Recent content by Portuga
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Equivalence between Bravais lattice definitions
Proof. To demonstrate that a Bravais lattice (Figure 1) can be considered as a set of points located by a linear combination of primitive vectors of the lattice with integer coefficients, a sequence of claims in increasing order of complexity can be adopted. First, what is shown for one octant...- Portuga
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- bravais lattice
- Replies: 0
- Forum: Advanced Physics Homework Help
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Incompatibility between ideal gas equations of state
Thank u very much!- Portuga
- Post #3
- Forum: Introductory Physics Homework Help
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Incompatibility between ideal gas equations of state
To solve this problem I used two equations: $$ PV=nRT, $$ where ##P## is the pressure, ##V##the volume, ##R##the gas constant, ##T##for temperature and is##n##the number of moles, related to the mass ##m## and molar mass ##M## by $$ n=\frac{m}{M}. $$ It will be also necessary consider the...- Portuga
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- Gas Ideal gas State
- Replies: 2
- Forum: Introductory Physics Homework Help
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I Parallel and perpendicular components
I followed the trick of Gram Schmidt ortogonalization method to build an ortogonal basis ##\left\{ u^{\prime},w^{\prime}\right\}## for ##W## and the strategy of decomposing ##v## as ##v_{\parallel}=v\ldotp u^{\prime}u^{\prime}+v\ldotp w^{\prime}w^{\prime}## and ##v_{\perp}=v-v_{\parallel}##. It...- Portuga
- Post #8
- Forum: Linear and Abstract Algebra
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I Parallel and perpendicular components
ok, thank you all very much!- Portuga
- Post #6
- Forum: Linear and Abstract Algebra
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I Parallel and perpendicular components
I should subtract first component, normalize and then subtract the other one, and normalize it?- Portuga
- Post #3
- Forum: Linear and Abstract Algebra
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I Parallel and perpendicular components
I first tried to use a method based on Gram Schmidt orthogonalization method: $$ v_{\parallel}=\left(v\ldotp\frac{u}{\left\Vert u\right\Vert }\right)\frac{u}{\left\Vert u\right\Vert }+\left(v\ldotp\frac{w}{\left\Vert w\right\Vert }\right)\frac{w}{\left\Vert w\right\Vert }, $$ and $$...- Portuga
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- Components Parallel Perpendicular
- Replies: 7
- Forum: Linear and Abstract Algebra
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Linear algebra problem with a probable typo
Yes, now I got the point.- Portuga
- Post #6
- Forum: Calculus and Beyond Homework Help
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Linear algebra problem with a probable typo
Oh, thanks!- Portuga
- Post #4
- Forum: Calculus and Beyond Homework Help
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Linear algebra problem with a probable typo
Well, my guess is that there is something wrong with the factors chosen, because ##\left\Vert \left(0,1,0\right)\right\Vert =1## and \begin{align} \left\Vert F\left(0,1,0\right)\right\Vert &=\left\Vert...- Portuga
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- Algebra Linear Linear algebra
- Replies: 5
- Forum: Calculus and Beyond Homework Help
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I Proof that two linear forms kernels are equal
Thank you very much! Sorry for the mess in latex.- Portuga
- Post #3
- Forum: Linear and Abstract Algebra
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I Proof that two linear forms kernels are equal
Attempt of a solution. By the Rank–nullity theorem, $$ \dim V=\dim Im_{F}+\dim\ker\left(F\right) \Rightarrow n=1+\dim\ker\left(F\right) \Rightarrow \dim\ker\left(F\right)=n-1. $$ Similarly, it follows that $$\dim\ker\left(G\right)=n-1.$$ This first part, for obvious reasons, is very clear. The...- Portuga
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- Forms Linear Proof
- Replies: 4
- Forum: Linear and Abstract Algebra
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A How Does the Non-Equality of Kernels Imply Their Sum Equals the Vector Space?
Oh my god! It was so simple! Thank you very much!- Portuga
- Post #3
- Forum: Linear and Abstract Algebra
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A How Does the Non-Equality of Kernels Imply Their Sum Equals the Vector Space?
This is actually a solved exercise from a Brazilian book on Linear Algebra. The author presented the following solution: The kernel and image theorem tells us that dimension ##\dim V=n=\dim\ker\left(F\right)+\dim \text{im}\left(F\right)=\dim\ker\left(G\right)+\dim\text{im}\left(G\right)##. As...- Portuga
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- Kernel
- Replies: 2
- Forum: Linear and Abstract Algebra
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Proving the existence of a real exponential function
Ok, thank you very much.- Portuga
- Post #6
- Forum: Calculus and Beyond Homework Help