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I am "struggling" with a basic question, and would like someone to help me to point out what's wrong in my logic.

Let A is a subset of vectors in V(4,3): A = {(0,1,2,1), (1,0,2,2), (1,1,1,0), (1,2,0,1)}

  • Is A linearly independent? Find the 2 bases for < A >.

My approach is that, consider: a(0,1,2,1) + b(1,0,2,2) + c(1,1,1,0) + d(1,2,0,1) = (0,0,0,0) Then we have a system of 4 equations:

b + c + d = 0
a + c + 2d = 0

2a + 2b + c = 0
a + 2b + d = 0

Solve for a, b, c and d. Then we got a = b = c = d = 0, Hence A is linearly independent.

However, the solution given by my math professor is that:

Since 1(0, 1, 2, 1) + 1(1, 0, 2, 2) + 2(1, 1, 1, 0) + 0(1, 2, 0, 1) = (0, 0, 0, 0), X is linearly dependent

To me, clearly it's not, as: 1 + 2 #0, 1 + 2 # 0, 2 + 2 + 2 # 0 and 1 + 2 # 0.

Can someone help me to understand why? And also, what is an easy way to determine if a set of vectors is linearly independent or not? As, my approach, solving a system equations can be very time consuming, so not very practical in the test or exam environment.
Many thanks.

Cooper Brian
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    I suspect the notation $V(4, 3)$ means vectors in $4$ dimensions with elements in $\Bbb F_3$. If that's the case, then $1+2 \equiv 0 \pmod 3$. – Robert Shore Sep 08 '23 at 00:15
  • And $(0,1,2,1)+(1,0,2,2)=(1,1,1,0)$. – Gerry Myerson Sep 08 '23 at 00:29
  • OMG, thank you so much Robert and Gerry. How silly I am! Now just ignore the solution, I am still not convinced that, because I solved the system of 4 equations and got a = b = c = d = 0. By the result, it must be independent. But why the correct answer is linear dependent? – Cooper Brian Sep 08 '23 at 01:10
  • When you solved it, did you ever divide something by 3? You can't do that in $\mathbb{F}_3$. – ConMan Sep 08 '23 at 04:20
  • Hi ConMan. No I didn't divide anything by 3 or so. I treated them as general algebra math. Only used substitution and adding/subtracting rows. Tried many times, even used calculator, but always got a = b = c = d = 0. Hence the confusion. – Cooper Brian Sep 08 '23 at 04:46

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