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Suppose $A$ and $B$ are $n$ by $n$ matrices, and $A B=I$. Prove from $\operatorname{rank}(A B) \leq$ $\operatorname{rank}(A)$ that the rank of $A$ is $n$. So $A$ is invertible and $B$ must be its two-sided inverse. Therefore $B A=I$ (which is not so obvious!).
Since $A$ is $n$ by $n, \operatorname{rank}(A) \leq n$ and conversely
$$
n=\operatorname{rank}\left(I_n\right)=\operatorname{rank}(A B) \leq \operatorname{rank}(A) .
$$
The rest of the problem statement seems to be “commentary,” and not further things to do.
(b) Find $A_1$ and $A_2$ so that $\operatorname{rank}\left(A_1 B\right)=1$ and $\operatorname{rank}\left(A_2 B\right)=0$ for $B=\left[\begin{array}{ll}1 & 1 \ 1 & 1\end{array}\right]$.
(a) That column $j$ of $B$ is a combination of previous columns of $B$ means precisely that there exist numbers $a_1, \ldots, a_{j-1}$ so that each row vector $\mathbf{x}=\left(x_i\right)$ of $B$ satisfies the linear relation
$$
x_j=\sum_{i=1}^{j-1} a_i x_i=a_1 x_1+\cdots+a_{j-1} x_{j-1}
$$
The rows of the matrix $A B$ are all linear combinations of the rows of $B$, and so also satisfy this linear relation. So, column $j$ is the same combination of previous columns of $A B$, as desired. Since a column is pivot column precisely when it is not a combination of previous columns, this shows that $A B$ cannot have previous columns and the rank inequality.
(b) Take $A_1=I_2$ and $A_2=0_2$ (or for a less trivial example $A_2=\left[\begin{array}{cc}1 & -1 \ 1 & -1\end{array}\right]$ ).
Explain why these are all false:
(a) The complete solution is any linear combination of $\mathbf{x}_p$ and $\mathbf{x}_n$.
(b) A system $A \mathbf{x}=\mathbf{b}$ has at most one particular solution.
(c) The solution $\mathbf{x}_p$ with all free variables zero is the shortest solution (minimum length $|x|$ ). Find a 2 by 2 counterexample.
(d) If $A$ is invertible there is no solution $\mathbf{x}_n$ in the nullspace.
(a) The coefficient of $\mathbf{x}_p$ must be one.
(b) If $\mathbf{x}_n \in \mathbf{N}(A)$ is in the nullspace of $A$ and $\mathbf{x}_p$ is one particular solution, then $\mathbf{x}_p+\mathbf{x}_n$ is also a particular solution.
(c) Lots of counterexamples are possible. Let’s talk about the 2 by 2 case geometrically: If $A$ is a 2 by 2 matrix of rank 1 , then the solutions to $A \mathbf{x}=\mathbf{b}$ form a line parallel to the line that is the nullspace. We’re asking that this line’s closest point to the origin be somewhere not along an axis. The line $x+y=1$ gives such an example.
Explicitly, let
$$
A=\left[\begin{array}{ll}
1 & 1 \
1 & 1
\end{array}\right] \quad \mathbf{b}=\left[\begin{array}{l}
1 \
1
\end{array}\right] \quad \text { and } \quad \mathbf{x}_p=\left[\begin{array}{ll}
\frac{1}{2} & \frac{1}{2}
\end{array}\right]
$$
Then, $\left|\mathbf{x}_p\right|=1 / \sqrt{2}<1$ while the particular solutions having some coordinate equal to zero are $(1,0)$ and $(0,1)$ and they both have $|\cdot|=1$.
(d) There’s always $\mathbf{x}_n=0$.
