# $AB$ is singular $\Leftrightarrow$ $A$ or $B$ is singular

Just for fun, I was solving few linear algebra problems. Hence, I came across the fact that-

1. $AB$ is singular $\Leftrightarrow$ $A$ or $B$ is(/are) singular.
2. $AB$ is not singular $\Leftrightarrow$ $A$ and $B$ both are not singular.

Notation: $A, B \in \mathbb{F}^{n\times n}$

My question is that let say we have proved the $2$nd fact, does it automatically imply the $1$st fact? Or, I have to prove it separately.

On a footnote, I have proved the second fact using Sylvester's rank inequality. Is there any other elegant way to do it?

Just for the preciseness, you can not use the fact that $\det(AB)=\det(A)\det(B)$ because you have to prove it before you use it, and for the proof let say you need the above mentioned facts.

• Since we do not have all the classical theorem (as I understand it) what is the definition of being singular here? (BTW Proposition 1 and 2 are strictly equivalent from a logical point of vue : $X\Leftrightarrow Y \text{ or }Z$ is logically equivalent to $not(X)\Leftrightarrow not(Y) \text{ and }not(Z)$ – Clément Guérin Dec 8 '15 at 8:39
• You are right : $(P\iff A \text{ or } B)\iff (\text{ no }P\iff\text{ no }A\text{ and }\text{ no }B)$ using de Morgan's laws. – Balloon Dec 8 '15 at 8:39
• @ClémentGuérin you can use the rank definition. Also, you can use the fact that determinant of a singular matrix is $0$. But as I mentioned, you can not use the result that $\det(AB)=\det(A)\det(B)$. – Rajat Dec 8 '15 at 8:42
• Do you have the rank equality? hence using "singular is equivalent to have a non-trivial element in the kernel"... – Clément Guérin Dec 8 '15 at 8:44
• Just a word about spelling: in English adjectives (like "singular") do not change form for the plural (not "matrices are singulars", but "matrices are singular"). And also: you can say "$A$ or $B$ is singular" without need to use "is/are". – Marc van Leeuwen Dec 8 '15 at 10:02

## 4 Answers

### On the logic part of the question...

Let $X$, $Y$, and $Z$ be propositions respectively that $AB$ is singular, $A$ is singular, and $B$ is singular. Statement 2 is:

\begin{equation}\neg X \leftrightarrow \neg Y \wedge \neg Z\end{equation}

Statement 1 is:

\begin{equation} X \leftrightarrow Y \vee Z \end{equation}

These are logically equivalent (simple application of De Morgan's laws).

### On the singular question...

Direction 1: If $B^{-1}$ exists and $A^{-1}$ exists then $B^{-1}A^{-1}$ is an inverse of $AB$.

Direction 2: If $(AB)^{-1}$ is an inverse of AB. Then:

• $(AB)^{-1}(AB) = I$

$((AB)^{-1}A)B = I$ hence inverse of $B$ exists (where $B^{-1} = (AB)^{-1}A$)

• $(AB)(AB)^{-1} = I$

$A(B(AB)^{-1}) = I$ hence inverse of $A$ exists

$A \in M(n,F)$ is singular iff any of the following equivalent conditions are met.

$A$ is not injective (has non-trivial kernel)

$A$ is not surjective(image rank less than domain rank)

$A$ has zero determinant

$A$ has $0$ as an eigenvalue

$A$ has no inverse

$A$ is a zero divisor

in validating the two logically equivalent statements you present concerning products, it is to some extent a matter of taste which route you take. for example if $AB$ has zero as an eigenvalue, let $v$ be an eigenvector, so $ABv=0$. clearly if $Bv \ne 0$ then $Bv$ is a non-trivial eigenvector for $A$ with the eigenvalue $0$

These statements are literally just contrapositives of each other. If you prove one, you get the other free of charge.

Assume that $AB$ is singular then there exists $v\in Ker(AB)$ with $v\neq 0$ (since the kernel is not trivial), it follows that $ABv=0$ then either $Bv=0$ and then $B$ is singular either $Bv\neq 0$ but $Bv$ is in the kernel of $A$ so $A$ is singular.

Assume that $B$ is singular, then take $v\in Ker(B)$ which is not trivial, clearly $ABv=0$ and $v$ is a non-trivial element in $Ker(AB)$ whence $AB$ is singular.

Assume that $A$ is singular then $rk(A)<n$ but $rk(AB)\leq rk(A)$ (trivial inequality) so that $rk(AB)<n$ and $AB$ is singular.

• Yes, I used the same method to prove my $2$nd fact. – Rajat Dec 8 '15 at 9:25