Vector Spaces and Coordinates Workshop

Definition Let $H$ be a subspace of a vector space $V$, and let $\mathbf{v}_1, \dots, \mathbf{v}_k$ be proposed basis vectors. They form a basis of $H$ exactly when three checks pass.

1. Membership. Each $\mathbf{v}_i$ lies in $H$.
2. Independence. The only solution of $c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k = \mathbf{0}$ is $c_1 = \cdots = c_k = 0$.
3. Spanning. Every vector of $H$ is a linear combination of $\mathbf{v}_1, \dots, \mathbf{v}_k$.

For a null space, the general solution of $A\mathbf{x} = \mathbf{0}$ written with its free variables proves spanning, and substituting each basis vector back into the equations confirms membership.

Definition Let $H$ be a subspace whose dimension $k$ is known from a separate argument. Then any $k$ linearly independent vectors in $H$ form a basis of $H$, and any $k$ vectors in $H$ that span $H$ form a basis of $H$. For a real $m \times n$ matrix $A$ of rank $r$, the transpose $A^{\top}$ also has rank $r$, so the column space and the row space both have dimension $r$. Rank-nullity applied to $A$ and to $A^{\top}$ gives the dimensions of the two null spaces. Together,

where $\operatorname{Row} A$ and $\operatorname{Null} A$ lie in $\mathbb{R}^n$, and $\operatorname{Col} A$ and $\operatorname{Null} A^{\top}$ lie in $\mathbb{R}^m$.

Definition To find a basis of $\operatorname{Col} A$, row reduce $A$ to an echelon form $U$ and note which columns contain pivots. The same columns of the original matrix $A$ form a basis of $\operatorname{Col} A$. This works because row operations do not change the solutions of $A\mathbf{x} = \mathbf{0}$, and such a solution is exactly a dependence relation $x_1\mathbf{a}_1 + \cdots + x_n\mathbf{a}_n = \mathbf{0}$ among the columns. So the columns of $A$ and of $U$ satisfy the same relations. In $U$ the pivot columns are independent and each non-pivot column is a combination of them, and therefore the same holds in $A$. Row operations can change the column space itself. They do not change the row space, so the nonzero rows of $U$ form a basis of $\operatorname{Row} A$.

Definition Let $\mathcal{B} = (\mathbf{b}_1, \dots, \mathbf{b}_n)$ be an ordered basis of $\mathbb{R}^n$ with basis matrix $B = \begin{bmatrix} \mathbf{b}_1 & \cdots & \mathbf{b}_n \end{bmatrix}$. If $[\mathbf{v}]_{\mathcal{B}} = (c_1, \dots, c_n)$, then

Rebuilding $\mathbf{v}$ this way checks a computed coordinate column. The $i$th coordinate always multiplies the $i$th basis vector, so the order of the basis is part of the data. In an abstract space there is no basis matrix to multiply. In $P_2$ with an ordered basis $(p_1, p_2, p_3)$, the reconstruction is $p = c_1 p_1 + c_2 p_2 + c_3 p_3$.

Definition For distinct real numbers $s_1$ and $s_2$, the evaluation map $E(p) = (p(s_1), p(s_2))$ from $P_2$ to $\mathbb{R}^2$ has the matrix

in the basis $(1, t, t^2)$ and the standard basis of $\mathbb{R}^2$. Row $i$ lists the values of $1$, $t$ and $t^2$ at $s_i$. The first two columns are independent because $s_2 - s_1 \ne 0$, so the rank is $2$ and the image is $\mathbb{R}^2$. Rank-nullity then gives nullity $3 - 2 = 1$, and since $(t - s_1)(t - s_2)$ is a nonzero polynomial that vanishes at both points, it spans $\ker E$. Replacing $\{s_1, s_2\}$ by any other set of two distinct points changes the matrix and the kernel, because the nonzero polynomials in $\ker E$ vanish only at $s_1$ and $s_2$. If the two points coincide, the rows are equal and the rank drops to $1$.

Definition The graph of a polynomial $p$ in $P_2$ is the set of points $(t, p(t))$ in the plane, while its coordinate column $[p]_{\mathcal{B}}$ in an ordered basis $\mathcal{B}$ of $P_2$ lies in $\mathbb{R}^3$. The polynomial is one vector of the three-dimensional space $P_2$. The graph shows that vector as a function, and the two axes of the drawing are not a basis of $P_2$. Two different members of $P_2$ can agree at two points, but not at three distinct points, because their difference is a nonzero polynomial of degree at most two and has at most two roots.