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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!

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

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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

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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.
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