Exercises — Lesson 5: Null Space, Rank, and Solving \(Ax=b\)
Exercise 5.1. [Hand] Solve by elimination, tracking pivots: \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix} x = \begin{pmatrix} 5 \\ 11 \end{pmatrix}\), then \(\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} x = \begin{pmatrix} 3 \\ 6 \end{pmatrix}\), then the same left side with right side \((3,7)\). Describe all solutions in each case (unique / infinitely many / none) and reconcile with rank and the three-step picture.
Solution.
First system.
\[ \left[\begin{array}{cc|c} 2 & 1 & 5 \\ 4 & 3 & 11 \end{array}\right] \;\to\; \left[\begin{array}{cc|c} 2 & 1 & 5 \\ 0 & 1 & 1 \end{array}\right] \]
\[ \begin{aligned} x_2 &= 1 \\ 2x_1 + x_2 &= 5 \\ 2x_1 &= 4 \\ x_1 &= 2 \end{aligned} \qquad\Longrightarrow\qquad x = \begin{pmatrix} 2 \\ 1 \end{pmatrix} \ \text{ unique}, \qquad \operatorname{null} \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \]
Second and third systems.
\[ \left[\begin{array}{cc|c} 1 & 2 & 3 \\ 2 & 4 & 6 \end{array}\right] \;\to\; \left[\begin{array}{cc|c} 1 & 2 & 3 \\ 0 & 0 & 0 \end{array}\right] \]
A particular solution and the homogeneous solutions:
\[ \begin{aligned} x_1 + 2x_2 &= 3 & x_1 + 2x_2 &= 0 \\ x_1 &= 3 - 2x_2 & x_1 &= -2x_2 \\ x_p &= \begin{pmatrix} 1 \\ 1 \end{pmatrix} \end{aligned} \]
\[ x = \begin{pmatrix} 1 \\ 1 \end{pmatrix} + x_h, \quad x_h \in \operatorname{span}\left( \begin{pmatrix} -2 \\ 1 \end{pmatrix} \right) \]
Infinitely many for \(\binom{3}{6}\), no solution for \(\binom{3}{7}\). Reconciling with rank:
\[ \operatorname{range} \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} = \operatorname{span}\left( \begin{pmatrix} 1 \\ 2 \end{pmatrix} \right), \qquad \operatorname{null} \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} = \operatorname{span}\left( \begin{pmatrix} -2 \\ 1 \end{pmatrix} \right) \]
Exercise 5.2. [Hand] For \(A = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \end{pmatrix}\), find a basis of \(\operatorname{null} A\) and a basis of the column space; verify rank–nullity.
Solution. \(A : \mathbb{R}^3 \to \mathbb{R}^2\), with row equations
\[ \begin{aligned} x_1 - x_3 &= b \\ x_2 + 2x_3 &= b \end{aligned} \qquad\qquad \dim \mathbb{R}^3 = 3 \]
\[ \begin{aligned} \operatorname{null} A &= \operatorname{span}\left( \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \right) &\qquad \operatorname{range} A &= \operatorname{span}\left( \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right) = \mathbb{R}^2 \\[4pt] \dim \operatorname{null} A &= 1 &\qquad \operatorname{rank} A &= \dim \operatorname{range} A = 2 \end{aligned} \]
Exercise 5.3. [Proof, \(\star\)] Prove the consequence for systems of the proposition that injectivity is equivalent to a trivial null space: \(Ax = b\) has at most one solution for every \(b\) if and only if \(Ax = 0\) has only the zero solution.
Proof.
(\(\Rightarrow\)) Suppose \(Ax = b\) has at most one solution. Let \(b = 0\). Since \(A\) is a linear map, we know \(Ax = 0\) when \(x = 0\). Thus since \(Ax = b\) can have at most one solution, \(Ax = 0\) only has the zero solution.
(\(\Leftarrow\)) Suppose \(Ax = 0\) has only the zero solution. That means \(\operatorname{null} A = \{0\}\), so \(\dim \operatorname{null} A = 0\).
