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\textbf{Exercises on column space and nullspace}
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\begin{problem} (3.1 \#30.  {\em Introduction to Linear Algebra:} Strang) Suppose $\vb S$ and $\vb T$ are two subspaces of a vector space $\vb V.$

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\noindent (a) \textbf{Definition:} The sum $\vb S + \vb T$ contains all sums $\vb s + \vb t$ of a vector $\vb s$ in $\vb S$ and a vector $\vb t$ in $\vb T.$ Show that $\vb S + \vb T$ satisfies the requirements (addition and scalar multiplication) for a vector space.

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\noindent(b) If $\vb S$ and $\vb T$ are lines in $\vb R^m$, what is the difference between $\vb S + \vb T$ and $\vb S \cup \vb T$? That union contains all vectors from $\vb S$ and $\vb T$ or both. Explain this statement: \textit{The span of $\vb S \cup \vb T$ is $\vb S + \vb T$.}  
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\noindent(a) Let $\vb s, \vb s'$ be vectors in $\vb S,$ let $\vb t, \vb t'$ be vectors in $\vb T,$ and let $c$ be a scalar. Then

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$$(\vb s + \vb t) + (\vb s' + \vb t') = (\vb s + \vb s') + (\vb t + \vb t') \text{ and } c(\vb s + \vb t) = c\vb s + c \vb t.$$

Thus $\vb S + \vb T$ is closed under addition and scalar multiplication; in other words, it satisfies the two requirements for a vector space.

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\noindent(b) If $\vb S$ and $\vb T$ are distinct lines, then $\vb S + \vb T$ is a plane, whereas $\vb S \cup \vb T$ is only the two lines. The span of $\vb S \cup \vb T$ is the set of all combinations of vectors in this union of two lines. In particular, it contains all sums $\vb s + \vb t$ of a vector $\vb s$ in $\vb S$ and a vector $\vb t$ in $\vb T,$ and these sums form $\vb S + \vb T.$ 

Since $\vb S + \vb T$ contains both $\vb S$ and $\vb T$, it contains $\vb S \cup \vb T$. Further, $\vb S + \vb T$ is a vector space. So it contains all combinations of vectors in itself; in particular, it contains the span of $\vb S \cup \vb T.$ Thus the span of $\vb S \cup \vb T$ is $\vb S + \vb T.$

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\begin{problem} (3.2 \#18.) The plane $x - 3y - z = 12$ is parallel to the plane $x - 3y - x = 0$. One particular point on this plane is $(12,0,0).$ All points on the plane have the form (fill in the first components)

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$$\threebyone x y z = \threebyone {} 0 0 + y \threebyone {} 1 0 + z \threebyone {} 0 1.$$ 

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\begin{solution} The equation $x = 12 + 3y + z$ says it all:

$$\threebyone x y z \left(=\threebyone {12 + 3y + z} y z \right) = \threebyone {\framebox{$12$}} 0 0 + y \threebyone {\framebox{$3$}} 1 0 + z \threebyone {\framebox{$1$}} 0 1.$$ 
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\begin{problem} (3.2 \#36.) How is the nullspace $\mathbf{N}(C)$ related to the spaces $\mathbf{N}(A)$ and $\mathbf{N}(B),$ if $C = \twobyone A B$?
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\begin{solution} $\mathbf{N}(C) = \mathbf{N}(A) \cap \mathbf{N}(B)$ contains all vectors that are in both nullspaces: 

$$C\vb x = \twobyone {A\vb x} {B \vb x} = 0 $$

\noindent if and only if $A\vb x = 0 $ and $B \vb x =0.$
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