Subspaces

A great way to understand vector spaces is to look at subspaces. A subspace of a vector space, $V,$ is a subset, $W \subseteq V,$ such that $W$ is also a vector space with the same operations as $V.$ There is a fast way to check that a subset is a subspace by checking just three properties. Finally, once we have the idea of a subspace, we can find a bounty of new vector space examples by looking at subspaces of vector spaces we already know - namely, the vector spaces listed in LinAlg5.html.

For some of the examples below, I am relying on the idea discussed in the lesson on vector spaces in the section Geometric Intuition, that a subspace of a vector space should have smaller dimension.

• Subspaces of $\mathbb{R}^n$
  
  • Since $\mathbb{R}^1$ is $1$ dimensional, it only has subspaces that are $1$ dimensional ("like" $\mathbb{R}^1$) or $0$ dimensional. Thus $\mathbb{R}^1$ only has two subspaces: the whole line, $\mathbb{R}^1,$ and the trivial subspace, $\left\{ \vec{0} \right\}.$
  • Since $\mathbb{R}^2$ is $2$ dimensional, it only has subspaces that are $0,$ $1,$ or $2$ dimensional. It has only one $2$ dimensional subspace, $\mathbb{R}^2 \subseteq \mathbb{R}^2,$ and only one $0$ dimensional subspace, $\left\{ \vec{0} \right\} \subseteq \mathbb{R}^2.$ But $\mathbb{R}^2$ has infinitely many subspaces that are $1$ dimensional. Every line through the origin is isomorphic to $\mathbb{R}^1$ and is a subspace of $\mathbb{R}^2.$
  • The subspaces of $\mathbb{R}^3$ are $\mathbb{R}^3,$ $\left\{ \vec{0} \right\},$ every line through the origin, and every plane through the origin.
  • The most important example of a subspace of $\mathbb{R}^n$ for differential equations is the following.
    
    
    The Kernel, or Null Space, of a matrix $A \in M_{m \times n}(\mathbb{R})$ is the set $$\textrm{ker}(A) \ = \ \textrm{Nul}(A) \ = \ \left\{ \, {\bf x} \ \in \ {\mathbb R}^n \middle\vert \ A {\bf x} \ = \ {\bf 0} \ \right\}$$ of all solutions ${\bf x}$ in ${\mathbb R}^n$ of the homogeneous equation $A {\bf x} \, = \, {\bf 0}.$

    Note: $\textrm{Nul}(A) \subseteq \mathbb{R}^n$ is always a subspace. In fact, every subspace of $\mathbb{R}^n$ is the null space of some matrix $A.$

    
    Let's prove that $\textrm{Nul}(A) \subseteq \mathbb{R}^n$ is a subspace.
    • Property Z: For $\vec{0} \in \mathbb{R}^n,$ $A \vec{0} = \vec{0} \in \mathbb{R}^m,$ so $\vec{0} \in \textrm{Nul}(A).$
    • Property AC: If $v_1, v_2 \in \textrm{Nul}(A),$ then $A v_1 = \vec{0}$ and $A v_2 = \vec{0}.$ So $A(v_1 + v_2) = Av_1 + Av_2 = \vec{0} + \vec{0} = \vec{0},$ thus $v_1 + v_2 \in \textrm{Nul}(A).$
    • Property SC: Let $\alpha \in \mathbb{R}$ and $v \in \textrm{Nul}(A).$ Then $A v = \vec{0}$ so $A(\alpha \, v) = \alpha \, Av = \alpha \, \vec{0}= \vec{0},$ thus $\alpha \, v \in \textrm{Nul}(A).$
    In the note in definition of $\textrm{Nul}(A),$ it says that every subspace of $\mathbb{R}^n$ is the null space of some matrix $A$. Let's see some examples.
    • Each 1 dimensional subspace of $\mathbb{R}^2$ is a line through the origin. The line with slope $m$ is the subspace $V = \left\{ [x,y] \in \mathbb{R}^2 \, \middle\vert \, y=mx \right\}$ which is the null space of $A = \left[ \begin{array}{cc} -m & 1 \end{array} \right].$ The line with infinite slope, the $y$-axis, is the null space of $A = \left[ \begin{array}{cc} 1 & 0 \end{array} \right].$
    • A 1 dimensional subspace of $\mathbb{R}^3$ is the $x$-axis. This subspace is $$V = \left\{ \ \left[ \begin{array}{c} x \\ 0 \\ 0 \\ \end{array} \right] \ \ \middle\vert \ \ x \in \mathbb{R} \right\} = \textrm{Nul} \left( \ \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right] \ \right).$$
    • A 2 dimensional subspace of $\mathbb{R}^3$ is the $xy$-plane. This subspace is $$V = \left\{ \ \left[ \begin{array}{c} x \\ y \\ 0 \\ \end{array} \right] \ \ \middle\vert \ \ x , y \in \mathbb{R} \right\} = \textrm{Nul} \left( \ \left[ \begin{array}{ccc} 0 & 0 & 1 \\ \end{array} \right] \ \right).$$
• Subspaces of $M_n(\mathbb{R})$
  
