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What If I have product of matrices , which is nonsingular,

then each matrix has to be non-singular?

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  • $\begingroup$ Are the matrices required to be square? $\endgroup$ – Cameron Buie Oct 11 '14 at 22:19
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    $\begingroup$ @CameronBuie Well, singular is meaningless for nonsquare, so...:) $\endgroup$ – Alan Oct 11 '14 at 22:24
  • $\begingroup$ @CameronBuie If the matrices are not square, then the matrix is necessarily singular. $\endgroup$ – hickslebummbumm Oct 11 '14 at 22:27
  • $\begingroup$ @hickslebummbumm: That depends on the convention, as Alan's comment points out. Still, it was a silly question for me to ask. $\endgroup$ – Cameron Buie Oct 12 '14 at 16:16
  • $\begingroup$ @Alan: That depends on the convention, as hickslebummbumm's comment points out. Still, it was a silly question for me to ask. $\endgroup$ – Cameron Buie Oct 12 '14 at 16:17
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Yes. If $A_1,\ldots,A_p$ are $p$ matrices such that

$$A=A_1\cdots A_p$$ is an invertible matrix then $\det (A)=\det(A_1)\cdots\det(A_p)\ne0$. Can you take it from here?

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  • $\begingroup$ is there any other way to prove this without concept of determinant? $\endgroup$ – aaa Oct 11 '14 at 22:21
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Suppose AB is invertible, so $(AB)C=C(AB)=I$ for some matrix C.

Then $A(BC)=I$ and $(CA)B=I$, so $A^{-1}=BC$ and $B^{-1}=CA$.

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Given: $A$ is non-singular and has factorization $A = BC$.

Suppose $C$ is singular, it follows that $\exists x$ s.t. $C x = 0$. But then $A x = B C x = B 0 = 0 \implies$ $A$ is singular which is a contradiction. $C$ cannot be singular.

Now suppose $C$ is not singular but $B$ is. Again, $\exists x$ s.t. $Bx = 0$. Now take $y = C^{-1} x$, it follows $A y = B C y = B C C^{-1} x = B x = 0$ so $A$ is singular which is a contradiction.

Conclusion: $B$ and $C$ must be non-singular in order for $A$ to be.

You can extend this prove to any factorization of $A$.

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Yes, the product of some matrices is nonsingular if and only if every one of them is nonsingular (nonsingular means invertible).

Analogous to:

the product of some numbers is $\ne 0$ if and every one of the numbers is $\ne 0$.

This a property specific to the monoid of linear transformations on a finite dimensional vector space over a field ( perhaps non-commutative).

Let's consider the example of linear maps of an infinite dimensional vector space $\mathbb{R}^{(\mathbb{N})}$. We have two linear maps $A$, $B$ with their actions on the basis $(e_n)_{n \in \mathbb{N}}$

\begin{eqnarray}A e_n &=& e_{n+1} \\ B e_n &=& \left\{ \begin{array}{rl} e_{n-1} & \text{if } n \ge 1,\\ 0 & \text{if } n = 0, \end{array} \right. \end{eqnarray}

We have $BA = \mathbb{1}$ but neither $A$ nor $B$ are invertible.

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