We can set up a matrix and use Gaussian elimination to figure out the dimension of the space they span.
( 1 2 3 0 1 2 0 0 1 ) \begin{pmatrix}
1 & 2 & 3 \\
0 & 1 & 2 \\
0 & 0 & 1 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 2 1 0 3 2 1 ⎠ ⎞ R 1 = R 1 − 2 R 2 R_1=R_1-2R_2 R 1 = R 1 − 2 R 2
( 1 0 − 1 0 1 2 0 0 1 ) \begin{pmatrix}
1 & 0 & -1 \\
0 & 1 & 2 \\
0 & 0 & 1 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 − 1 2 1 ⎠ ⎞
R 1 = R 1 + R 3 R_1=R_1+R_3 R 1 = R 1 + R 3
( 1 0 0 0 1 2 0 0 1 ) \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 2 \\
0 & 0 & 1 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 0 2 1 ⎠ ⎞
R 2 = R 2 − 2 R 3 R_2=R_2-2R_3 R 2 = R 2 − 2 R 3
( 1 0 0 0 1 0 0 0 1 ) \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{pmatrix} ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞ Since the rank of the matrix is 3, then the set {(1,2,3),(0,1,2),(0,0,1) of vectors generates or span R 3 . R^3. R 3 .
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