If $A$ is 5 by 3 and $B$ is 3 by 5 (with dependent columns), is $AB = I$ impossible?

Question

Let me first introduce the problem. This is part of the quiz problem from MIT's 18.06 course (Spring 2012 semester, quiz 1, problem 3). My question is related to (b) but (a) is mentioned in the solution to (b). So I included (a) as well.

(a) Suppose three matrices satisfy $AB = C$. If the columns of $B$ are dependent, show that the columns of $C$ are dependent.

Solution
The columns of $B$ being dependent means by definition that there is a vector $x \neq 0$ such that $Bx = 0$. But then we also have $$Cx = (AB)x = A(Bx) = A(0) = 0.$$ which means that the columns of $C$ are dependent.

(b) If $A$ is 5 by 3 and $B$ is 3 by 5, show using part (a) or otherwise that $AB = I$ is impossible.

Solution
The columns of B are dependent, since these are five vectors in $\mathbb R^3$, and $5>3$. Thus, by part (a), the columns of $AB$ must be dependent. However, columns of $I$ are independent, so $AB$ can never equal $I$. [Note: Switching the order matters here. One can indeed find a $3\times 5$ matrix $A$, and a $5\times 3$ matrix $B$ such that $AB = I$ is the $3 \times 3 $ identity - hence any "proof" that is insensitive to the order of $A$ and $B$ must be flawed.]

My question is related to the text inside the angled bracket ("Note: Switching..."). I don't understand how "One can indeed find a $3\times 5$ matrix $A$, and a $5\times 3$ matrix $B$ such that $AB = I$ is the $3 \times 3 $ identity."

The following is my reasoning.

In this reverse scenario, $A$ is $3 \times 5 $ and $B$ is $5 \times 3 $. So $C = AB$ is $3 \times 3$. Because $B$ has dependent columns, its rank $r_B < 3$. $AB = C$ means that each row of $C$ is a linear combination of the rows of $B$ (coefficient being the corresponding row of $A$). Then the dimension of the row space of $C$ is at most that of $B$ (i.e., $r_B$). Because a dimension of a row space is equal to a rank, the rank of $C$, $r_C \leq r_B < 3$. Because $C$ has 3 columns, $r_C < 3$ implies that the columns of $C$ are dependent.

Is my reasoning incorrect?

Thank you in advance!

Answer 1

In your reasoning

Because $B$ has dependent columns, its rank $r_B < 3$.

is not generally true. In the original proof, since $B$ is a $3 \times 5$ matrix, it is legit to the columns of $B$ are dependent. However, in the later setting, $B$ is a $5 \times 3$ matrix, this is not necessarily true.

In fact, whether $B$ is full column rank determines whether there exists such a $3 \times 5$ matrix $A$ such that $AB = I_3$.

It may be somehow confusing when we have different definitions for the matrices $A$ and $B$ in the two statements. We can unify and conclude as follows:

Suppose $B$ is a $3 \times 5$ matrix. Then

1. For any $5 \times 3$ matrix $A$, the $5 \times 5$ matrix $AB$ cannot be the identity matrix $I_5$.
2. There may or may not exist a $5 \times 3$ matrix $A$ such that the $3 \times 3$ matrix $BA$ is the identity matrix $I_3$. The existence depends on whether $\text{rank}(B) = 3$.