10.30 Subspaces

1. Which of the following subsets of \(\mathbb{R}^n\) \((n \geq 3)\) are subspaces of \(\mathbb{R}^n\)?

   a. \(S=\{a=(a_1,\ldots,a_{n})\in\mathbb{R}^{n}:a_1\geq0\}\)

   Not a subspace since \(\left(-1)(1,1,\ldots,1\right)\notin S\).

   b. \(S = \{ a = (a_1, \ldots, a_n) \in \mathbb{R}^n : a_1 + 2a_2 = a_3 \}\)

   • If \(u \in S\), let \(\alpha \in \mathbb{R} \implies u_1 + 2(u_2) = u_3 \implies \alpha u_1 + 2(\alpha u_2) = \alpha u_3 \implies \alpha u \in S\).
   • \(0 \in S\): yes, since \(0 + 2 \cdot 0 = 0\).
   • If \(u, v \in S \implies u_1 + 2u_2 = u_3\) and \(v_1 + 2v_2 = v_3 \implies (u_1 + v_1) + 2(u_2 + v_2) = u_3 + v_3\).

   c. \(S = \{ a = (a_1, \ldots, a_n) \in \mathbb{R}^n : a_2 = a_3^2 \}\)

   Not a subspace since \((1, 1, \ldots, 1) \in S\) (since \(1 = 1^2\)) but \((-1)(1, \ldots, 1) \notin S\) (since \(-1\neq(-1)^2\)).

   d. \(S = \{ a = (a_1, \ldots, a_n) \in \mathbb{R}^n : a_1 a_2 = 0 \}\)

   Not a subspace since \((1, 0, 1, \ldots, 1) \in S\) and \((0, 1, 1, \ldots, 1) \in S\) but \((1, 0, 1, \ldots, 1) + (0, 1, 1, \ldots, 1) = (1, 1, 2, \ldots, 2) \notin S\).

   e. \(S = \{ a = (a_1, a_2, \ldots, a_n) \in \mathbb{R}^n : a_2 \in \mathbb{Q} \}\)

   Not a subspace since \((1, 1, 1, \ldots, 1) \in S\) but \(\pi (1, 1, \ldots, 1) \notin S\) (since \(\pi \notin \mathbb{Q}\)).

2. Let \(V = \{ f : \mathbb{R} \to \mathbb{R} \}\) (functions from \(\mathbb{R}\) to \(\mathbb{R}\)) real vector space. Which of the subsets of \(V\) are subspaces of \(V\)?

   a. \(S=\{f\in V:f(x^2)=f(x)^2\ \forall x\in\mathbb{R}\}\)

   Not a subspace. Consider \(f(x) = 1\) and \(g(x) = x\):

   • \(f(x^2)=1=(f(x))^2\)
   • \(g(x^2)=x^2=(g(x))^2\)

   So \(f + g \notin S\) since \((f + g)(x) = x + 1\), but \((x + 1)^2 \neq x^2 + 1\).

   b. \(S = \{ f \in V : f(0) = f(1) \}\)

   • \(0 \in S\): \(0(0) = 0 = 0(1)\).
   • If \(f, g \in S \implies f(0) = f(1)\) and \(g(0) = g(1)\), then \((f+g)(0) = f(0) + g(0) = f(1) + g(1) = (f+g)(1) \implies f + g \in S\).
   • If \(f \in S\), let \(\alpha\in\mathbb{R}\implies f(0)=f(1)\Rightarrow\left.(\alpha f\right)(0)=(\alpha f)(1)\implies\alpha f\in S\).