  • The space of diagonal matrices, $D_n,$ is the subspace of square matrices that have zeroes everywhere off the main diagonal. As a set they can be written as $$D_n = \left\{ \, A \in M_n(\mathbb{R}) \, \middle\vert \, \left[ A \right]_{\, i,\, j} = 0 \textrm{ for } i \neq j \, \right\} = \left\{ \, \, \left[ \begin{array}{cccc} a_{\, 1, \, 1} & 0 & \cdots & 0 \\ 0 & a_{\, 2, \, 2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & a_{\, n, \, n} \\ \end{array} \right] \, \, \middle\vert \, a_{\, i , \, i} \in \mathbb{R} \right\}. $$
  • The space of lower triangular matrices, $LT_n,$ is the subspace of square matrices that have zeroes everywhere above the main diagonal. As a set they can be written as $$LT_n = \left\{ \, A \in M_n(\mathbb{R}) \, \middle\vert \, \left[ A \right]_{\, i,\, j} = 0 \textrm{ for } i < j \, \right\} = \left\{ \, \, \left[ \begin{array}{cccc} a_{\, 1, \, 1} & 0 & \cdots & 0 \\ a_{\, 2, \, 1} & a_{\, 2, \, 2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ a_{\, n, \, 1} & a_{\, n, \, 2} & \dots & a_{\, n, \, n} \\ \end{array} \right] \, \, \middle\vert \, a_{\, i , \, j} \in \mathbb{R} \right\}. $$
  • The space of upper triangular matrices, $UT_n,$ is the subspace of square matrices that have zeroes everywhere below the main diagonal. As a set they can be written as $$UT_n = \left\{ \, A \in M_n(\mathbb{R}) \, \middle\vert \, \left[ A \right]_{\, i,\, j} = 0 \textrm{ for } i > j \, \right\} = \left\{ \, \, \left[ \begin{array}{cccc} a_{\, 1, \, 1} & a_{\, 1, \, 2} & \cdots & a_{\, 1, \, n} \\ 0 & a_{\, 2, \, 2} & \cdots & a_{\, 2, \, n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & a_{\, n, \, n} \\ \end{array} \right] \, \, \middle\vert \, a_{\, i , \, j} \in \mathbb{R} \right\}. $$
  • Notice that $D_n$ is actually a subspace of $LT_n$ and $UT_n,$ as well as a subspace of $M_n(\mathbb{R}).$ In fact, $D_n = LT_n \cap UT_n.$ In general, the intersection of any two subspaces is also a subspace. Can you prove it?
• Subspaces of $C(\mathbb{R})$
  Remember our nested list of vector spaces?

  $$\mathcal{P}_0 \subseteq \mathcal{P}_1 \subseteq \mathcal{P}_2 \subseteq \cdots \subseteq \mathcal{P}_n \subseteq \cdots \subseteq \mathcal{P} \subseteq C^{\, \infty}(\mathbb{R}) \subseteq \cdots $$ $$\cdots \subseteq C^{\, k}(\mathbb{R}) \subseteq \cdots \subseteq C^{\, 2}(\mathbb{R}) \subseteq C^{\, 1}(\mathbb{R}) \subseteq C^{\, 0}(\mathbb{R}) = C(\mathbb{R})$$

  Now we see that each subset is actually a subspace of any other vector space in the list in which it is contained. For example, $\mathcal{P}_3 \subseteq C^{\, 17}(\mathbb{R})$ is a subspace.
• Subspaces from differential equations
  
  • If $L$ represents an $n$th order, homogeneous O.D.E., then $V_L \subseteq C(\mathbb{R})$ is a subspace. If $A \in M_n(\mathbb{R}),$ then $V_A \subseteq C^{\, \infty}(\mathbb{R},\mathbb{R}^n)$ is a subspace.
  • If $L[\,y\,] = y^{(n)} = 0$, then $V_L = \mathcal{P}_{n-1}$. For example, two fundamental solutions to $L[\,y\,] = y^{\, \prime \prime} = 0$ are $y_1(t) = 1$ and $y_2(t) = t$. Thus $V_L = \left\{ c_1 \cdot 1 + c_2 \cdot t \, \middle\vert \, c_1, c_2 \in \mathbb{R} \right\} = \mathcal{P}_1$.
  • Consider the following three subsets of $C^{\, 1}(\mathbb{R})$:
    • $W_1 = \left\{ \, f \, \middle\vert \, f^{\, \prime}(1) = 0 \, \right\}$
    • $W_2 = \left\{ \, f \, \middle\vert \, f^{\, \prime}(1) = 1 \, \right\}$
    • $W_3 = \left\{ \, f \, \middle\vert \, f^{\, \prime}(t) = f(t) \, \right\}$
    As similar as they make look in description, they are all quite different. The subset $W_2$ is not a subspace. One can see immediately that it fails Property Z. It is a good exercise to show that it fails Property AC and Property SC as well. Both $W_1$ and $W_2$ are subspaces, so they are similar in that respect. Again, it is a very good exercise to prove this. They have different dimensions, though. In fact, $$W_3 = V_{\left\{y^{\,\prime}-y\,=\,0 \right\}} = \left\{ c \cdot e^t \, \middle\vert \, c \in \mathbb{R} \right\}$$ so it is 1 dimensional, $W_3 \cong \mathbb{R}^1.$ However, $W_1$ is infinite dimensional.
  • Returning to $A \in M_n(\mathbb{R}),$ we have $\textrm{Nul}(A) \subseteq V_A.$ For example, $$\textrm{if } \ A = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} \right], \textrm{ then } \textrm{Nul}(A) = \left\{ \ \left[ \begin{array}{c} 0 \\ c \\ \end{array} \right] \ \ \middle\vert \ \ c \in \mathbb{R} \right\}$$ and $$V_A = \left\{ \ c_1 \left[ \begin{array}{c} e^t \\ 0 \\ \end{array} \right] + c_2 \left[ \begin{array}{c} 0 \\ 1 \\ \end{array} \right] \ \ \middle\vert \ \ c_1 , c_2 \in \mathbb{R} \right\}, \textrm{ thus } \textrm{Nul}(A) \subseteq V_A.$$