In vector space V 4 V_4 V 4 any four linearly independent vectors form a basis. Let us extend
{ ( 1 , 1 , 1 , 1 ) , ( 1 , 2 , 1 , 2 ) } \{(1, 1, 1, 1), (1, 2, 1, 2)\} {( 1 , 1 , 1 , 1 ) , ( 1 , 2 , 1 , 2 )} to { ( 1 , 1 , 1 , 1 ) , ( 1 , 2 , 1 , 2 ) , ( 0 , 0 , 1 , 0 ) , ( 0 , 0 , 0 , 1 ) } \{(1, 1, 1, 1), (1, 2, 1, 2), (0,0,1,0),(0,0,0,1)\} {( 1 , 1 , 1 , 1 ) , ( 1 , 2 , 1 , 2 ) , ( 0 , 0 , 1 , 0 ) , ( 0 , 0 , 0 , 1 )} .
Since ∣ 1 1 1 1 1 2 1 2 0 0 1 0 0 0 0 1 ∣ = ∣ 1 1 1 1 0 1 0 1 0 0 1 0 0 0 0 1 ∣ = 1 ≠ 0 , \left|\begin{array}{cccc}
1 & 1 & 1 & 1\\
1 & 2 & 1 & 2\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{array}
\right|=
\left|\begin{array}{cccc}
1 & 1 & 1 & 1\\
0 & 1 & 0 & 1\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{array}
\right|=1\ne 0, ∣ ∣ 1 1 0 0 1 2 0 0 1 1 1 0 1 2 0 1 ∣ ∣ = ∣ ∣ 1 0 0 0 1 1 0 0 1 0 1 0 1 1 0 1 ∣ ∣ = 1 = 0 , we conclude that the vectors
( 1 , 1 , 1 , 1 ) , ( 1 , 2 , 1 , 2 ) , ( 0 , 0 , 1 , 0 ) , ( 0 , 0 , 0 , 1 ) (1, 1, 1, 1), (1, 2, 1, 2), (0,0,1,0),(0,0,0,1) ( 1 , 1 , 1 , 1 ) , ( 1 , 2 , 1 , 2 ) , ( 0 , 0 , 1 , 0 ) , ( 0 , 0 , 0 , 1 ) are linearly independent, and thus form a basis of V 4 . V_4. V 4 .
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