Exercises 2C — Axler, Linear Algebra Done Right, 4th Edition
Exercise 1. Show that the subspaces of \(\mathbb{R}^2\) are precisely \(\{0\}\), all lines in \(\mathbb{R}^2\) containing the origin, and \(\mathbb{R}^2\).
Proof. We know from the reading and exercises all three of these subsets of \(\mathbb{R}^2\) are subspaces of \(\mathbb{R}^2\), so what we want to show is that these are the only ones. First note \(\dim \mathbb{R}^2 = 2\), \(\dim \{0\} = 0\), and for any nonzero \(u \in \mathbb{R}^2\), \(\dim \{au : a \in \mathbb{R}\} = 1\). So every subspace in \(\mathbb{R}^2\) has dimension either \(0\), \(1\), or \(2\). Let \(U\) be a subspace of \(\mathbb{R}^2\).
Case: \(\dim U = 0 \ \Rightarrow\ U = \{0\}.\)
Case: \(\dim U = 2 \ \Rightarrow\ U = \mathbb{R}^2.\)
Case: \(\dim U = 1 \ \Rightarrow\ \) basis for \(U\) is a nonzero vector in \(U\), call this \(u\).
\[ \begin{aligned} &\Rightarrow\ U \text{ is the line containing this vector and its scalar multiples} \\ &\Rightarrow\ U = \{au : a \in \mathbb{R}\} = \text{line passing through origin.} \qquad\square \end{aligned} \]
Exercise 2. Show that the subspaces of \(\mathbb{R}^3\) are precisely \(\{0\}\), all lines in \(\mathbb{R}^3\) containing the origin, all planes in \(\mathbb{R}^3\) containing the origin, and \(\mathbb{R}^3\).
Proof. Similar to exercise 1, all of these are subspaces, and we want to show these are the only ones. Note every subspace in \(\mathbb{R}^3\) has dimension either \(0\), \(1\), \(2\), or \(3\). Let \(U\) be a subspace of \(\mathbb{R}^3\).
Case: \(\dim U = 0 \ \Rightarrow\ U = \{0\}.\)
Case: \(\dim U = 3 \ \Rightarrow\ U = \mathbb{R}^3.\)
Case: \(\dim U = 1 \ \Rightarrow\ \) basis for \(U\) is any nonzero vector in \(U\), call it \(u\).
\[ \Rightarrow\ U = \{au : a \in \mathbb{R}\} = \text{line passing through origin containing } u. \]
Case: \(\dim U = 2 \ \Rightarrow\ \) basis for \(U\) is any two linearly independent vectors in \(U\), call these two vectors \(u, v\).
\[ \Rightarrow\ U = \{au + bv : a, b \in \mathbb{R}\} = \text{plane passing through origin containing both } u \text{ and } v. \qquad\square \]
Exercise 10. Suppose \(m\) is a positive integer. For \(0 \le k \le m\) let \[ p_k(x) = x^k(1-x)^{m-k}. \] Show \(p_0,\dots,p_m\) is a basis of \(\mathcal{P}_m(\mathbf{F})\).
Proof. This list of vectors is of length \((m+1)\), which is \(\dim \mathcal{P}_m(\mathbf{F})\). So we just have to show this list is linearly independent for it to be a basis. We will do it via induction on \(k\), \(0 \le k \le m\), where we’ll show \(p_0,\dots,p_k\) is linearly independent.
Base: \(k = 0\). \[ p_0(x) = x^0(1-x)^m = (1-x)^m. \] \(p_0\) is not the zero polynomial and thus list containing only \(p_0\) is linearly independent.
Induction Step: Assume \(p_0,\dots,p_{k-1}\) is linearly independent for \(1 \le k \le m\). We want to show that \(p_0,\dots,p_{k-1}, p_k\) is also linearly independent if \(p_0,\dots,p_{k-1}\) is. Suppose
\[ a_0 p_0(x) + a_1 p_1(x) + \cdots + a_k p_k(x) = 0,\qquad a_i \in \mathbf{F},\ i = 0,\dots,k. \]
Note \(p_j(x) = x^j (1-x)^{m-j}\) for \(j = 0,\dots,m\). Then
\[ a_0 (1-x)^m + a_1 x (1-x)^{m-1} + \cdots + a_k x^k (1-x)^{m-k} = 0. \]
Let’s evaluate at \(x = 0\). Then we know \(a_0 = 0\).
\[ \Rightarrow\ a_1 x (1-x)^{m-1} + \cdots + a_k x^k (1-x)^{m-k} = 0. \]
Divide by \(x\), no division by \(0\) concern since its polynomial identity.
\[ \Rightarrow\ a_1 (1-x)^{m-1} + a_2 x (1-x)^{m-2} + \cdots + a_k x^{k-1}(1-x)^{m-k} = 0. \]
Again evaluate at \(x = 0\). Then \(a_1 = 0\). Repeating this process you can see that \(a_0 = a_1 = \cdots = a_k = 0\).
\[ \Rightarrow\ p_0, p_1,\dots,p_k \text{ is linearly independent.} \]
\[ \Rightarrow\ \text{By induction } p_0,\dots,p_m \text{ is linearly independent.} \]
\[ \Rightarrow\ p_0,\dots,p_m \text{ is a basis of } \mathcal{P}_m(\mathbf{F}). \qquad\square \]
Note: These polynomials are called Bernstein polynomials and are used to approximate continuous functions on \([0,1]\).
