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\section*{Lecture 13:  Exam 1 review}

This lecture is a review for the exam.  The majority of the exam is on
what we've learned about rectangular matrices.

\subsection*{Sample question 1}

Suppose $\vb u$, $\vb v$ and $\vb w$ are non-zero vectors in $\R^7$.  They span a subspace of $\R^7$.  What are the possible dimensions of that vector space?

\smallskip

The answer is $1$, $2$ or $3$.  The dimension can't be higher because a basis for this subspace has at most three vectors.  It can't be $0$ because the vectors are non-zero.

\subsection*{Sample question 2}

Suppose a $5$ by $3$ matrix $\mb R$ in reduced row echelon form has $r=3$ pivots.

\begin{enumerate}
\item What's the nullspace of $\mb R$?

Since the rank is $3$ and there are $3$ columns, there is no
combination of the columns that equals $\vb 0$ except the trivial one.
$N(\mb R) = \{\vb 0\}$.

\item Let $\mb B$ be the $10$ by $3$ matrix $\twobyone {\mb R} {2 \mb R}$.  What's the reduced row echelon form of $\mb B$?

Answer:  $\twobyone {\mb R} {\mb 0}$.

\item What is the rank of $\mb B$?

Answer:  3.

\item What is the reduced row echelon form of $\mb C = \twobytwo {\mb R} {\mb R} {\mb R} {\mb 0}$?

When we perform row reduction we get:
$$\twobytwo {\mb R} {\mb R} {\mb R} {\mb 0} \longrightarrow 
\twobytwo {\mb R} {\mb R} {\mb 0} {- \mb R} \longrightarrow 
\twobytwo {\mb R} {\mb 0} {\mb 0} {- \mb R} \longrightarrow 
\twobytwo {\mb R} {\mb 0} {\mb 0} {\mb R}.$$
Then we might need to move some zero rows to the bottom of the matrix.

\item What is the rank of $\mb C$?

Answer:  $6$.

\item What is the dimension of the nullspace of $\mb C^T$?

$m = 10$ and $r=6$ so $\mbox{dim } N(\mb C^T) = 10-6 = 4$.
\end{enumerate}

\subsection*{Sample question 3}

Suppose we know that $\mb A \vb x = \small \threebyone 242$ and
that: $$\vb x = \threebyone 200 + c\threebyone 110 + d \threebyone
001$$ is a complete solution.

Note that in this problem we don't
know what $\mb A$ is.

\begin{enumerate}

\item What is the shape of the matrix $\mb A$?

Answer:  $3$ by $3$, because $\vb x$ and $\vb b$ both have three components.

\item What's the dimension of the row space of $\mb A$?

From the complete solution we can see that the dimension of the nullspace of $\mb A$ is $2$, so the rank of $\mb A$ must be $3-2=1$.

\item What is $\mb A$?

Because the second and third components of the particular solution $\small \threebyone 200$ are
zero, we see that the first column vector of $\mb A$ must be $\small
\threebyone 121$.

Knowing that $\small \threebyone 001$ is in the nullspace tells us that the third column of $\mb A$ must be $\vb 0$.  The fact that  $\small \threebyone 110$ is in the nullspace tells us that the second column must be the negative of the first.  So,
$$\mb A = \threebythree 1 {-1} 0 2 {-2} 0 1 {-1} 0.$$

If we had time, we could check that this $\mb A$ times $\vb x$ equals $\vb b$.

\item For what vectors $\mb b$ does $\mb A \vb x = \vb b$ have a
  solution $\vb x$?

This equation has a solution exactly when $\vb b$ is in the column space of $\mb A$, so
when $\vb b$ is a multiple of $\small \threebyone 121$.  This makes
sense; we know that the rank of $\mb A$ is $1$ and the nullspace is
large.

In contrast, we might have had $r = m$ or $r = n$.

\end{enumerate}

\subsection*{Sample question 4}

Suppose: $$\mb B  = \mb C \mb D = \threebythree 110010101 \left[ \begin{array}{rrrr}
1 & 0 & -1 & 2\\
0 & 1 & 1 & -1\\
0 & 0 & 0 & 0
\end{array}\right].$$

Try to answer the questions below without performing
this matrix multiplication $\mb C \mb D$.

\begin{enumerate}

\item Give a basis for the nullspace of $\mb B$.

