Proof verification : Assume $A$ is a $n×m$ matrix, and $B$ is $m×n$. Prove that $AB$, an $n×n$ matrix is not invertible, if $n>m$.

Question

Proof verification : Assume $A$ is a $n×m$ matrix, and $B$ is $m×n$. Prove that $AB$, an $n×n$ matrix, is not invertible, if $n>m$.

So, I have to proof : $(A∈M_{n×m}∧B∈M_{m×n}∧n>m)⇒(¬∃(AB)^{-1}:[AB(AB)^{-1}=I])$.

For the sake of contradiction, I assume $∃(AB)^{-1}:[AB(AB)^{-1}=I]$, So, since $(AB)^{-1}=B^{-1}A^{-1}$, it must be true that $(AB)(B^{-1}A^{-1})=I$, but when using the associative property $A(BB^{-1})A^{-1}=I$, $BB^{-1}$ cannot be $I$ because for $B^{-1}$ to exist, $B$ has to be a square matrix in first place, and since $n>m$ also means that $n≠m$, it leads to a contradiction. $$∴¬∃(AB)^{-1}:[AB(AB)^{-1}=I]$$

Is this proof ok? my book suggests to show that there is a vector $x≠0$ such that $ABx=0$, but I didn't figure out how to prove it that way, so I came up with this one. Is it fine?

Thanks in advance.

Answer 1

The hint actually makes the whole proof way simpler.

Consider $$ Bx=0. $$ Since the number of variables is bigger than the number of equations there is a non-zero vector $\hat x$ such that $B\hat x=0$.

Now consider $ABx=0$. Since $AB\hat x=A(B\hat x)=0$ and hence $\hat x$ also solves $(AB)x=0$, which implies that $AB$ is not invertible.

Answer 2

There is a serious (in fact, I would say fatal) error in your proof.

The identity $(AB)^{-1}=B^{-1}A^{-1}$ only holds when $A$ and $B$ are invertible. But in your case $A$ is not square and therefore cannot be invertible; and likewise for $B$.

You have sort of said this afterwards, but you don't seem to have realised that it has destroyed your whole argument.

Answer 3

As the matrix rank and column rank of a matrix are same, for a $m\times n$ matrix $M$, $$\mathrm{rank}(M)\le\min\{m,n\}$$. So $\mathrm{rank}(B)\le m<n$. But $AB$ is just linear combinations of the rows of $B$, and hence number of independent rows in $AB$ is $\le$ the number of independent rows of $B$. Hence $$\mathrm{rank}(AB)\le\mathrm{rank}(B)\le m<n$$. But $AB$ is an $n\times n$ matrix. Hence not all the rows of $AB$ are independent. Hence it is not invertible.