SUBSPACES

By John Meth

• It is not necessary to read any proofs unless otherwise noted.
• Everywhere that Beezer uses $\mathbb{C}$, pretend that it is $\mathbb{R}$.
• I will use Beezer's notation Property Z, Property AC, and Property SC in these notes and on quizzes and exams.

• Subsection S: Subspaces
  
  • Focus on Definition S and Example SC3.
• Subsection TS: Testing Subspaces
  
  • Focus on Theorem TSS (not the proof), Example SP4, Example NSC2Z, Example NSC2A, and Example NSC2S.
  • Skip Theorem NSMS and Example RSNS. We will return to these in a later lesson.
• Skip Subsection TSS and Subsection SC.

• When proving Property AC or Property SC, you must prove the property holds for all vectors $u, v \in W$ or for all scalars and vectors $\alpha \in \mathbb{R}, u \in W$. You do not get to make an assumption about which particular vector or scalar you are dealing with. For example, the subset $V$ in Question 1 is a subspace. I will assert that $V$ is a subspace, but I will not prove all three conditions since I have only been asked to prove Property AC, as follows:
  
  

  Solution 1:

  The subset $V = \left\{ [x,y] \in \mathbb{R}^2 \, \middle\vert \, y=5x \right\} \subset \mathbb{R}^2$ is a subspace of $\mathbb{R}^2.$ If $u, v \in V$ then $u = [x_1, 5 x_1]$ and $v = [x_2, 5 x_2].$ Thus $$u+v = [x_1, 5 x_1] + [x_2, 5 x_2] = [x_1+x_2, 5 x_1 + 5 x_2] = [(x_1+x_2),5(x_1+x_2)]$$ which is again a vector in $V.$ Thus $V$ has Property AC.

  
  
• When disproving Property AC or Property SC, you need to create a counterexample that shows the property does not hold. That means you do get to make an assumption about which particular vector or scalar you are dealing with. In fact, it is your goal to create a particular vector or scalar that shows the property does not hold. For example, the subset $W$ in Question 2 is not a subspace. I will create one counterexample to show that Property AC does not hold:
  
  

  Solution 2:

  The subset $W = \left\{ [x,y] \in \mathbb{R}^2 \, \middle\vert \, x^2=y^2 \right\} \subset \mathbb{R}^2$ is not a subspace of $\mathbb{R}^2.$ For example, $u = [1,1] \in W,$ since $(1)^2 = (1)^2,$ and $v = [-1,1] \in W,$ since $(-1)^2 = (1)^2,$ but $u+v=[-1,1]+[1,1]=[0,2] \notin W$ since $(0)^2 \neq (2)^2.$

  
  
• When proving or disproving Property Z, our method is different. Since this property is about exactly one vector, $\vec{0}$, we don't have much room for creativity. You either prove that $\vec{0}$ is in $W$ or that it is not.

• 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.

    Definition 1: The Kernel, or Null Space, of an $m \times n$ matrix $A$ is the set $$\text{ker}(A)\ = \ \textrm{Nul}(A) \ = \ \left\{ \, {\bf x}\ \in\ {\mathbb R}^n \middle\vert \, A {\bf x} \ = \ {\bf 0}\, \right\}$$ 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 $$UT_n = \left\{ \, A \in M_n(\mathbb{R}) \, \middle\vert \, \left[ A \right]_{\, i,\, j} = 0 \textrm{ for } i > j \, \right\} = \left\{ \, \, \left[ \begin{array}{cccc} a_{\, 1, \, 1} & a_{\, 1, \, 2} & \cdots & a_{\, 1, \, n} \\ 0 & a_{\, 2, \, 2} & \cdots & a_{\, 2, \, n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & a_{\, n, \, n} \\ \end{array} \right] \, \, \middle\vert \, a_{\, i , \, j} \in \mathbb{R} \right\}. $$
  • You are asked to prove that $UT_n$ is a subspace of $M_n(\mathbb{R})$ in Exercise S.T20 from Beezer. This exercise is listed in the homework section.
  • 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 $$V_A = \left\{ \ c_1 \left[ \begin{array}{c} e^t \\ 0 \\ \end{array} \right] + c_2 \left[ \begin{array}{c} 0 \\ 1 \\ \end{array} \right] \ \ \middle\vert \ \ c_1 , c_2 \in \mathbb{R} \right\}, \textrm{ thus } \textrm{Nul}(A) \subseteq V_A.$$