# Prove if the product of $k$ matrices $A_1$ … $A_k$ is nonsingular, then each matrix $A_i$ is nonsingular.

I'm having trouble proving this without using determinants. I know how to prove it with the product of just two matrices, but I'm not sure how to generalize this to a product of k matrices. Is there a way to do this proof without determinants?

To clarify the question, each matrix is an $$n$$ by $$n$$ matrix.

For example, I know that if the determinant of the product is nonzero, then all of the determinants of the individual matrices must also be nonzero.

I also know that if a product of two matrices $$A$$ and $$B$$, $$AB$$ is nonsingular, then there exists a matrix $$C$$ so that $$C(AB) = I$$ and $$(AB)C = I$$, and so $$(CA)B = I$$ and $$A(BC) = I$$, so $$A$$ and $$B$$ are both invertible, and thus nonsingular.

I'm looking for a way to generalize this. Or just any other way to prove this without determinants.

Thanks so much!

• Take your argument for two and induct. – Randall Dec 5 '18 at 4:20
• Induction is how you would prove this for all $k \geq 2$. I would also suggest proving the contrapositive of the statement – JavaMan Dec 5 '18 at 4:32

Since each $$A_j : \mathbb F^n \to \mathbb F^n$$ ($$\mathbb F$$ can be $$\mathbb R$$ or $$\mathbb C$$), $$A_j$$ is invertible if and only if $$A_j$$ is injective by Rank-Nullity Theorem.

Now if the product $$A_1 \dots A_k$$ is invertible, then $$A_1 \dots A_k$$ is injective and it follows for every $$j$$, $$A_j$$ must be injective. So $$A_j$$ is invertible for each $$j$$.

Another way to present it: A matrix $$A$$ is invertible if and only if image of $$R^n$$ by $$A$$ is equal to $$R^n$$. It follows that if a given $$A_i$$ is singular, then the product is singular

It's already stated in comment, but I think it deserve an answer.

You say you are able to do it with 2 matrix $$A$$ and $$B$$. So, if $$k=2$$, you're done. Now, use induction to prove it's true for every $$k$$.

First, let P(k)="if the product of $$k$$ matrices $$A_1$$ ... $$A_k$$ is nonsingular, then each matrix $$A_i$$ is nonsingular"

To use induction, you need to prove:

1. $$\exists k \in \mathbb N, P(k)$$
2. $$\forall n \in \mathbb N, P(n) \implies P(n+1)$$

In your case, this is how you do it:

1. You can show that P(2) is true as you did in the question body, but you can even start with P(1), which read as "If $$A_1$$ is nonsingular, then $$A_1$$ is non singular".
2. Suppose P(n) and try to proove that P(n+1) holds. To do this, consider $$A=\prod_{i \in \{1\ldots n\}}{A_i}$$ and $$B=A_{n+1}$$, and use the same technic as in your question. You'll end up with 2 nonsingular matrix. One being $$A_{n+1}$$, and the other one being a nonsingular product of $$n$$ matrix. But since you supposed $$P(n)$$, this means that every matrix in the product is nonsingular itself.