Subspace Question: Determine if H is a Subspace in Vector Space

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Homework Statement



Determine if the sets are a subspace of the real vector space:

Prof is kinda hard to hear and doesn't explain stuff that well, can I get some help with this one?


Homework Equations



H = {[a,b,c,d] exist in 4-space| 4a+2b-8c+2d = 3a-5b+6d = b-6c-2d = 0}
H = {[a,b,c] exist in 3-space| c = 5b-4a; (a^2) = bc}


The Attempt at a Solution


Not sure whether linear dependence/indepence pertains to whether the sets are a subspace or not, mainly just tried getting the matrices into R.R.E.F.
 
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Linear independence and linear dependence don't enter into this at all. You have two different sets (it would be good to give them different names).

There is a definition that says what it means for a subset of a vector space to be a subspace of the vector space. Do you know this definition?
 
I know the ones about having the zero vector, any scalar multiple of a vector must be included in the space and if two vectors, the sum of the two must be in the space, but other than that nothing.
 
That's all you need. For each of your two problems,

Verify that 0 is a vector in the set.
For any arbitray vectors u and v in set H, verify that u + v is also in H.
For any scalar c and any arbitray vector u, verify that cu is also in H.

You will need to use the definitions of the sets in your problems. For example, in your second problem, you can write a vector u as (x, y, z), and the coordinates have to satisfy z = 5y - 4x and x2 = yz. Similar conditions would apply in your first problem.
 
Alright, heh. Simple enough. One more thing...there's one question that gives a matrix A with vectors v1 = [1,2,-3,1,-1] v2 = [-1,-1,4,0,2] and v3 = [1,3,-2,2,0]. It asks to give a matrix B such that the column space of A is the same as the null space of B. So what I was getting out of this is that you need to determine the column space of A for some vector v = [a,b,c,d,e] in order to determine what to match up with the null space of B, but I'm kind of getting lost in the midst of the differences between the two.
 
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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