If $A$ is an $m \times n$ matrix and $B$ is an $n \times m$ matrix and $n

Question

If $A$ is an $m \times n$ matrix and $B$ is an $n \times m$ matrix and $n < m$, then the product $AB$ is not invertible.

This exercise is from the first chapter sixth section of Hoffman and Kunze's Linear Algebra book. Meaning it has not gone over rank, determinants or linear maps which all the other answers to this exercise has relied on. The things that have been discussed up until this point are:

    - Systems of Linear Equations 
    - Matrices
    - Elementary Row Operations
    - Row-Reduced Echelon Matrices
    - Matrix Multiplication
    - Invertible Matrices

I'm sorry to say that I don't really have much to say about what I've already tried and I would appreciate any hints that would point me in the right direction using the previously stated tools. Thank you.

Edit: My question is different to the one Dietrich Burde linked due to the answers in that question relies on things not yet covered in the book.

Edit 2: I want to clarify something in my post. I am not interested in solutions that use the concepts of rank, determinants or linear maps seeing as I'm sure the authors were smart enough to realize that they cannot make an exercise which is impossible to do without using linear algebra concepts that haven't yet been introduced in their introductory linear algebra textbook.

Please review what the textbook says about invertible matrices. You should try to apply one of the results about them to show $AB$ cannot be invertible. — hardmath, Apr 30 '20 at 13:31
Does this answer your question? If $A$ is an $m\times n$ matrix, $B$ is an $n\times m$ matrix and $n< m$, then $AB$ is not invertible. — Dietrich Burde, Apr 30 '20 at 13:33

score 0 · Answer 1 · answered Apr 30 '20 at 13:38

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Hint:$\operatorname{rank}(AB)\le\max(\operatorname{rank}A,\operatorname{rank}B)$.

$AB$ is invertible if and only its rank is … What do you know about the ranks of $A$ and $B\,$?

answered Apr 30 '20 at 13:38

Bernard

175,478

1

The book hasn't yet gone over what rank is, which I have implied in my post. – Dippset Apr 30 '20 at 13:41
Do you know what the rank of a linear map is? – Bernard Apr 30 '20 at 13:43
I do not, the book has not gone over linear maps yet, or rank for that matter. – Dippset Apr 30 '20 at 13:50
You've seen row reduction, and do not know about the rank of a matrix?? – Bernard Apr 30 '20 at 13:52
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That is correct, I searched for the word rank in the book and it is defined in the chapter where vector spaces are introduced. – Dippset Apr 30 '20 at 13:55
Do you know that matrices represent linear maps? – Bernard Apr 30 '20 at 13:57
If you read my post you can see that I specifically stated that the book has not yet gone over rank, determinants nor linear maps. – Dippset Apr 30 '20 at 14:00

score 0 · Accepted Answer · answered Apr 30 '20 at 15:49

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Hard to answer definitively without knowing exactly what's in the book, but here's a possibility involving matrix equations. Consider the two matrix equations: \begin{align*} Ax &= b \tag{1} \\ ABx &= b \tag{2} \end{align*} If $AB$ is invertible, then given any $b$, $(2)$ has a (unique) solution. Let $x_0$ be this solution. Then, $(1)$ has at least one solution as well: $x = Bx_0$. This means that, given any $b$, $(1)$ has a solution, which means that every column vector $b \in \Bbb{R}^m$ belongs in the columnspace of $A$. But, $A$ has $n < m$ columns, which is too few to span all of $\Bbb{R}^m$. Thus, we have a contradiction.

answered Apr 30 '20 at 15:49

user780985

666

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Span isn’t either a concept introduced yet. But how about this: we know that Bx = 0 has a non trivial solution which implies that ABx = 0 has a non trivial solution, thus AB is not invertible. – Dippset Apr 30 '20 at 15:58
@Dippset That works.too. If it makes sense with respect to the book so far, even better. – user780985 Apr 30 '20 at 16:10

If $A$ is an $m \times n$ matrix and $B$ is an $n \times m$ matrix and $n

2 Answers2