Linear Algebra · Worked Example 6
Show that {v₁, v₂, v₃} is a Basis for ℝ³
Problem. Let \( v_1 = (1,2,1) \), \( v_2 = (2,9,0) \), and \( v_3 = (3,3,4) \).
Show that the set \( S = \{v_1, v_2, v_3\} \) is a basis for \( \mathbb{R}^3 \).
Proof
A set of 3 vectors forms a basis for \( \mathbb{R}^3 \) if the vectors are linearly independent.
This can be verified by checking whether the determinant of the matrix formed by these vectors
(as rows or columns) is non-zero.
Let \(A\) be the matrix with the given vectors as rows:
\[ A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 9 & 0 \\ 3 & 3 & 4 \end{bmatrix} \]Step 1 — Evaluate the determinant of A
Expanding along the third column (since it contains a zero):
\[ \begin{aligned} |A| &= 1 \cdot \begin{vmatrix} 2 & 9 \\ 3 & 3 \end{vmatrix} - 0 \cdot \begin{vmatrix} 1 & 2 \\ 3 & 3 \end{vmatrix} + 4 \cdot \begin{vmatrix} 1 & 2 \\ 2 & 9 \end{vmatrix} \\ &= 1 \cdot (6 - 27) - 0 + 4 \cdot (9 - 4) \\ &= 1 \cdot (-21) + 4 \cdot (5) \\ &= -21 + 20 \\ &= -1 \end{aligned} \]Since \( |A| = -1 \neq 0 \), the matrix \(A\) is non-singular.
Conclusion
Since the determinant is non-zero, the vectors \(v_1, v_2,\) and \(v_3\) are linearly independent. Any set of 3 linearly independent vectors in the 3‑dimensional space \( \mathbb{R}^3 \) forms a basis.
The set \( S = \{v_1, v_2, v_3\} \) forms a basis for \( \mathbb{R}^3 \).
Tutorial Video: Show That a Set of Vectors is a Basis for R³
Note: replace REPLACE_WITH_VIDEO_ID with your actual YouTube video ID for this topic.
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