Question 1095299: Demonstrate that vectors (1,0,0),(0,1,0),(0,0,1),(0,1,1) are not an efficient spanning set by showing that an arbitrary vector in R3 can be expressed in more than one way as a linear combination of these vectors.
I have (a,b,c)=a(1,0,0)+b(0,1,0)+c(0,0,1)+0(0,1,1)
I need one more example with different scalers.
Answer by ikleyn(52778)   (Show Source):
You can put this solution on YOUR website! .
Let me make writing more understandable for this problem.
Let i = (1,0,0) be the first given vector,
j = (0,1,0) be the second given vector,
and k = (0,0,1) be the third given vector.
We also have m = (0,1,1) as the fourth given vector.
Let c = a*i + b*j + c*k be an arbitrary vector of your 3D space.
Then it has TWO different presentations.
One is this c = a*i + b*j + c*k. (1)
The second is this: c = a*i + b*m + (c-b)*k. (2)
(Check it by considering its by-component parts.)
These two presentations of the same vector give you the example you are looking for.
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