Consider the vector space of polynomials over \(\mathbb{Q}\) spanned by \(p_1(x) = x^2 + x + 1\), \(p_2(x) = x^2 + 2x\), \(p_3(x) = x^2 + 2\), \(p_4(x) = x - 1\). Find the dimension of this vector space.
Solution
1. Coordinate representation in standard basis:
1.1. Represent polynomials in the standard basis \(\{x^2, x, 1\}\) of the polynomial subspace \(\mathbb{Q}_{\le 2}[x]\):\[ p_1(x) = (1, 1, 1), \quad p_2(x) = (1, 2, 0), \quad p_3(x) = (1, 0, 2), \quad p_4(x) = (0, 1, -1). \]
coordinate isomorphism \(\mathbb{Q}_{\le 2}[x] \cong \mathbb{Q}^3\).
1.2. Form the \(4 \times 3\) matrix \(A\) whose rows are the coordinate vectors of \(p_1, p_2, p_3, p_4\):\[ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 2 \\ 0 & 1 & -1 \end{pmatrix}. \]
definition of row space.
2. Row reduction and rank computation:
2.1. Perform elementary row operations on \(A\):- \(R_2 \leftarrow R_2 - R_1\): row becomes \((0, 1, -1)\).- \(R_3 \leftarrow R_3 - R_1\): row becomes \((0, -1, 1)\).- \(R_4\): \((0, 1, -1)\).
row operations preserve row space.
2.2. Further reduce:- \(R_3 \leftarrow R_3 + R_2\): row becomes \((0, 0, 0)\).- \(R_4 \leftarrow R_4 - R_2\): row becomes \((0, 0, 0)\).The row echelon form of \(A\) is:\[ \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. \]
Gaussian elimination.
2.3. There are exactly 2 non-zero pivot rows in the row echelon form, so \(\operatorname{rank}(A) = 2\).Explicitly:\[ p_2(x) = p_1(x) + p_4(x), \qquad p_3(x) = p_1(x) - p_4(x). \]
\((x^2 + x + 1) + (x - 1) = x^2 + 2x\) and \((x^2 + x + 1) - (x - 1) = x^2 + 2\).
3. Conclusion:The spanning set \(\{p_1, p_2, p_3, p_4\}\) reduces to the linearly independent basis \(\{x^2 + x + 1, x - 1\}\).Thus the dimension of the subspace is \(\dim(\operatorname{span}\{p_1, p_2, p_3, p_4\}) = 2\). Q.E.D.
1 and 2.