If $\mathbf{ABC}$ non-singular prove that $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ non-singular too

Question

I am interested in the following exercise, and I have tried to solve it with the following way. Firstly, could you please check the correctness of the given answer. Secondly, can you give an alternative answer?

$\textbf{Exercise}$: If the product $\mathbf{M} = \mathbf{ABC}$ of three square matrices is invertible, then $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ are invertible.

$\textbf{Answer}$:

Part 1: If $\mathbf{C}$ is singular, there is $\mathbf{x} \neq 0$ such that $\mathbf{Cx} = 0 \iff \mathbf{ABCx} = 0 \iff \mathbf{Mx} = 0$. This comes in contradiction with the fact the $\mathbf{M}$ is non-singlular by the exercises definition. Thus $\mathbf{C}^{-1}$ exists. Using the last statement, we may write $\mathbf{MC}^{-1} = \mathbf{AB}$. Knowing that $\mathbf{M}$ and $\mathbf{C}^{-1}$ are invertible, we are interested to prove the invertibily of $\mathbf{MC}^{-1}$, so as to continue with similar way with the prof of matrix $\mathbf{B}$ invertibility. We may have:

$$\mathbf{J} = \mathbf{MC}^{-1} \iff \mathbf{M}^{-1}\mathbf{J} = \mathbf{C}^{-1} \iff \mathbf{CM}^{-1}\mathbf{J} = \mathbf{I}$$

This means that matrix $\mathbf{MC}^{-1}$ has a left inverse given by $\mathbf{J}^{-1} = \mathbf{CM}^{-1}$.

Part 2: Based on the last statement, and similarly with the invertibility prof we followed for $\mathbf{C}$, If $\mathbf{B}$ is singular, there is $\mathbf{x} \neq 0$ such that $\mathbf{Bx} = 0 \implies \mathbf{ABx} = 0 \implies \mathbf{MC}^{-1}\mathbf{x} = 0 \implies \mathbf{J}\mathbf{x} = 0$. This comes in contradiction with the fact the $\mathbf{J}$ is non-singlular. Thus $\mathbf{B}^{-1}$ is invertible.

Part 3: Finally, we may write $\mathbf{A} = \mathbf{MC}^{-1}\mathbf{B}^{-1} = \mathbf{J}\mathbf{B}^{-1}$. Matrix $\mathbf{A}$ is non-singular, because:

$$(\mathbf{JB}^{-1})^{-1}\mathbf{JB}^{-1} = \mathbf{I}~~\text{and}~~\mathbf{JB}^{-1}(\mathbf{JB}^{-1})^{-1} = \mathbf{I}$$

Thank you!

^{PS: Changes have been made taking into account users comments. I hope the post is improved.}

Answer 1

Your proof is unclear when you say that $J$ is invertible because $JJ^{-1}=J^{-1}J=I$. Thats the definition of being invertible, not the proof!

As an alternative proof, note that $0\neq\det(ABC)=\det(A)\det(B)\det(C)$.

EDIT

Your proof can be fixed to be valid: $J$ is indeed invertible, and you can prove it (e.g., by writing the explicit form of $J^{-1}$ in terms of $M$ and $C$)

Answer 2

You're misusing $\iff$, to say the least.

You start correctly: if $\bf C$ is not invertible, then $\bf Cx=0$ for some $\bf x\ne0$; then $\bf Mx=ABCx=AB0=0$, against the invertibility of $\bf M$. Therefore $\bf C$ is invertible.

The rest is confuse. You can just go on the same way, because the product of invertible matrices is invertible, so $\bf AB=MC^{\rm-1}$ is invertible and you can do the same as before, starting with $\bf By=0$.

Actually, you get an easier proof by noting that you can establish first the result for a product of two matrices.

If $\bf AB$ is invertible, then $\bf A$ and $\bf B$ are invertible.

Indeed, if $\bf B$ is not invertible, then $\bf Bz=0$ for some $\bf z\ne0$. Therefore $\bf ABz=0$, contradicting the invertibility of $\bf AB$. Hence $\bf B$ is invertible and so also is $\bf A=(AB)B^{\rm-1}$.

Now apply the result to $\bf M=ABC=(AB)C$, getting that $\bf C$ and $\bf AB$ are invertible.

Answer 3

In part $1$, you shouldn't write $\iff$ but $\implies$. You deduce that $C$ is invertible which implies that $C^{-1}$ is invertible. Your justification why $J$ is invertible doesn't make sense - you can't even write $J^{-1}$ without justifying before why $J$ is invertible. Similarly for parts $2$ and $3$.

You can correct your argument by noting that $J$ is the product of two invertible matrices and justifying why the product of two invertible matrices is invertible.

Answer 4

I'll leave the "firstly" case to others.

Secondly: For some integer $n$, $A,B$ and $C$ are $n\times n$ matrices over some field $K$. Because matrix multiplication is associative, it's enough to prove that

$AB$ is singular iff either $A$ or $B$ is singular.

Using the $K$-valued $n\times n$ determinant, $\operatorname{det}$, this is clear, because

• $\operatorname{det}(AB) = \operatorname{det}(A)\operatorname{det}(B)$, and
• $\operatorname{det}(A) = 0$ iff $A$ is singular.

Note that the statement above is equivalent to $$ \text{$AB$ is non-singular $ iff$ both $A$ and $B$ are non-singular. } $$

Now, if $ABC$ is non-singular, then $AB$ and $C$ are non-singular, hence $A, B, C$ are all non-singular.