Chapter 14 Quiz — Row Space, Left Null Space, and Rank-Nullity
Twelve quick checks on the four fundamental subspaces, rank, and the rank-nullity theorem. Try each before opening the answer. Most are conceptual; a couple need a little arithmetic.
Q1. Name the four fundamental subspaces and say which ambient space each lives in.
Answer
**Column space** $C(A)$ and **left null space** $N(A^{\mathsf{T}})$ both live in the *output* space $\mathbb{R}^m$. **Row space** $C(A^{\mathsf{T}})$ and **null space** $N(A)$ both live in the *input* space $\mathbb{R}^n$. The column space is what $A$ reaches; the null space is what it destroys; the row space is the part of the input it acts on faithfully; the left null space is the part of the output it can never reach.Q2. Why is the row space of $A$ denoted $C(A^{\mathsf{T}})$?
Answer
Because transposing $A$ turns its rows into columns, so the **columns of $A^{\mathsf{T}}$ are the rows of $A$**, and the span of those columns is $C(A^{\mathsf{T}})$. The row space of $A$ literally *is* the column space of $A^{\mathsf{T}}$ — so every theorem about column spaces transfers for free.Q3. State the rank-nullity theorem, and say what is on the right-hand side.
Answer
For any $m \times n$ matrix $A$: $\operatorname{rank}(A) + \dim N(A) = n$. The right-hand side is $n$, **the number of columns** (the dimension of the input space) — *not* $m$, and *not* the rank. A common error is to write $m$; the right side is always the dimension of the space the map acts *on*.Q4. A basis for the row space comes from the RREF rows, but a basis for the column space comes from the original columns of $A$. Why the difference?
Answer
Row operations **preserve the row space** (each new row is a combination of old rows, and the operation is reversible), so the cleaner RREF rows span the same space. But row operations **change the column space** (they move the columns to new positions in $\mathbb{R}^m$), so you must use the *original* pivot columns of $A$. What row reduction *does* preserve is the dependence relations among the columns — which is why the *positions* of the pivots are still trustworthy.Q5. A matrix is $5 \times 3$ with rank $2$. Give the dimensions of all four fundamental subspaces.
Answer
With $m = 5$, $n = 3$, $r = 2$: $\dim C(A) = r = 2$; $\dim N(A) = n - r = 1$; $\dim C(A^{\mathsf{T}}) = r = 2$; $\dim N(A^{\mathsf{T}}) = m - r = 3$. Check: input pair $2 + 1 = 3 = n$ ✓, output pair $2 + 3 = 5 = m$ ✓.Q6. Why is it surprising that row rank always equals column rank, and what explains it?
Answer
Surprising because the rows live in $\mathbb{R}^n$ and the columns live in $\mathbb{R}^m$ — different spaces, often of different sizes — so there is no obvious reason the count of independent rows should match the count of independent columns. The explanation: **both counts equal the number of pivots** in the RREF, and a matrix has exactly one set of pivot positions. One pivot count, two readings.Q7. What does it mean for $\mathbf{y}$ to be in the left null space $N(A^{\mathsf{T}})$, stated in terms of the rows of $A$?
Answer
$\mathbf{y} \in N(A^{\mathsf{T}})$ means $A^{\mathsf{T}}\mathbf{y} = \mathbf{0}$, equivalently $\mathbf{y}^{\mathsf{T}}A = \mathbf{0}$ — the combination of the **rows** of $A$ weighted by the entries of $\mathbf{y}$ is zero. The left null space records the **dependence relations among the rows**. (For a graph incidence matrix, these are the loops.)Q8. A matrix $A$ has full column rank ($r = n$). What is its nullity, and what does that say about solutions of $A\mathbf{x} = \mathbf{b}$?
Answer
Nullity $= n - r = 0$, so $N(A) = \{\mathbf{0}\}$. The map is one-to-one: $A\mathbf{x} = \mathbf{b}$ has **at most one** solution (a unique solution when $\mathbf{b} \in C(A)$, and none otherwise). Full column rank kills the null space and removes all free parameters.Q9. Are the row space and null space orthogonal? Are the column space and left null space orthogonal? In which spaces do these relationships live?
Answer
**Yes to both.** The row space $C(A^{\mathsf{T}})$ is orthogonal to the null space $N(A)$ inside the *input* space $\mathbb{R}^n$; the column space $C(A)$ is orthogonal to the left null space $N(A^{\mathsf{T}})$ inside the *output* space $\mathbb{R}^m$. In each space the two subspaces are **orthogonal complements** — perpendicular, meeting only at the origin, with dimensions summing to fill the whole space. (This is the Part IV preview.)Q10. True or false: a wide matrix ($n > m$) always has a nonzero null space.
Answer
**True.** If $n > m$ then $r \le m < n$, so $\dim N(A) = n - r \ge n - m > 0$ — the null space is always nontrivial. (Consequence: a wide homogeneous system $A\mathbf{x} = \mathbf{0}$ always has nontrivial solutions, and a wide system $A\mathbf{x} = \mathbf{b}$, *when solvable*, has infinitely many solutions.) But beware: this does **not** mean every $\mathbf{b}$ is reachable — solvability still requires $\mathbf{b} \in C(A)$.Q11. The system $A\mathbf{x} = \mathbf{b}$ turns out to be inconsistent (no solution). What does that say about $\mathbf{b}$ relative to the four subspaces?
Answer
$\mathbf{b} \notin C(A)$ — it lies outside the column space. Equivalently, $\mathbf{b}$ has a **nonzero component in the left null space** $N(A^{\mathsf{T}})$ (since $C(A)$ and $N(A^{\mathsf{T}})$ are orthogonal complements of $\mathbb{R}^m$). The augmented-rank test detects it: $\operatorname{rank}[A \mid \mathbf{b}] > \operatorname{rank}(A)$. The best you can do is project $\mathbf{b}$ onto $C(A)$ — the least-squares solution of Chapters 17 and 19.Q12. Why does rank-nullity count as a genuine theorem and not just the arithmetic identity $r + (n - r) = n$?