ x = \sqrt{ \det \left( AB + I \right) } $$

As quick inspection for small matrices suggests that this $x$ is a polynomial of the elements of $A$ and $B$, for example for $3 \times 3$ matrices we find

$$ x = 1 - a_{12} b_{12} - a_{13} b_{13} - a_{23} b_{23} $$

and I checked this analytically for matrices up to $6 \times 6$.

Since Ore extensions of domains are Ore extensions, this question can be reformulated as follows: Since the relation look almost like those of a skew polynomial ring, can this ring in fact be written as a skew polynomial ring?Consider the $k$-algebra $R=k\langle x_i, c_i\rangle_{i=1,\ldots,n}$, $k$ not of characteristic 2,

subject to the following relations:

$x_ix_j = x_jx_i$ for all $i, j$

$x_ic_i = -c_i x_i$

$x_ic_j = c_j x_i$ for all $i\neq j$

$c_ic_j = -c_j c_i$ for all $i\neq j$

I haven't found a counterexample yet.