Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given $A,B\in R^{n\times n}$ such that A is singular, and B is non-singular. Is $(AB)$ always singular? If so, how do I prove it?

share|cite|improve this question
up vote 8 down vote accepted

The easiest way to see this is by looking at the determinant, since $\det(AB) = \det(A)\det(B)$ and a matrix $A$ is singular iff $\det(A) = 0$.

share|cite|improve this answer
It's so simple... For some reason, I thought the proof would be harder... – Paul Jun 28 '12 at 9:27

Without determinants: if $A$ and $AB$ are invertible, then

$$\begin{cases} \big((AB)^{-1}A\big)B=(AB)^{-1}(AB)=I \\[8pt] B=A^{-1}(AB)\implies B\big((AB)^{-1}A\big)=I, \end{cases}$$

hence $B$ is invertible, and in particular $B^{-1}=(AB)^{-1}A$.

(Applicable to arbitrary multiplicative monoids with identity, including rings under multiplication.)

share|cite|improve this answer
So the same result still holds over non-commutative coefficient fields (or rings); the determinant approach does not work there. – Marc van Leeuwen May 19 at 8:26

Yes, it is singular, since the following holds: $$ \det(AB)=\det(A)\det(B) $$

share|cite|improve this answer
Of course!! :) I completely forgot about that:) Thanks for reminding me. – Paul Jun 28 '12 at 9:25

I wished to put forth something that does not involve determinants:

$A, B \in \mathbb{R}^{n\times n}$. If you consider $C=A\times B$, The $i^{th}$column of $C$ is a linear combination of the columns of $A$ with the $i^{th}$ column of $B$ as the weights.

If $A$ were singular, the columns of $A$ would not be independent and hence the columns of $C (=A\times B)$ would not be independent and hence $C$ would have dependent rows and hence would be singular.

On the other hand, if $B$ were singular, We can consider the row-picture of the multiplication suggesting that the product of A and B is actually the linear sum of the rows of B. If the rows of B are not independent, then the rows of $C$ are not independent.

share|cite|improve this answer

If you think of the matrix in terms of being a linear transformation on $\mathbb{R}^n$, then a nonsingular matrix has full rank. A singular matrix diminishes rank. Once you diminish rank, there is no way back. Hence the product of any square matrix with a singuluar matrix is singular.

share|cite|improve this answer
That's exactly what I was thinking of, intuitively... I just needed a way to justify it analytically. – Paul Jun 28 '12 at 12:27
That is not just intuitive. With a little care, it is not hard to structure an argument along these lines. – ncmathsadist Jun 28 '12 at 14:01

Here is another way to show this without determinants. Since $B$ is non-singular $B(AB)B^{-1}=BA$ will be singular iff $AB$ is singular. But if $x\ne 0$, $Ax=0$ then $BAx=0$ also, so $BA$ and hence $AB$ is singular.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.