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\textbf{Exercises on independence, basis, and dimension}
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\begin{problem} (3.5 \#2.  {\em Introduction to Linear Algebra:} Strang) Find the largest possible number of independent vectors among: 
$$ \vb v_1= \fourbyone 1 {-1} 0 0, \vb v_2 = \fourbyone 1 0 {-1} 0, \vb v_3 =\fourbyone 1 0 0 {-1},$$ $$\vb v_4 = \fourbyone 0 1 {-1} 0, \vb v_5 = \fourbyone 0 1 0 {-1}  \text{ and } \vb v_6 = \fourbyone 0 0 1 {-1}.$$

\end{problem}

\solution{ Since $\vb v_4 = \vb v_2 - \vb v_1, \vb v_5 = \vb v_3 - \vb v_1,$ and $\vb v_6 = \vb v_3 - \vb v_2,$ the vectors $\vb v_4, \vb v_5,$ and $\vb v_6$ are dependent on the vectors $\vb v_1, \vb v_2$ and $\vb v_3$. To determine the relationship between the vectors $\vb v_1, \vb v_2$ and $\vb v_3$  we apply row reduction to the matrix $[\vb v_1 \hspace{4pt} \vb v_2 \hspace{4pt} \vb v_3]$:
\begin{small}
$$\left [ \begin{array}{rrr} 
1 & 1 & 1 \\
-1 & 0 & 0 \\
0& -1 & 0 \\
0& 0 & -1\end{array} \right ] \longrightarrow 
\left [ \begin{array}{rrr} 
1 & 1 & 1 \\
0 & 1 & 1 \\
0& -1 & 0 \\
0& 0 & -1\end{array} \right ] \longrightarrow
\left [ \begin{array}{rrr} 
1 & 1 & 1 \\
0 & 1 & 1 \\
0& 0 & 1 \\
0& 0 & -1\end{array} \right ] \longrightarrow \left [ \begin{array}{rrr} 
1 & 1 & 1 \\
0 & 1 & 1 \\
0& 0 & 1 \\
0& 0 & 0\end{array} \right ].$$
\end{small}

As there are three pivots, the vectors $\vb v_1, \vb v_2,$ and $\vb v_3$ are independent. Therefore the largest number of independent vectors among the given six vectors is \textbf{three.}
}

\begin{problem} (3.5 \#20.) Find a basis for the plane $x - 2y + 3z = 0$ in $\mathbb{R}^3$. Then find a basis for the intersection of that plane with the $xy$ plane. Then find a basis for all vectors perpendicular to the plane. 
\end{problem} %issue; this is not the question in the book

\solution{ This plane is the nullspace of the matrix
$$A = \threebythree 1 {-2} 3 0 0 0 0 0 0.$$
The special solutions to $A\vb x=\vb 0$ are $$\vb v_1 = \threebyone 2 1 0  \text{ and }\vb v_2 = \threebyone {-3} 0 1.$$
These form a basis for the nullspace of $\mb A$ and thus for the plane. 

The intersection of this plane with the $xy$ plane contains $\vb v_1$ and does not contain $\vb v_2$; the intersection must be a line.  Since $\vb {v_1}$ lies on this line it also provides a basis for it. 

Finally, we can use ``inspection'' or the cross product to find the vector 
$$\vb v_3 = \threebyone 1 {-2} 3,$$ 
which is perpendicular to both $\vb v_1$ and $\vb v_2$.  It is therefore perpendicular to the plane. Since the space of vectors perpendicular to a plane in $\mathbb{R}^3$ is one-dimensional, $\vb v_3$ serves as a basis for that space. 
}

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