Linear Algebra: Vector space axioms

preet
Messages
96
Reaction score
0

Homework Statement


One of the fundamental axioms that must hold true for a set of elements to be considered a vector space is as follows:
1*x = x

I was given a particular space: The set of all polynomials of degree greater than or equal to three, and zero, and asked to evaluate whether or not it was a vector space or not. The one that doesn't hold true is 1*x=x (according to the solutions), but I don't understand why. I can't find a situation where the above axiom holds true. Could anyone help me out?

Thanks
Preet
 
Physics news on Phys.org
I'd check the solutions again. The problem is closure under vector addition.
 
The set of vectors and the vector addition must form an abelian group. That is true for those polynomials. However, your polynomial space lacks the "null/zero vector"; since we know that the scalar zero times any vector from your set should give the zero vector. Since the zero vector is not an element of the space, it follows that the polynomials' space plus vector addition plus scalar multiplication do not yield a vector space.

Daniel.
 
Daniel, the original post says that zero is an element of the set. The problem is indeed vector addition, since x^3 + (-x^3 + 1) = 1, for exampl.
 
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...
Back
Top