kerf--prophetes.ai

kerf

▪ I. kerf (kɜːf) Forms: α. 1 cyrf, 4 kyrf, 5 kirf, 6 kyrfe. β. 4–7 kerfe, (4–5 -ff(e), 4– kerf, (9 dial. curf, kurf). γ. See carf. [OE. cyrf, app.:—*kurƀi-, f. *kurƀ- (cf. ON. kurf-r chip, kyrfa to cut), ablaut-form of *kerƀ, stem of OE. ceorfan to carve. Hence ME. kirf, giving later kerf and karf; ... Oxford English Dictionary

prophetes.ai

Kerf (disambiguation)

Kerf is the width of a saw cut. Kerf may also refer to: Kerf, molten metal and metal oxide blown out when metal is cut by an oxy-gas torch: see Oxy-fuel welding and cutting Kerf (sculpture wikipedia.org

en.wikipedia.org

Show there exist p such as $Kerf \varsubsetneqq Kerf^2\varsubsetneqq ...\varsubsetneqq Kerf^p=Kerf^{p+1}$ Let E a finite dimensional vector space ($dimE = n \in \mathbb{N^*}$) and $f\in \mathcal{L}(E)$ How to show th...

(I assume that, besides, you know how to prove that for any $k$, $Kerf^k \subset Kerf^{k+1}$ ).

prophetes.ai

Prove or disprove if $f:G\to K$ is group homomorphism then necessarily $|kerf|\le |imf|$ I wrote a counterexample: $$G=(\Bbb Z_5,+_{mod 5})=\\{0,1,2,3,4\\}$$ $$K=(\Bbb Z_3,+_{mod 3})=\\{0,1,2\\}$$ $$e_k=0,f(a)=3a$$ so...

Yeah, your counterexample looks good. In general if you have groups $G$ and $H$ you can consider the morphism $\varphi: G\rightarrow H$ defined by $\varphi(g)=e_H$ for all $g\in G$. The image is $\\{1_H\\}$ and the kernel is $G$. So this is a valid counterexample whenever $G$ is not trivial.

prophetes.ai

Finding some non zero endomoprhism $f$ satisfying $Kerf = Imf$ > My example: Find some non zero endomoprhism $f$ such that $Kerf = Imf$. * Maybe this question has been already answered somewhere on this page, but I...

For example: $f: \mathbb R^2 \to \mathbb R^2$, $(a, b) \mapsto (b, 0)$. $\ker f = \mathrm{Im}\, f = \\{(a, 0) | a \in \mathbb R\\}$. Or did I misunderstand the question?

prophetes.ai

Yes/No : Is the quotient $\ker(g) / \ker(f)$ is zero-dimensional? > let $V$ be a real finite dimensional vector space and $f,g$ are nonzero linear functional on $V$ real vector space,Assume that $Ker(f)\subset Ker(g)$...

The kernel of a non-zero linear form on an $n$-dimensional vector space is of dimension $n-1$ by the rank theorem. Hence both your kernels have the same dimension. As one kernel is included in the other they are equal. So the quotient zero.

prophetes.ai

Please explain how to kerf birch!! special birch kerfing trick revealed!

going to be curving for today is a single 10-inch si enclosure and enjoy guys now here's my actual piece now we've got the depth all set I've got my kerf drawn out make another video here shortly to show you guys how to figure out the math for that but for now enjoy yourself so we've got our first set of kerf

prophetes.ai

A basis for $\{(x-y,y-z,z-x) : (x,y,z)\in \mathbb R^3\}$ The original question : Prove that $F: \mathbb R^3 \to \mathbb R^3$ such that $F(x,y,z)=(x-y,y-z,z-x)$ is a linear-map and find a basis for $imF$ and $kerF$. ...

\begin{align} im(F)&=\\{(x-y,y-z,z-x) \mid x,y,z\in\mathbb{R}\\}\\\ &=\\{x(1,0,-1)+y(-1,1,0)+z(0,-1,1) \mid x,y,z\in\mathbb{R}\\}\\\ &=span\\{(1,0,-1),(-1,1,0),(0,-1,1)\\}\\\ &=span\\{(1,0,-1),(-1,1,0)\\}\\\ \end{align} Show that $\\{(1,0,-1),(-1,1,0)\\}$ is linearly independent and you've got yours...

prophetes.ai

Homomorphism property I have a problem with understanding one of homomorphism properties. Given $f$ - is a homomorphism, $g \in G$ and $h = f(g)$, then $f^{-1}(h) = g \cdot Kerf$. Can anyone please prove it?

First you need to know that since $f$ is not necessarily bijective. $f^{-1}(h)$ is not an element, but is the pre-image of $h$ which is a set. So to prove equality of set we need to show that they are subset of each other. First, let $x\in f^{-1}(h)$, which means $f(x)\in \\{h\\}$. Therefore, $f(x)=...

prophetes.ai

Image equal to kernel Define the general formula of linear transformation $f:\mathbb{R}^{4}\rightarrow \mathbb{R}^{4}$ such that $Imf=Kerf$. Any ideas?

**Hint** : Use the well known rank-nullity theorem $\text{dim(Ker(}f\text{))+dim(Im(}f\text{))=dim(}\mathbb{R}^4)$ to establish that (since the kernel is equal to the image) the dimension of both the kernel and the image is $2$. Can you pick it up from here?

prophetes.ai

Endomorphism of a rings I was trying to prove that a surjective Endomorphism $f:A \to A$ of a noetherian ring is also injective. I would like to know why this argument is not correct? $A/\rm{Ker}f \cong \rm{Im}f=A \Ri...