The matrix $\mb B$ is $3$ by $4$, so $N(\mb B) \subseteq \R^4$.
Because $\mb C = \small \threebythree 110010101$ is invertible, the
nullspace of $\mb B$ is the same as the nullspace of $\mb D = \left[
  \small \begin{array}{rrrr}
1 & 0 & -1 & 2\\
0 & 1 & 1 & -1\\
0 & 0 & 0 & 0
\end{array}\right]$.  Matrix $\mb D$ is in reduced form, so its special solutions form a basis for $N(\mb D) = N(\mb B)$:
$$
\left[\begin{array}{r}
1 \\
-1\\
1 \\
0 
\end{array}\right],
\left[\begin{array}{r}
-2 \\
1\\
0 \\
1 
\end{array}\right].$$

These vectors are independent, and if time permits we can multiply to
check that they are in $N(\mb B)$.

\item Find the complete solution to $\mb B \vb x = \small \threebyone 101$.

We can now describe any vector in the nullspace, so all we need to do
is find a particular solution.  There are many possible particular
solutions; the simplest one is given below.

One way to solve this is to notice that $\mb C \small \threebyone 100 =
\small \threebyone 1 0 1$ and then find a vector $\vb x$ for which
$\mb D \vb x = \small \threebyone 100$.  Another approach is to
notice that the first column of $\mb B = \mb C \mb D$ is $\small \threebyone
101$.  In either case, we get the complete solution:
$$\vb x = \left[ \begin{array}{r} 1 \\ 0 \\ 0 \\ 0\end{array}\right] + c
\left[\begin{array}{r}
1 \\
-1\\
1 \\
0 
\end{array}\right] +
d \left[\begin{array}{r}
-2 \\
1\\
0 \\
1 
\end{array}\right].$$

Again, we can check our work by multiplying.

\end{enumerate}

\subsection*{Short questions}

There may not be true/false questions on the exam, but it's a good
idea to review these:

\begin{enumerate}

\item Given a square matrix $\mb A$ whose nullspace is just $\{ \vb 0
  \}$, what is the nullspace of $\mb A^T$?

\smallskip

$N(\mb A^T)$ is also $\{ \vb 0 \}$ because $\mb A$ is square.

\item Do the invertible matrices form a subspace of the vector space
  of $5$ by $5$ matrices?

\smallskip

No.  The sum of two invertible matrices may not be invertible.  Also,
$\mb 0$ is not invertible, so is not in the collection of invertible
matrices.

\item True or false:  If $\mb B^2 = \mb 0$, then it must be true that $\mb B = \mb 0$.

\smallskip

False.  We could have $\mb B = \twobytwo 0100$.

\item True or false: A system $\mb A \vb x = \vb b$ of $n$ equations
  with $n$ unknowns is solvable for every right hand side $\vb b$ if the
  columns of $\mb A$ are independent.

\smallskip

True. $\mb A$ is invertible, and $\vb x = \mb A^{-1} \vb b$ is a
(unique) solution.


\item True or false:  If $m = n$ then the row space equals the column space.

\smallskip

False.  The dimensions are equal, but the spaces are not.  A good example to look at is $\mb B = \twobytwo 0100$.

\item True or false: The matrices $\mb A$ and $- \mb A$ share the same four spaces.

\smallskip

True, because whenever a vector $\vb v$ is in a space, so is $- \vb v$.

\item True or false: If $\mb A$ and $\mb B$ have the same four subspaces, then $\mb A$ is a multiple of $\mb B$.

\smallskip

A good way to approach this question is to first try to convince
yourself that it isn't true -- look for a counterexample.  If $\mb A$
is $3$ by $3$ and invertible, then its row and column space are both
$\R^3$ and its nullspaces are $\{\vb 0\}$.  If $\mb B$ is any other
invertible $3$ by $3$ matrix it will have the same four subspaces, and
it may not be a multiple of $\mb A$.  So we answer ``false''.

It's good to ask how we could truthfully complete the statement ``If $\mb A$ and $\mb B$ have the same four subspaces, then ...''

\item If we exchange two rows of $\mb A$, which subspaces stay the same?

\smallskip

The row space and the nullspace stay the same.

\item Why can't a vector $\vb v = \small \threebyone 123$ be in 
  the nullspace of $\mb A$ and also be a row of $\mb A$?

\smallskip

Because if $\vb v$ is the $n^{\mbox{th}}$ row of $\mb A$, the
$n^{\mbox{th}}$ component of the vector $\mb A \vb v$ would be $14$,
not $0$.  The vector $\vb v$ could not be a solution to $\mb A \vb v =
\vb 0$.

In fact, we will learn that the row space is perpendicular to the nullspace.

\end{enumerate}


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