Chapter 13 Quiz — Column Space and Null Space
Twelve quick checks on the column space, the null space, and what they say about $A\mathbf{x} = \mathbf{b}$. Try each before opening the answer. Most are conceptual; a couple need a little arithmetic.
Q1. In one sentence, what is the column space $C(A)$, and what question about $A\mathbf{x} = \mathbf{b}$ does it answer?
Answer
$C(A)$ is the **span of the columns of $A$** — equivalently, the set of all outputs $A\mathbf{x}$. It answers *existence*: $A\mathbf{x} = \mathbf{b}$ has a solution **if and only if** $\mathbf{b} \in C(A)$ (the target is reachable).Q2. What is the null space of a matrix, in one sentence, and what does it tell you geometrically?
Answer
$N(A)$ is the set of all solutions of $A\mathbf{x} = \mathbf{0}$ — equivalently, all input vectors the transformation **collapses to zero**. Geometrically it is the set of directions the matrix crushes; the bigger it is, the more information $A$ destroys.Q3. For an $m \times n$ matrix, in which space does $C(A)$ live, and in which does $N(A)$ live?
Answer
$C(A) \subseteq \mathbb{R}^m$ (the **output** space — each column has $m$ entries), and $N(A) \subseteq \mathbb{R}^n$ (the **input** space — a solution $\mathbf{x}$ has $n$ entries). They generally live in different spaces and have different dimensions.Q4. When you find a basis for $C(A)$ from the RREF, do you use the pivot columns of $R$ or of the original $A$? Why?
Answer
The pivot columns of the **original $A$**. Row operations change the column space (they mix each column's entries), so columns of $R$ are generally *not* in $C(A)$. Use $R$ only to *locate* the pivot positions, then read those columns from $A$.Q5. How do you get a basis for the null space from the RREF?
Answer
Find the **special solutions**: for each free variable, set it to $1$ and the other free variables to $0$, then solve for the pivot variables. There is one special solution per free variable, and together they form a basis for $N(A)$.Q6. $A$ row-reduces to an RREF with pivots in columns 1 and 4, and $A$ has 6 columns. What is $\dim C(A)$ and $\dim N(A)$?
Answer
There are $2$ pivots, so $\dim C(A) = \operatorname{rank}(A) = 2$. There are $6 - 2 = 4$ free variables, so $\dim N(A) = 4$. Note $2 + 4 = 6 = $ number of columns (rank–nullity, Chapter 14).Q7. True or false: if $A\mathbf{x} = \mathbf{b}$ has two different solutions, it has infinitely many.
Answer
**True.** If $\mathbf{x}_1 \neq \mathbf{x}_2$ both solve it, then $\mathbf{x}_1 - \mathbf{x}_2$ is a *nonzero* vector in $N(A)$. So the null space is nontrivial, and $\mathbf{x}_1 + c(\mathbf{x}_1 - \mathbf{x}_2)$ is a solution for *every* scalar $c$ — infinitely many. A linear system never has exactly two solutions.Q8. Describe the complete solution set of a consistent system $A\mathbf{x} = \mathbf{b}$ in terms of $N(A)$.
Answer
It is **one particular solution plus the entire null space**: $\{\mathbf{x}_p + \mathbf{x}_n : \mathbf{x}_n \in N(A)\}$. Geometrically, it is the flat $N(A)$ slid over to pass through $\mathbf{x}_p$ — a shifted copy of the null space.Q9. A square matrix $A$ has $N(A) = \{\mathbf{0}\}$. What can you conclude about $C(A)$ and about invertibility?
Answer
For a *square* matrix, trivial null space forces **full column space**: $C(A) = \mathbb{R}^n$, every $\mathbf{b}$ is reachable, and $A$ is **invertible**. The two properties (one-to-one and onto) coincide for square matrices — this is the invertibility story of Chapter 9 in subspace language.Q10. Why does a wide matrix ($n > m$, more columns than rows) always have a nontrivial null space?
Answer
The rank is at most the number of rows, $r \le m < n$, so there are at least $n - m \ge 1$ free variables — hence at least one nonzero special solution. Equivalently: more column vectors than the output space can hold independently, so some combination must reach $\mathbf{0}$.Q11. A special solution of $A\mathbf{x} = \mathbf{0}$ is $(-2, 1, 0)$. What dependence relation among the columns $\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$ does it encode?
Answer
It says $-2\mathbf{a}_1 + 1\mathbf{a}_2 + 0\mathbf{a}_3 = \mathbf{0}$, i.e. $\mathbf{a}_2 = 2\mathbf{a}_1$ — the second column is twice the first. Every null-space vector is a recipe for combining the columns to zero; that *is* a linear-dependence relation (Chapter 6).Q12. How can you test, with numpy, whether $\mathbf{b}$ is in the column space of $A$ — without solving the system?