Linear algebra- what is a vector space?

preluderacer
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I looked up what is a vector space online and it always gives like formula or long explanations. In a couple sentences can you tell me exactly what a vector space is? I know it has to go through the origin, but what else is true about a vector space?
 
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preluderacer said:
I looked up what is a vector space online and it always gives like formula or long explanations. In a couple sentences can you tell me exactly what a vector space is? I know it has to go through the origin, but what else is true about a vector space?

I suggest you to look at http://mathworld.wolfram.com/VectorSpace.html.
I don't know what you mean by "it has to go through the origin". A vector space must contain an element (called vector) that is neutral under addition. In other words, there must exist a vector say \vec 0 such that for any vector \vec x in the vector space, the following hold: \vec x + \vec 0 = \vec x.
 
fluidistic said:
I suggest you to look at http://mathworld.wolfram.com/VectorSpace.html.
I don't know what you mean by "it has to go through the origin". A vector space must contain an element (called vector) that is neutral under addition. In other words, there must exist a vector say \vec 0 such that for any vector \vec x in the vector space, the following hold: \vec x + \vec 0 = \vec x.

It just means the 0 vector must be in that set (from memory...)
 
If the following axioms are true for all objects u, v, and w in V and all scalars c and k then V is called a vector space and the objects in V are called vectors.

(a) u + v is in V
(b) cu is in V
(c) u + v = v + u
(d) u + (v + w) = (u + v) + w
(e) 0 in V, such that for all u in V we have u + 0 = 0 + u = u.
(f) For every u in V there is -u such that u + (-u) = 0.
(g) c(u + v) = cu + cv
(h) (c + k)u = cu + ku
(i) c(ku) = (ck)u
(j) 1u = u
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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