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If $A$ is an $m\times n$ matrix which can be row reduced Echelon to $X$ which has 1 or more zero rows and $B$ is an $n\times p$ matrix Is it true that that $AB$ can be row reduced to a matrix with 1 or more zero rows ?

When I asked this question this is the theorem that I wanted to prove for exercise 10 for Hoffman and kunze

  1. Prove the following generalization of Exercise 6. If $A$ is an $m \times n$ matrix, B is an $n \times m$ matrix and $n < m$, then $AB$ is not invertible.

But the statement in the linked question was wrong but if this statement was correct then it is easy to come up with an elementary proof for this exercise.

Mathematics enjoyer
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    I added a few words in the title so that it is easier for readers to see what is the subject. – Vincent Jan 12 '24 at 13:35

2 Answers2

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Yes, your claim is true. Reducing to echelon form is the same as choosing a new basis, so writing $A$ and $B$ in this basis, say $\tilde{A}$ and $\tilde{B}$ (i.e. such that at least the final row of $\tilde{A}$ is $0$), it is clear from the definition of multiplication that $\tilde{A}\tilde{B}$ will have its final row be $0$. Now, it should be clear that $\tilde{A}\tilde{B}= \tilde{(AB)}$, and so the echelon form of $\tilde{(AB)}$ is the same as the row echelon form of $AB$.

SomeCallMeTim
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Generalization of the problem This can be understood by assuming this multiplication of $AB$ where $A$ is a fixed Matrix of order $m×n$ as a linear transformation $$T:(\mathbb{R^{n×p}}) \rightarrow \mathbb{R^{m×p}}$$ As $$T(B) = AB $$ The proof for the following would be too lengthy, I'm giving you hint, Let $B1$ be the basis of $R^{n×p}$ and $B2$ be the basis of $R^{m×p}$ then the basis Matrix of this transformation $[T]_{B1}^{B2}$ will be a diagonal Matrix for some bases $B1$ & $B2$ (obv. Not for all $B1$ & $B2$) of order $mp×np$ having the Matrix $A$ (as block Matrix) at its diagonal.

Then the rank of this transformation will be $p$ times the rank of A. Now if one of the row of $A$ is zero then $p$ rows of $[T]_{B1}^{B2}$ will be zero, Which will result in $1$ row of $AB$ will be zero.

Clearly speaking

Rank of T = $p×$Rank$A$

So your claim is true.

You can deduce the following result too that

If A is not invertible then AB will also not invertible.

Gajjze
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