Are these vectors also a basis of the subspace? So I did a Linear Algebra test today and had to decide a basis for the subspace 
$ [(1,2,3,2),(0,1,8,5),(-2,-4,-6,-4)]\;in\;\Bbb R^4 $. The correct answer acording to the solution was $ \{(1,2,3,2),(0,1,8,5)\} $ which I agree to, but does that also mean it is the only solution? I came up with $ \{(13,-8,1,0),(8,-5,0,1)\} $ after Gaussian elimination and thought that was another way to go at it. Is my answer wrong? 
 A: Other answers have already pointed out that you made an error, but to address your question: No, there is no one solution, but a solution that just consists of some of the given vectors (like the given) are preferable in most settings.
A: Unfortunately is wrong because you can't write $(1,2,3,2)$ as a linear combination of the vectors of the basis you have given.
A: To see that your proposed basis is incorrect, let's try to write $\{13,-8,1,0\}$ in the desired subspace. We want $$a\times \{1,2,3,2\}+b\times \{0,1,8,5\}=\{13,-8,1,0\}$$  But this instantly gives $$\{a,2a+b,3a+8b,2a+5b\}=\{13,-8,1,0\}$$  and it is easy to see that this system is not solvable.
A: Your question itself is incomplete- show this is a basis for [b]what[/b] subspace?  I suspect you mean the subspace it spans.  Since the only requirements that a set if be a basis for a given space is that they span the space and are linearly independent, if we are talking about the subspace they span, then it is only necessary to determine if they are linearly independent.  
Personally, I prefer not to use "Gaussian elimination"- instead use the basic definition of "linearly independent".  Look for a, b, c, such that a(1, 2, 3, 1)+ b(0, 1, 8, 5)+ c(-2, -4, -6, -4)= (a- 2c, 2a+ b- 4c, 3a+ 8b- 5c, a+ 5b- 4c)= (0, 0, 0, 0) so a- 2c= 0, 2a+ b- 4c= 0, 3a+ 8b- 5c= 0, a+ 5b- 4c= 0.  From a- 2c= 0, a= 2c  Replacing a by 2c in the other three equations, 2(2c)+ b- 4c= b= 0, 3(2c)+ 8b- 5c= 8b+ c= 0, and since b= 0, c= 0, a= 2c= 0.  Therefore, this set of vectors [b]is[/b] linearly independent and is a basis for its span. 
