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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?

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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 $\det(A) = 0$.

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  • $\begingroup$ It's so simple... For some reason, I thought the proof would be harder... $\endgroup$ – Paul Jun 28 '12 at 9:27
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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.)

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  • $\begingroup$ So the same result still holds over non-commutative coefficient fields (or rings); the determinant approach does not work there. $\endgroup$ – Marc van Leeuwen May 19 '16 at 8:26
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Yes, it is singular, since the following holds: $$ \det(AB)=\det(A)\det(B) $$

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  • $\begingroup$ Of course!! :) I completely forgot about that:) Thanks for reminding me. $\endgroup$ – Paul Jun 28 '12 at 9:25
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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.

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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.

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

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