Now suppose for the sake of contradiction \(\exists\, b\) such that \(Ax = b\) has at least two solutions. Call these \(x_1\) and \(x_2\), with \(x_1 \neq x_2\). Now let \(x_3 = x_1 - x_2\) and note \(x_3 \neq 0\). Then
\[ A x_3 = A(x_1 - x_2) = A x_1 - A x_2 = b - b = 0, \]
so \(x_3 \in \operatorname{null} A\) and \(x_3 \neq 0\). This contradicts our assumption that \(Ax = 0\) has only the zero solution. Thus \(Ax = b\) can have at most one solution. \(\square\)
Exercise 5.4. [Proof] Prove that \(\operatorname{null} T\) and \(\operatorname{range} T\) are subspaces, where \(T\) is a linear map from a vector space \(V\) to a vector space \(W\).
Proof.
\(\operatorname{null} T\). Since \(T\) is linear, clearly \(T0 = 0\), where the left \(0\) is from \(V\). So \(0 \in \operatorname{null} T\).
Let \(x_1, x_2 \in \operatorname{null} T\). Note if \(\operatorname{null} T = \{0\}\), addition and scalar multiplication are closed, so assume \(\operatorname{null} T\) has more than just the zero vector, so \(x_1 \neq x_2\). Let \(\alpha, \lambda \in \mathbb{F}\). Then
\[ T(\alpha x_1 + \lambda x_2) = \alpha T x_1 + \lambda T x_2 = 0 + 0 = 0, \]
so \(\operatorname{null} T\) is closed under addition and scalar multiplication. \(\square\)
\(\operatorname{range} T\). Since \(T\) is linear, clearly \(T0 = 0\), where the right \(0\) is from \(W\). So \(0 \in \operatorname{range} T\).
First note if \(\operatorname{range} T = \{0\} \subseteq W\), then clearly addition and scalar multiplication are closed. So assume \(\operatorname{range} T\) has at least two elements; pick any \(y_1, y_2\) and note \(y_1 \neq y_2\). Thus \(\exists\, x_1, x_2 \in V\), which may or may not be equal, where \(T x_1 = y_1\) and \(T x_2 = y_2\). Let \(\alpha, \lambda \in \mathbb{F}\). Then
\[ T(\alpha x_1 + \lambda x_2) = \alpha T x_1 + \lambda T x_2 = \alpha y_1 + \lambda y_2, \]
so \((\alpha y_1 + \lambda y_2) \in \operatorname{range} T\), so \(\operatorname{range} T\) is closed under addition and scalar multiplication. \(\square\)
Exercise 5.5. [Proof] Suppose \(T : V \to V\) is linear and bijective. Prove that its inverse function \(T^{-1}\) is linear. (Apply \(T\) to both sides of what you want to show.)
Proof. Let \(x_1, x_2 \in V\) and \(\alpha, \lambda \in \mathbb{F}\); \(V\) is a vector space.
We want to show \(T^{-1}(\alpha x_1 + \lambda x_2) = \alpha T^{-1} x_1 + \lambda T^{-1} x_2\).
Let’s apply \(T\) to both sides of the equation individually and see if they are the same:
\[ \begin{aligned} T\bigl(T^{-1}(\alpha x_1 + \lambda x_2)\bigr) &= \alpha x_1 + \lambda x_2 \\[4pt] T\bigl(\alpha T^{-1} x_1 + \lambda T^{-1} x_2\bigr) &= \alpha T T^{-1} x_1 + \lambda T T^{-1} x_2 = \alpha x_1 + \lambda x_2 \end{aligned} \]
Because \(T\) is linear and bijective, that means \(T^{-1}(\alpha x_1 + \lambda x_2) = \alpha T^{-1} x_1 + \lambda T^{-1} x_2\): by definition of injectivity, two things mapping to the same value means the two things are equal. So \(T^{-1}\) satisfies homogeneity and additivity, so \(T^{-1}\) is linear. \(\square\)