   So \(S\) is a subspace.

   c. \(S = \{ f \in V : f(3) = 1 + f(-5) \}\)

   \(f(x) = \begin{cases} 1 & x = 3 \\ 0 & x \neq 3 \end{cases}\) and \(g(x)=\begin{cases}1 & x=3\\ 0 & x\neq3\end{cases}\)

   \(\Rightarrow f, g \in S\). But note that \((f + g)(3) = 2\) but \(1 + (f + g)(-5) = 1\)

   So \(f+g \notin S\). So \(S\) is not subspace (you can also see that \(0 \notin S\))

   d. \(S = \{ f \in V : f(-1) = 0 \}\)

   \(0 \in S\) since \(0(-1) = 0\)

   \(f,g \in S \Rightarrow f(-1) = 0, g(-1) = 0 \Rightarrow (f+g)(-1) = 0 + 0 = 0 \Rightarrow f+g \in S\)

   \(\alpha \in \mathbb{R}, f \in S \Rightarrow (\alpha f)(-1) = \alpha f(-1) = \alpha \cdot 0 = 0 \Rightarrow \alpha f \in S\)

   So \(S\) is a subspace.

3. Prove the following statements:

   1. The only subspaces of \(\mathbb{R}\) are \(\mathbb{R}\) and \(\{0\}\).

      Since \(\{0\}\) is a subspace of any vector space, then \(\{0\}\) is contained

      Now suppose there is another vector space \(V\) such that \(V\neq\{0\}\subseteq\mathbb{R}\)

      If \(V=\R\), we are done

      If \(V\subset \R\), then \(\exists v\neq0,v\in V\) and \(\exists w\in \R, w\notin V\)

      Then since \(V\) is a vector space, then let \(\lambda=\frac{w}{v}\), then \(\lambda v=w\in V\)

      Contradiction!

      Therefore the only subspaces of \(\mathbb{R}\) are \(\{0\}\) and \(\mathbb{R}\). 2. The only subspaces of \(\mathbb{R}^2\) are \(\mathbb{R}^2\), \(\{(0,0)\}\) and \(M=\{\lambda(a,b):a,b\in\mathbb{R},\lambda\in\mathbb{R}\}\)

      Let \(V\) be a subspace of \(\mathbb{R}^2\) \(\Rightarrow(0,0) \in V\) by definition. We have 2 cases.

      1. \(V\) has only one element \(\Rightarrow V\) must be \(V = \{(0,0)\}\)

      2. \(V\) has more than one element \(\Rightarrow\exists v=\left(a,b\right)\in V\) with \(v\neq(0,0)\)

         Since \(V\) is a subspace, \(\lambda v=\lambda\left(a,b\right)\in V\) for all \(\lambda \in \mathbb{R}\), then \(M\subseteq V\)

         If \(M=V\), we are done

         If \(M\subset V\), then \(\exists w\in V,w\neq \lambda v\)

         Since we have \(v,w\in V\) such that \(w\neq\lambda v\)

         Thus \(\forall x,y\in\mathbb{R},(x,y)\in\mathbb{R}^2\), we have \((x,y)=\lambda v+\mu w\) for \(\lambda,\mu\in\mathbb{R}\)

         This is possible since \(w\neq\lambda v\) for \(\lambda \in \mathbb{R}\)

         Therefore \(V=\mathbb{R}^2\).

      Thus the only subspaces of \(\mathbb{R}^2\) are \(\mathbb{R}^2\), \(\{(0,0)\}\) and \(M=\{\lambda(a,b):a,b\in\mathbb{R},\lambda\in\mathbb{R}\}\)

   Strategy: We can start from the trivial subspace, and consider other cases step by step.

   We can have a new vector when we assume a subspace is different from previous one

   Trick: Use linear combination to achieve the whole space, use scalar to achieve the subspace

4. Let \(U_1\) and \(U_2\) be subspaces of \(V\) (\(F\)-vector space). Prove that \(U_1 \cup U_2\) is a subspace of \(V \Leftrightarrow U_1 \subseteq U_2\) or \(U_2 \subseteq U_1\).

   Proof

   \(\Leftarrow\)​) If \(U_1 \subseteq U_2 \Rightarrow U_1 \cup U_2 = U_2\) and since \(U_2\) is a subspace, so is \(U_1 \cup U_2\).

   If \(U_2 \subseteq U_1 \Rightarrow U_1 \cup U_2 = U_1\), which is a subspace.

