11.1 Subspaces

1. Are the following subsets \(S\) of the vector space \(V\) subspaces?

   • (a) \(S = \{f : \mathbb{R} \to \mathbb{R} : f \text{ is a periodic function}\}\) where \(V = \{f : \mathbb{R} \to \mathbb{R} : f \text{ is a function}\}\).

   (a) Let's check 3 things:

   1. \(0 \in S \ \checkmark\)

      Since \(0(x) = 0 = 0(x + T)\), \(\exists \ T \in \mathbb{R}\). 2. \(\forall v,w\in S\),

      • \(f(x) = f(x + T_1) \ \forall T_1 \in \mathbb{R}\)
      • \(g(x) = g(x + T_2) \ \forall T_2 \in \mathbb{R}\)

      Then \((f+g)(x)=f(x)+g(x)=f(x+kT_1)+g(x+tT_1)\) but \(T_1\) may be not integer, then it doesn't hold 3. \(\forall v\in S,\forall\lambda\in\mathbb{R}\), \(\lambda v\in S\ \checkmark\)

      • \((af)(x)=af(x)=af(x+T_1)=(af)(x+T_1)\ \forall T_1\in\mathbb{R}\)

   Thus, \(S\) is not a subspace. * (b) \(S = \{A \in M_n(\mathbb{R}) : AB = BA \text{ for some } B \in M_n(\mathbb{R})\}\) where \(V = M_n(\mathbb{R})\).

   (b) Let's check 3 things

   1. \(0 \in S \ \checkmark\)

      Since \(0 \cdot 0 = 0 \cdot 0 = 0\) for \(A = 0\), \(\exists B = 0\) 2. \(\forall v, w \in S\), \(v + w \in S \ \checkmark\)

      \(A_1B_1=B_1A_1\), \(A_2B_2=B_2A_2\) \(,B_1,B_2=I_{n}\) \((A_1 + A_2) \cdot B = B \cdot (A_1 + A_2)\)

      Then \(A_1+A_2\in S\) when \(B = I_n\) 3. \(\forall v \in S\), \(\exists \lambda \in \mathbb{R}\), \(\lambda v\in S\ \checkmark\)

      Since \(AB = BA\), then \(a(AB)=(aI_{n})(AB)=(aI_{n})(BA)=a(BA)\)

      Thus \(S\) is a subspace

2. Prove or give a counterexample: if \(U_1\), \(U_2\), and \(W\) are subspaces of \(V\) such that \(U_1 + W = U_2 + W\), then \(U_1 = U_2\).

   No! Let \(V = \mathbb{R}^3\), \(W = \{(x, 0, 0) : x \in \mathbb{R}\}\)

   \(U_1 = \{(x, y, 0) : x, y \in \mathbb{R}\}\) \(U_2 = \{(0, y, 0) : y \in \mathbb{R}\}\)

   Then \(U_1 + W = U_2 + W = \{(x, y, 0) : x, y \in \mathbb{R}\}\)

   But \(U_1\neq U_2\)

3. Consider the following subspace of \(\mathbb{F}^5\):
   ​\(U = \{(x, y, x + y, x - y, 2x) \in \mathbb{F}^5 : x, y \in \mathbb{F}\}.\)
   Find three subspaces \(W_1\), \(W_2\), \(W_3\) of \(\mathbb{F}^5\) (none of them equal to \(\{0\}\)) such that \(\mathbb{F}^5 = U \oplus W_1 \oplus W_2 \oplus W_3\)​

   Since we know \((x, y, z, w, u) = (x, y, x + y, x - y, 2x) + (0, 0, z - x - y, w - x + y, u - 2x) \in U + W\)​\(\Rightarrow \mathbb{F}^5 = U + W\)

   Then we can let \(W_1=\{(0,0,0,0,u-2x)\in\mathbb{F}^5:x,u\in\mathbb{F}\}.\)

   \(W_2=\{(0,0,0,w-x+y,0)\in\mathbb{F}^5:x,y,w\in\mathbb{F}\}.\)

   \(W_3=\{(0,0,z-x-y,0,0)\in\mathbb{F}^5:x,y,z\in\mathbb{F}\}.\)

   Then we have \((x,y,z,w,u)=(x,y,x+y,x-y,2x)+(0,0,0,0,u-2x)+(0,0,0,w-x+y,0)+(0,0,z-x-y,0,0)\in U+W_1+W_2+W_3\)​\(\Rightarrow\mathbb{F}^5=U+W_1+W_2+W_3\)

   Then we need check \(0\) can be written in a unique way of \(U+W_1+W_2+W_3\)​

   Thus \((0,0,0,0,0)=(x,y,x+y,x-y,2x)+(0,0,0,0,u-2x)+(0,0,0,w-x+y,0)+(0,0,z-x-y,0,0)\)​

   Then \(x=y=0\), then we have \((0,0,0,0,0)=(0,0,0,0,0)+(0,0,0,0,u)+(0,0,0,w,0)+(0,0,z,0,0)\)​

   Thus \(u=0,w=0,z=0\) is unique

   Thus \(\mathbb{F}^5 = U \oplus W_1 \oplus W_2 \oplus W_3\)

4. Let \(S_1\) and \(S_2\) be subsets of a vector space \(V\) and denote the subspaces spanned by \(S_1\) and \(S_2\) as \(\operatorname{Span}(S_1)\) and \(\operatorname{Span}(S_2)\) respectively. Show that if \(S_1 \subseteq S_2\), then \(\operatorname{Span}(S_1) \subseteq \operatorname{Span}(S_2)\).

   In particular, if \(S_1 \subseteq S_2\) and \(\operatorname{Span}(S_1) = V\), deduce that \(\operatorname{Span}(S_2) = V\).

   Use definition \(span(v_1,\ldots,v_m)= \{a_1v_1+\cdots+a_mv_m:a_1,\ldots,a_m\in F\}\)

   Then take a linear combination in \(Span(S_1)\), then prove it is also in the \(Span(S_2)\)​

5. Consider the real vector space \(M_2(\mathbb{R})\). Determine whether the following matrices are in the span of
   ​\(S = \left\{ \begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \right\}:\)​

   • (a) \(\begin{pmatrix}1 & 2\\ -3 & 4\end{pmatrix}\).

   \(\begin{aligned} \begin{pmatrix} 1 & 2 \\ -3 & 4 \end{pmatrix} &= a \begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix} + b \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} + c \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \end{aligned}\)​

   \(\begin{cases} 1 = a + c \\ 2 = b + c \\ -3 = -a \\ 4 = b \end{cases} \implies \begin{cases} c = -2 \\ a = 3 \\ b = 4 \end{cases}\)​

   Yes! * (b) \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).
   ​\(\begin{aligned} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} &= a \begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix} + b \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} + c \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \end{aligned}\)
   ​\(\begin{cases} 1 = a + c \\ 0 = b + c \\ 0 = -a \\ 1 = b \end{cases} \implies \begin{cases} c = 1 \\ c = -1 \\ a = 0 \\ b = 1 \end{cases}\)​

   Contradiction!

   No!

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