Suppose $A,B$ are square matrices of size $n\times n$. Can $AB$ be invertible, even though both $A$ and $B$ are singular (not invertible)?

And if not, does it follow that if $A_1 \times A_2 \times ... \times A_n$ is the invertible product of square matrices, then all factors $A_i$ are invertible?

  • $\begingroup$ Are these square matrices? $\endgroup$
    – hardmath
    Nov 17, 2014 at 21:34
  • $\begingroup$ @hardmath yes, sorry, I had to specify that. All matrices are of the same shape and are square matrices. $\endgroup$
    – wvxvw
    Nov 18, 2014 at 7:49
  • $\begingroup$ I'll edit that in. It's good to have the complete problem statement in the body of the Question, even if it repeats some of what the title/subject line says. $\endgroup$
    – hardmath
    Nov 18, 2014 at 11:17
  • $\begingroup$ Note that the second part, while true, says something stronger than what one gets by induction on the first part. In the first part you assume neither of $A,B$ is invertible (both are singular), and ask if the product $AB$ can be invertible (it cannot). You need something stronger for the second part, namely that if either of $A,B$ is singular (not invertible), then the product is also singular. $\endgroup$
    – hardmath
    Nov 18, 2014 at 11:30
  • $\begingroup$ @hardmath thanks. $\endgroup$
    – wvxvw
    Nov 18, 2014 at 17:38

3 Answers 3


$$\det(A_1\cdot A_2\cdot\ldots\cdot A_n)\neq0\quad\Leftrightarrow\quad \det A_1\cdot\det A_2\cdot\ldots\cdot\det A_n\neq0\quad\Leftrightarrow\quad\det A_k\neq 0\quad \forall k$$

  • $\begingroup$ Thank you, this is very educational! $\endgroup$
    – wvxvw
    Nov 17, 2014 at 21:17
  • 1
    $\begingroup$ This works only if the matrices are square. $\endgroup$ Nov 18, 2014 at 4:57
  • 3
    $\begingroup$ @darijgrinberg Usually when one talks about invertible matrices, one refers to square matrices. Non-square matrix have no inverse: they can have a right-inverse or a left-inverse, but not both. $\endgroup$
    – Dario
    Nov 18, 2014 at 6:50

Suppose $A$ and $B$ are such that $(AB)^{-1} = C$. Then $ABC = I$, so $A$ has inverse $BC$. Likewise, $CAB = I$, so $B$ has inverse $CA$.

(This relies on knowing that any left inverse is also a right inverse and vice versa.)

  • $\begingroup$ I'm going to accept Adhvaitha's answer because it came first, but thank you too! $\endgroup$
    – wvxvw
    Nov 17, 2014 at 21:16

If $A$ and $B$ are not invertible, then $AB$ is also not-invertible. More generally, $$\text{rank}(AB) \leq \max(\text{rank}(A), \text{rank}(B))$$ And the answer to your second question is yes as well by inducting the above argument.

  • $\begingroup$ Oh, thanks! I couldn't find a way to prove invertibility of a matrix given by a large product, where I can only prove that factors are invertible, but not in that order. Now I can sleep soundly :) $\endgroup$
    – wvxvw
    Nov 17, 2014 at 21:13
  • $\begingroup$ In fact, rank(AB) is less than or equal to minimum of rank(A) and rank(B). $\endgroup$
    – Learning
    Dec 10, 2021 at 6:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .