Find a basis for the subspace of \(\mathbb R^4\) spanned by \[ (1,1,1,1),\qquad (3,4,6,7),\qquad (5,6,8,9). \]
Solution
- It will exactly be the row space of \[ A = \left(\begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 3 & 4 & 6 & 7 \\ 5 & 6 & 8 & 9 \end{array}\right), \] where we note that \(R_3 = 2R_1 + R_2\) but \(R_1 \neq \lambda R_2\) and so the first two rows span the correct subspace. We can also compute the RREF, which has the same rowspace, \[\tilde A = \left(\begin{array}{rrrr} 1 & 0 & -2 & -3 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 0 & 0 \end{array}\right)\] from which we find that \(\vector v_1 = \thevector{1,0,-2,-3}\) and \(\vector v_2 = \thevector{0,1,3,4}\) also do the job. \(\qed\)
Solution
Goal: Find a basis for the subspace \(W = \operatorname{span}\{(1,1,1,1), \, (3,4,6,7), \, (5,6,8,9)\} \subseteq \mathbb{R}^4\).
1. Let \(v_1 = (1,1,1,1)\), \(v_2 = (3,4,6,7)\), and \(v_3 = (5,6,8,9)\). The subspace \(W\) equals the row space \(\operatorname{Row}(A)\) of the matrix:\[A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 3 & 4 & 6 & 7 \\ 5 & 6 & 8 & 9 \end{pmatrix}.\]
By definition of the row space of a matrix.
2. The reduced row echelon form (RREF) of \(A\) is:\[R = \begin{pmatrix} 1 & 0 & -2 & -3 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 0 & 0 \end{pmatrix}.\]Proof:
2.1. Apply elementary row operations: \(R_2 \gets R_2 - 3R_1\) and \(R_3 \gets R_3 - 5R_1\):\[\begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 4 \\ 0 & 1 & 3 & 4 \end{pmatrix}.\]
2.2. Apply \(R_3 \gets R_3 - R_2\) and \(R_1 \gets R_1 - R_2\):\[\begin{pmatrix} 1 & 0 & -2 & -3 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 0 & 0 \end{pmatrix}.\]This matrix is in RREF.
3. Elementary row operations preserve the row space, so \(\operatorname{Row}(A) = \operatorname{Row}(R)\).
Standard linear algebra theorem: each elementary row operation is invertible and preserves linear combinations of the rows.
4. The nonzero rows of \(R\), namely \(w_1 = (1, 0, -2, -3)\) and \(w_2 = (0, 1, 3, 4)\), form a basis for \(W\).Proof:
4.1. \(w_1\) and \(w_2\) span \(\operatorname{Row}(R) = W\) since the third row is zero.
4.2. \(w_1\) and \(w_2\) are linearly independent because their leading 1s occur in distinct columns (columns 1 and 2): \(c_1 w_1 + c_2 w_2 = (c_1, c_2, -2c_1+3c_2, -3c_1+4c_2) = (0,0,0,0) \implies c_1 = 0\) and \(c_2 = 0\).
4.3. Alternatively, the first two rows \(\{v_1, v_2\}\) of the original matrix also form a basis for \(W\) since \(\rank(A) = 2\) and \(v_1, v_2\) are linearly independent with \(v_3 = 2v_1 + v_2\).Hence \(\dim(W) = 2\) and \(\{(1,0,-2,-3), (0,1,3,4)\}\) is a basis for \(W\). Q.E.D.