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\textbf{Exercises on orthogonal vectors and subspaces}
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\begin{problem} (4.1 \#7.  {\em Introduction to Linear Algebra:} Strang)
For every system of $m$ equations with no solution, there are numbers $y_1, . . . , y_m$ that multiply the equations so they add up to $0 = 1$. This is called {\em Fredholm's Alternative}:

\quote{Exactly one of these problems has a solution:\\ $\mb A\vb x = \vb b$ OR  $\mb A^T \vb y = \vb 0$ with $\vb y^T \vb b = 1.$}

If $\vb b$ is not in the column space of $\mb A$ it is not orthogonal
to the nullspace of $\mb A^T$. Multiply the equations $x_1 - x_2 =
1$, $x_2 - x_3 = 1$ and $x_1 - x_3 = 1$ by numbers $y_1$, $y_2$ and
$y_3$ chosen so that the equations add up to $0 = 1$.
\end{problem}

\solution{
Let $y_1  = 1$, $y_2  = 1$ and $y_3  = -1$.  Then the left-hand side of the 
sum of the equations is:
$$(x_1 - x_2) + (x_2 - x_3) - (x_1 - x_3) = x_1 - x_2 + x_2 - x_3 + x_3 - x_1  = 0,$$
and the right-hand side verifies that $\vb y^T \vb b = 1$:
$$1 + 1 - 1 = 1.$$
}


\begin{problem} (4.1\#32.)
Suppose I give you  four nonzero vectors $\vb r$, $\vb n$, $\vb c$ and  $\vb l$  in $\R^2$. 
\begin{enumerate}[a)]
\item What are  the  conditions  for  those  to be bases  for  the  four  fundamental  subspaces $C(\mb A^T)$, $N(\mb A)$, $C(\mb A)$, and $N(\mb A^T)$ of a $2$ by $2$ matrix? 
\item What is one possible matrix $\mb A$? 
\end{enumerate}
\end{problem}

\solution{
\begin{enumerate}[a)]
\item In order for $\vb r$ and $\vb n$ to be bases for $N(\mb A)$ and
  $C(\mb A^T)$, we must have $$\vb r \cdot \vb n = 0,$$ as the row space
  and null space must be orthogonal.  Similarly, in order for $\vb c$
  and $\vb l$ to form bases for $C(\mb A)$ and $N(\mb A^T)$ we need
  $$\vb c \cdot \vb l = 0,$$ as the column space and the left nullspace
  are orthogonal.  In addition, we need:
 $$\dim N(\mb A)+\dim C(\mb A^T) = n \quad \text{and} \quad \dim N(\mb A^T )+\dim C(\mb A) = m;$$ however, in this
  case $n = m = 1$, and as the four vectors we are given are nonzero
  both of these equations reduce to $1 + 1 = 2$, which is automatically
  satisfied.  

\item One possible such matrix is $\mb A = \vb c\vb r^T$.

Note that
  each column of $\mb A$ will be a multiple of $\vb c$, so it will
  have the desired column space.  On the other hand, each row of $\mb A$
  will be a multiple of $\vb r$, so $\mb A$ will have the desired row
  space.  The nullspaces don't need to be checked, as any matrix with
  the correct row and column space will have the desired nullspaces (as
  the nullspaces are just the orthogonal complements of the row and
  column spaces).
\end{enumerate}
}

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