Vector Spaces & Fundamental Subspaces

A vector space is a set closed under addition and scalar multiplication; a subspace is a nonempty subset containing zero and closed under linear combinations. A matrix $A \in \mathbb{R}^{m \times n}$ induces four fundamental subspaces: the column space $C(A) \subseteq \mathbb{R}^m$, the null space $N(A) \subseteq \mathbb{R}^n$, the row space $C(A^T) \subseteq \mathbb{R}^n$, and the left null space $N(A^T) \subseteq \mathbb{R}^m$. The rank-nullity theorem gives the dimension relations $\dim C(A) = \dim C(A^T) = r = \text{rank}(A)$, $\dim C(A) + \dim N(A) = n$, and $\dim C(A) + \dim N(A^T) = m$; moreover $N(A)$ is the orthogonal complement of $C(A^T)$ in $\mathbb{R}^n$, and $N(A^T)$ the orthogonal complement of $C(A)$ in $\mathbb{R}^m$. Intuition: the row space carries where information comes from, the null space is the set of directions crushed to zero, and the two orthogonally split the input space.

deciding whether $Ax = b$ is solvable and the dimension of its solution set; the projection geometry of least squares; the theoretical foundation of SVD, pseudoinverse and rank analysis; principal directions in PCA live in the row space; often the opening question of derivation-focused interviews.

it translates "does a solution exist / is it unique" into subspace containment: $Ax = b$ is solvable iff $b \in C(A)$, and every solution equals a particular solution plus a vector from $N(A)$, so the solution is unique iff $N(A) = \{0\}$ (full column rank, $r = n$); the orthogonal decompositions $\mathbb{R}^n = C(A^T) \oplus N(A)$ and $\mathbb{R}^m = C(A) \oplus N(A^T)$ supply the geometric framework for approximating unsolvable systems (least squares).

By hand: for a concrete matrix (e.g. 2×3), write bases and dimensions of all four fundamental subspaces and verify rank-nullity.

State the dimension relations $\dim C(A) + \dim N(A) = n$ and $\dim C(A) + \dim N(A^T) = m$; which ambient space does each term live in?

Prove $N(A^T) = C(A)^\perp$ and show every $y \in \mathbb{R}^m$ splits uniquely into a $C(A)$ part plus an $N(A^T)$ part.

Derive why a unique solution to $Ax = b$ is equivalent to $N(A) = \{0\}$; demonstrate no-solution, unique-solution, and infinite-solution cases with 2×2 examples.

Why is $\text{rank}(A) = \text{rank}(A^T)$? Derive it fully from $N(A) = C(A^T)^\perp$ and the rank-nullity theorem.