Showing Linear Subspace & Affine Subspace of Vector a in R^n

tk1234
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Let a be a fixed nonzero vector in R^n

a. Show that the set S of all vectors x such at ax=0 is a subspace of R^n.
b. show that if k is a nonzero real number, then the set A of all vectors x such that ax=k is an affine subspace of R^n, but not a linear subspace.how do i even show this.. I am so confused. help please! thanks.
 
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tk1234 said:
Let a be a fixed nonzero vector in R^n

a. Show that the set S of all vectors x such at ax=0 is a subspace of R^n.
b. show that if k is a nonzero real number, then the set A of all vectors x such that ax=k is an affine subspace of R^n, but not a linear subspace.


how do i even show this.. I am so confused. help please! thanks.

Try?? Pretty please? Tell me what you might need to prove S is a subspace. You can look it up in the book if you want to. What's the definition of a subspace?
 
i know that S needs to be closed under addition and numerical multiplication.. I am just not sure how to show this?!
 
tk1234 said:
i know that S needs to be closed under addition and numerical multiplication.. I am just not sure how to show this?!

If x1 is in S then a.x1=0. If x2 is in S then a.x2=0. (I'm using '.' for dot product). Can you give me some reason why that would mean x1+x2 is in S? What would you have show to prove x1+x2 is in S? That would give you closure under addition.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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