Exercise 18. Suppose \(V\) is finite-dimensional with \(\dim V = n \ge 1\). Prove that there exists one-dimensional subspaces \(V_1,\dots,V_n\) of \(V\) such that \[ V = V_1 \oplus \cdots \oplus V_n. \]
Proof. Let \(v_1,\dots,v_n\) be a basis of \(V\). Now let \(V_i = \operatorname{span}(v_i)\), \(i = 1,\dots,n\). Note \(\dim V_i = 1\). Since \(\{v_i\}\) is a basis,
\[ V = V_1 + \cdots + V_n. \]
Now consider \(w_1 + \cdots + w_n = 0\), \(w_i \in V_i\), \(i = 1,\dots,n\). Since \(V_i = \operatorname{span}(v_i)\), \(w_i = a_i v_i\), \(a_i \in \mathbf{F}\), \(i = 1,\dots,n\). Thus we have
\[ a_1 v_1 + \cdots + a_n v_n = 0. \]
Since \(v_1,\dots,v_n\) is linearly independent, \(a_1 = \cdots = a_n = 0\), which implies \(w_1 = \cdots = w_n = 0\). This implies
\[ V = V_1 \oplus \cdots \oplus V_n. \qquad\square \]
Exercise 19. Prove or give counterexample: \(V_1, V_2, V_3\) are subspaces of finite-dimensional vector space \(V\).
\[ \dim(V_1 + V_2 + V_3) = \dim V_1 + \dim V_2 + \dim V_3 - \dim(V_1 \cap V_2) - \dim(V_1 \cap V_3) - \dim(V_2 \cap V_3) + \dim(V_1 \cap V_2 \cap V_3). \]
Solution. False. Counterexample: Let \(V = \mathbb{R}^2\).
\[ V_1 = \operatorname{span}(\{(1,0)\}),\quad V_2 = \operatorname{span}(\{(0,1)\}),\quad V_3 = \operatorname{span}(\{(1,1)\}). \]
\[ \begin{aligned} \dim V_1 &= \dim V_2 = \dim V_3 = 1 \\ \dim(V_1 \cap V_2) &= 0 \\ \dim(V_1 \cap V_3) &= 0 \\ \dim(V_2 \cap V_3) &= 0 \\ \dim(V_1 \cap V_2 \cap V_3) &= 0 \\ \dim(V_1 + V_2 + V_3) &= 2 \\ \Rightarrow\ 2 &\ne 3. \qquad\square \end{aligned} \]
Exercise 20. If \(V_1, V_2, V_3\) are subspaces of a finite-dimensional vector space, then
\[ \begin{aligned} \dim(V_1 + V_2 + V_3) = {}& \dim V_1 + \dim V_2 + \dim V_3 \\ & - \tfrac{1}{3}\bigl[\dim(V_1 \cap V_2) + \dim(V_1 \cap V_3) + \dim(V_2 \cap V_3)\bigr] \\ & - \tfrac{1}{3}\bigl[\dim((V_1 + V_2) \cap V_3) + \dim((V_1 + V_3) \cap V_2) + \dim((V_2 + V_3) \cap V_1)\bigr]. \end{aligned} \]
Proof. Note \(\dim(U + W) = \dim U + \dim W - \dim(U \cap W)\) for subspaces \(U, W\) of \(V\).
\[ \begin{aligned} \Rightarrow\ \dim((V_1 + V_2) + V_3) &= \dim(V_1 + V_2) + \dim V_3 - \dim((V_1 + V_2) \cap V_3) && [1] \\ \Rightarrow\ \dim((V_1 + V_3) + V_2) &= \dim(V_1 + V_3) + \dim V_2 - \dim((V_1 + V_3) \cap V_2) && [2] \\ \Rightarrow\ \dim((V_2 + V_3) + V_1) &= \dim(V_2 + V_3) + \dim V_1 - \dim((V_2 + V_3) \cap V_1) && [3] \\ \Rightarrow\ \dim(V_1 + V_2) &= \dim V_1 + \dim V_2 - \dim(V_1 \cap V_2) && [4] \\ \Rightarrow\ \dim(V_1 + V_3) &= \dim V_1 + \dim V_3 - \dim(V_1 \cap V_3) && [5] \\ \Rightarrow\ \dim(V_2 + V_3) &= \dim V_2 + \dim V_3 - \dim(V_2 \cap V_3) && [6] \end{aligned} \]
Adding \([1], [2], [3]\) together and substituting \([4], [5], [6]\) gives
\[ \begin{aligned} 3 \dim(V_1 + V_2 + V_3) = {}& 3\bigl[\dim V_1 + \dim V_2 + \dim V_3\bigr] \\ & - \bigl[\dim(V_1 \cap V_2) + \dim(V_1 \cap V_3) + \dim(V_2 \cap V_3)\bigr] \\ & - \bigl[\dim((V_1 + V_2) \cap V_3) + \dim((V_1 + V_3) \cap V_2) + \dim((V_2 + V_3) \cap V_1)\bigr]. \end{aligned} \]
Dividing both sides by \(3\) gives the desired equation. \(\qquad\square\)