I am trying to figure out how to identify a spanning set in R^3 from the 4 examp

Question

I am trying to figure out how to identify a spanning set in R^3 from the 4 examples below. I'm looking for the these problems to be worked out (that means going STEP BY STEP -- do not assume that I know a detail and skip over it) so that I can understand how to identify a spanning set in R^3. I am not looking for definitions or other gobbledygook regarding this. I don't want to know any "tricks" or "tips" or "shortcuts", I have to know the foundation before I can start trying to apply shortcuts. What I do want is to see them fully worked out from start to finish, step by step. Really, it's that simple. Thanks!

Explanation / Answer

Since you want to know the foundations and the step by step solution, I'll teach you the same for problem a). You can then apply this to the rest. Step 1) We know that (1,0,0)T, (0,1,0)T and (0,0,1)T form a spanning set of vectors for R^3. Hence vectors in a) will form a spanning set of vectors IF and only IF their span contains the vectors (1,0,0)T, (0,1,0)T and (0,0,1)T a_ We now check if they can span (1,0,0)T (1,0,0)T = a(1,2,1)T + b(1,3,1)T + c(1,1,1)T for some a,b,c giving us the equations a+b+c=1, 2a+3b+c=0 and a+b+c=0 these equations can never be simultaneously satisfied as a+b+c cannot be 0 and 1 simultaneously. Hence Vectors in a) do not form a spanning set of vectors in R^3. I'll check for a set of vectors which do span R^3 c) a(-2,1,0)T+b(2,-1,1)T+c(2,1,3)T = (1,0,0)T -2a+2b+2c=1 -a-b+c=0 b+3c=0 c=-1/12 b = 3/12 a = -4/12 a(-2,1,0)T+b(2,-1,1)T+c(2,1,3)T = (0,1,0)T -2a+2b+2c=0 -a-b+c=1 b+3c=0 for a=-2/6 b=-3/6 a=1/6 a(-2,1,0)T+b(2,-1,1)T+c(2,1,3)T = (0,0,1)T -2a+2b+2c=0 -a-b+c=0 b+3c=1 c=1/3, b=0 a=1/3 hence these form a spanning set of vectors for R^3

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