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11.1 Subspaces

  1. Consider the real vector space \(V = \{ f : \mathbb{R} \rightarrow \mathbb{R} : f \text{ is a function} \}\) and the subsets \(U_1, U_2\) of \(V\) given by \(U_1 = \{ f : \mathbb{R} \rightarrow \mathbb{R} : f \text{ is an even function} \}\) and \(U_2 = \{ f : \mathbb{R} \rightarrow \mathbb{R} : f \text{ is an odd function} \}\). Prove that \(V = U_1 \oplus U_2\).

    (i) \(U_1\) and \(U_2\) are subspaces.

    • \(U_1\):

    • \(f, g \in U_1 \Rightarrow (f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x) \Rightarrow f + g \in U_1\)

    • \(\lambda \in \mathbb{R}, f \in U_1 \Rightarrow (\lambda f)(-x) = \lambda f(-x) = \lambda f(x) = (\lambda f)(x) \Rightarrow \lambda f \in U_1\)
    • \(U_2\): exercise.

    (ii) \(V = U_1 + U_2\): let \(f \in V\), \(f(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}\).

    Let \(g(x) = \frac{f(x) + f(-x)}{2}\) and \(h(x) = \frac{f(x) - f(-x)}{2}\). Note that

    • \(g(-x) = \frac{f(-x) + f(x)}{2} = g(x) \Rightarrow g \in U_1\)
    • \(h(-x) = \frac{f(-x) - f(x)}{2} = -\frac{f(x) - f(-x)}{2} = -h(x) \Rightarrow h \in U_2\)

    Since \(f(x) = g(x) + h(x)\), then \(V = U_1 + U_2\).

    (iii) \(U_1 \cap U_2 = \{ 0 \}\).

    Let \(f \in U_1 \cap U_2 \Rightarrow \begin{cases} f(-x) = f(x) & (f \in U_1) \\ f(-x) = -f(x) & (f \in U_2) \end{cases}\)

    \(\Rightarrow f(x) = -f(x) \Rightarrow f(x) = 0 \quad \forall x \in \mathbb{R}\). Therefore \(f = 0\). So \(U_1 \cap U_2 = \{ 0 \}\).

    So we have \(V = U_1 \oplus U_2\).

    (iii)another way directly followed by definition

    If \(0=u_1\left(x\right)+u_2\left(x\right)\), then \(u_1\left(x\right)=u_2\left(x\right)=0\)​

    Let take \(u_1\in U_1\), then \(u_1\left(x)=u_1\left(-x\right)\right.\), let take \(u_2\in U_2\), then \(u_2\left(x)=-u_2\left(-x\right)\right.\)

    Since \(U_1\) is a subspace, then \(u_1(x)+(-u_1(x))=0\), then \(2u_1(x)=0\Rightarrow u_1(x)=0\)

    Since \(0=u_1\left(x\right)+u_2\left(x\right)\), then \(u_2(x)=0\)

  2. 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 a subspace \(W\) of \(\mathbb{F}^5\) such that \(\mathbb{F}^5 = U \oplus W\).

    Note that only the first 2 coordinates in \(U\) are arbitrary. So we can take \(W = \{ (0, 0, z, w, u) \in \mathbb{F}^5 : z, w, u \in \mathbb{F} \}.\)

    Why? Because we need to use \(U\) and \(W\) to construct \(F^5\). We already can ensure the first 2 is good, thus we consider the remaining 3

    Let's check that \(\mathbb{F}^5 = U \oplus W\).

    • \(U\) and \(W\) are subspaces (check).

    \(\mathbb{F}^5 = U + W\)

    Let \((x, y, z, w, u) \in \mathbb{F}^5\)

    We need to write \((x, y, z, w, u) = (x, y, x + y, x - y, 2x) + (0, 0, a, b, c) \quad \text{where } (x, y, x + y, x - y, 2x) \in U \text{ and } (0, 0, a, b, c) \in W,\)

    and find these values of \(a\), \(b\), and \(c\).

    We need \(\begin{cases} z = x + y + a \\ w = x - y + b \\ u = 2x + c \end{cases} \Rightarrow \begin{cases} a = z - x - y \in \mathbb{F} \\ b = w - x + y \in \mathbb{F} \\ c = u - 2x \in \mathbb{F} \end{cases}\)

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

    • \(U \cap W = \{0\}\). Let \((x, y, z, w, u) \in U \cap W\).

    • \(z = x + y\), \(w = x - y\), \(u = 2x\) since \((x, y, z, w, u) \in U\).

    • \(x = y = 0\) since \((x, y, z, w, u) \in W\).

    Therefore, \(x = y = z = w = u = 0\). So \(U \cap W = \{0\}\).

    So \(\mathbb{F}^5 = U \oplus W\).

  3. Is the vector \((3, -1, 0, -1)\) in the subspace of \(\mathbb{R}^4\) spanned by the vectors \((2, -1, 3, 2)\), \((-1, 1, 1, -3)\), and \((1, 1, 9, -5)\)?

    We just need to check if we can write
    ​\((3, -1, 0, -1) = a(2, -1, 3, 2) + b(-1, 1, 1, -3) + c(1, 1, 9, -5)\) for \(a, b, c \in \mathbb{R}\).

    Then \(\begin{cases}3 = 2a - b + c \\ -1 = -a + b + c \\ 0 = 3a + b + 9c \\ -1 = 2a - 3b - 5c\end{cases}\)

    \(\Rightarrow\begin{cases}b=2a+c-3\\ a=b+c+1=2a+c-3+c+1=2a+2c-2\\ 0=3a+2a+c-3+9c=5a+10c-3\\ -1=2a-6a-3c+9-5c=-4a-8c+9\end{cases}\)​

    \(\Rightarrow\begin{cases}b=2a+c-3\\ a=-2c+2\\ 0=-10c+10+10c-3\Rightarrow0=7\quad\text{contradiction}\\ -1=8c-8-8c+9\Rightarrow-1=1\quad\text{contradiction}\end{cases}\)​

    Therefore \(a, b, c\) can't exist.

  4. Let \(W\) be the set of all \((x_1, x_2, x_3, x_4, x_5) \in \mathbb{R}^5\) which satisfy the following set of linear equations: \(\begin{cases} 2x_1 - x_2 + \frac{4}{3} x_3 - x_4 = 0 \\ x_1 + \frac{2}{3} x_3 - x_5 = 0 \\ 9x_1 - 3x_2 + 6x_3 - 3x_4 - 3x_5 = 0 \end{cases}\)

    Find a finite set of vectors in \(\mathbb{R}^5\) which spans \(W\).

    Note that \(W\) can be written as \(Ax=0\) where \(A = \begin{pmatrix} 2 & -1 & \frac{4}{3} & -1 & 0 \\ 1 & 0 & \frac{2}{3} & 0 & -1 \\ 9 & -3 & 6 & -3 & -3 \end{pmatrix}\)
    So \(W\) is a subspace since it is all the solutions to \(Ax = 0\) (homogeneous system of linear equations).

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Definition: Let \(S\) be a set of vectors in a vector space \(V\). The subspace spanned/generated by \(S\) is the intersection of all subspaces of \(V\) which contain \(S\).

Theorem: The subspace spanned by \(S\) is the set of all linear combinations of elements in \(S\).

If \(S = \{s_1, \dots, s_n\}\), then \(\langle S \rangle = \{a_1 s_1 + \dots + a_n s_n : a_i \in \mathbb{F} \}\)

where \(\langle S \rangle\) is the subspace spanned by \(S\).

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