Your argument cannot be correct, since it doesn't use the noetherian hypothesis; this is because for a non-noetherian ring the statement is wrong: Consider the polynomial ring $A = \mathbb Z[x_1,x_2,x_3,\dotsc]$ in infinitely many variables. It is clearly not noetherian. Now, consider the ring homom...

prophetes.ai

Quotient group of $C^$ by Circle group is isomorphic to $R^+$ $C^$ is the set of complex numbers except $0$ and $S'$ is the circle group. I have to show the quotient group $C^*/S'$ is isomorphic to $R^+$. Let's defi...

$\ker f=\\{z\in\mathbb{C}^*\,|\,\lvert z\rvert=1\\}=S^1$.

prophetes.ai

Prove that if A is simple every R-module endomorphism is either the zero map or an isomorphism. How can I prove that: "Prove that if R has an identity and A is a nonzero unitary R-module and if A is simple every R-mod...

It's right, but it can be made a bit simpler. First, a simple module $A$ is nonzero. If $f\colon A\to A$ is a homomorphism and $x\in A$, $x\ne0$, then there are two cases: 1. $x\in \ker f$ 2. $x\notin\ker f$ In case 1, $\ker f\ne\\{0\\}$, so $\ker f=A$ and $f$ is the zero map. In case 2, $\ker f\ne ...

prophetes.ai

How do we get the canonical cokernel-kernel decomposition in a pre-abelian category? In a pre-abelian category, every morphism $f: A \to B$ has a canonical decomposition: $$ A \to coker(kerf) \to ker(cokerf) \to B $$...

We have the commutative diagram !enter image description here Then, by universal property of the limit $\ker(\mathrm{coker}\, f)$, we have arrows such that !enter image description here Then, by universal property of the colimit $\mathrm{coker}(\ker f)$, we have an arrow !enter image description her...

prophetes.ai

Determine a linear application How can I determine a linear application $f: \mathbb{R}^2 \to \mathbb{R}^2 $ such that $Kerf = span[ (-1, -5) $? By the definition of Ker, we know that $ kerf = [ (x,y) \in \mathbb{R}^2...

Hint: try to find a linear map $\Bbb R^2\to\Bbb R$ with that kernel. Then to make an example $\Bbb R^2\to\Bbb R^2$, you can compose with a (almost any) linear map $\Bbb R\to\Bbb R^2$ so that the composition has the same kernel.

prophetes.ai

kerf

Answers

MindMap

Sources

kerf

Kerf (disambiguation)

Show there exist p such as $Kerf \varsubsetneqq Kerf^2\varsubsetneqq ...\varsubsetneqq Kerf^p=Kerf^{p+1}$ Let E a finite dimensional vector space ($dimE = n \in \mathbb{N^*}$) and $f\in \mathcal{L}(E)$ How to show th...

Prove or disprove if $f:G\to K$ is group homomorphism then necessarily $|kerf|\le |imf|$ I wrote a counterexample: $$G=(\Bbb Z_5,+_{mod 5})=\\{0,1,2,3,4\\}$$ $$K=(\Bbb Z_3,+_{mod 3})=\\{0,1,2\\}$$ $$e_k=0,f(a)=3a$$ so...

Finding some non zero endomoprhism $f$ satisfying $Kerf = Imf$ > My example: Find some non zero endomoprhism $f$ such that $Kerf = Imf$. * Maybe this question has been already answered somewhere on this page, but I...

Yes/No : Is the quotient $\ker(g) / \ker(f)$ is zero-dimensional? > let $V$ be a real finite dimensional vector space and $f,g$ are nonzero linear functional on $V$ real vector space,Assume that $Ker(f)\subset Ker(g)$...

Please explain how to kerf birch!! special birch kerfing trick revealed!

A basis for $\{(x-y,y-z,z-x) : (x,y,z)\in \mathbb R^3\}$ The original question : Prove that $F: \mathbb R^3 \to \mathbb R^3$ such that $F(x,y,z)=(x-y,y-z,z-x)$ is a linear-map and find a basis for $imF$ and $kerF$. ...

Homomorphism property I have a problem with understanding one of homomorphism properties. Given $f$ - is a homomorphism, $g \in G$ and $h = f(g)$, then $f^{-1}(h) = g \cdot Kerf$. Can anyone please prove it?

Image equal to kernel Define the general formula of linear transformation $f:\mathbb{R}^{4}\rightarrow \mathbb{R}^{4}$ such that $Imf=Kerf$. Any ideas?

Endomorphism of a rings I was trying to prove that a surjective Endomorphism $f:A \to A$ of a noetherian ring is also injective. I would like to know why this argument is not correct? $A/\rm{Ker}f \cong \rm{Im}f=A \Ri...

Quotient group of $C^$ by Circle group is isomorphic to $R^+$ $C^$ is the set of complex numbers except $0$ and $S'$ is the circle group. I have to show the quotient group $C^*/S'$ is isomorphic to $R^+$. Let's defi...

Prove that if A is simple every R-module endomorphism is either the zero map or an isomorphism. How can I prove that: "Prove that if R has an identity and A is a nonzero unitary R-module and if A is simple every R-mod...

How do we get the canonical cokernel-kernel decomposition in a pre-abelian category? In a pre-abelian category, every morphism $f: A \to B$ has a canonical decomposition: $$ A \to coker(kerf) \to ker(cokerf) \to B $$...

Determine a linear application How can I determine a linear application $f: \mathbb{R}^2 \to \mathbb{R}^2 $ such that $Kerf = span[ (-1, -5) $? By the definition of Ker, we know that $ kerf = [ (x,y) \in \mathbb{R}^2...