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Mathematics 23 Online
OpenStudy (anonymous):

Cómo determinar si V = (x,y,z) en R : z= x + y , es un subespacio vectorial de R?

OpenStudy (anonymous):

i guess it will be more helpful if you post the question in english, please.. if you can

OpenStudy (anonymous):

I think this translates to, "how can we determine if \(V=(x,y,z)\in\mathbb R,z=x+y\), where \(V\) is a subspace".

OpenStudy (anonymous):

So, what are the three properties, and how do we test for them?

OpenStudy (agreene):

"How do you determine if V = (x,y,z) in R: z=x+y is a 'subvector' of R" or in other words, if it is a vector that is contained in R.

OpenStudy (agreene):

^ a better translation.

OpenStudy (anonymous):

I"m going to sleep; be back in the morning, so cheers!

OpenStudy (turingtest):

tratare a explicarlo en espanol, aunque no puedo usar acentos: hay dos cosas para determinar si W es un subespacio vectorial de V 1) todos los vectores en W son cerrados en relacion con adicion vectorial 2) todos los vectores en W son cerrados en relacion con multiplicacion escalar entonces, escribe las formas de los vectores en W\[\vec u=\langle x,y,x+y\rangle\]\[\vec v=\langle a,b,a+b\rangle\]es \[\vec v+\vec u\]un elelmento de W? y, es\[c\vec v\]donde \(c\) es un escalar un elemento de W? (disculpa por mi espanol, aun estoy aprendiendo)

OpenStudy (anonymous):

Oh man, @TuringTest , you are the awesomest.

OpenStudy (turingtest):

thanks, but most people here speak multiple languages. I barely speak 2.

OpenStudy (anonymous):

I speak 2: English, and internet English :D

OpenStudy (anonymous):

In any case, the important part is you managed to be helpful to someone speaking in another language.

OpenStudy (turingtest):

thanks, I try :)

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