   \(\Rightarrow\)) Let assume \(U_1\subseteq U_2\), the other cases is similar

   Let's take \(u_1\in U_1\text{ or }u_2\in U_2\), N.T.P. \(u_1\in U_2\text{ or }u_2\in U_1\)

   Since \(U_1\cup U_2\) is subspace, then \(u_1+u_2\in U_1\cup U_2\)

   1. \(u_1+u_2\in U_1\)​

      Since \(u_1\in U_1\) and \(U_1\) is a subspace, then \(-u_1\in U_1\)​

      Then we have \(u_2\in U_1\)​

      Also since \(u_2\in U_2\) by hypothesis, then \(U_2\subseteq U_1\)​ 2. \(u_1+u_2\in U_2\)​

      Since \(u_2\in U_2\) and \(U_2\) is a subspace, then \(-u_2\in U_2\)

      Then we have \(u_1\in U_2\)

      Also since \(u_1\in U_1\) by hypothesis, then \(U_1\subseteq U_2\)​

   Thus we finish the proof

   Strategy: Only by definition of inclusion, take \(u_1\in U_1\), to prove \(u_1\in U_2\)​

   We know the reason that why \(U_1\cup U_2\) is not a subspace, because it satisfies the \(0\) closure and scalar, but addition closure doesn't hold

   Think i have \(u_1\in U_1\) and \(U_1\cup U_2\) is a subspace. Then naturally, \(u_1\in U_1\cup U_2\), but i need \(u_1\in U_2\).

   Trick: Use addition closure

   Then we need \(u_1+u_2\in U_1\cup U_2\), then we consider \(\in U_1\) or \(U_2\)​

   Conclusion: The "or" in the problem sometimes means that you need to consider case by case instead of considering just one case, and the other is just similar

5. Prove that the operation of addition of the subspaces of a vector space \(V\) is commutative and associative. That is: if \(U_1, U_2, U_3\) are subspaces of \(V\), then:

   • Commutative: \(U_1 + U_2 = U_2 + U_1\)
   • Associative: \(U_1 + (U_2 + U_3) = (U_1 + U_2) + U_3\)

   ---

   • Commutative:

   (\(\subseteq\)​)Let \(x \in U_1 + U_2 \Rightarrow x = u_1 + u_2\) for \(u_1 \in U_1\) and \(u_2 \in U_2\). By commutativity of \(V\), \(x = u_1 + u_2 = u_2 + u_1 \in U_2 + U_1\).

   (\(\supseteq\)​)Let \(x \in U_2 + U_1 \Rightarrow x = u_2 + u_1\) for \(u_2 \in U_2\) and \(u_1 \in U_1\). Therefore, \(x \in U_1 + U_2\). * Associative:

   (\(\subseteq\))Let \(x \in U_1 + (U_2 + U_3) \Rightarrow x = u_1 + v_1\) for \(u_1 \in U_1\) and \(v_1 \in U_2 + U_3\). Since \(v_1 \in U_2 + U_3 \Rightarrow v_1 = u_2 + u_3\) for \(u_2 \in U_2, u_3 \in U_3\). Therefore \(x=u_1+v_1=u_1+(u_2+u_3)=(u_1+u_2)+u_3\in(U_1+U_2)+U_3\).

   (\(\supseteq\))Let \(x \in (U_1 + U_2) + U_3 \Rightarrow x = (u_1 + u_2) + u_3\) for \(u_1 \in U_1, u_2 \in U_2, u_3 \in U_3\) \(\Rightarrow x = u_1 + (u_2 + u_3) \in U_1 + (U_2 + U_3)\).

6. Let \(U\) be a subspace of \(V\), what is \(U + V\)?

   Let's show that \(U + V = V\):

   (\(\subseteq\))Let \(x \in U + V \Rightarrow x = u + v\) for \(u \in U, v \in V\). Since \(U \subseteq V\), we have \(u \in V\)

   Therefore \(x = u + v \in V\) since \(V\) is a vector space. So \(U + V \subseteq V\).

   (\(\supseteq\)​)Let \(x \in V\). Note that \(x = 0 + x\) and since \(U\) is a subspace \(0 \in U\). So therefore \(x = 0 + x \in U + V\)

   Therefore \(V \subseteq U + V\)

   So \(U + V = V\)

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