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Let $A$ and $B$ be square matrices, where the product $AB$ is invertible. Is $A$ necessarily invertible?

(HINT: You cannot assume that $(AB)^{-1}= B^{-1} A^{-1}$, since we do don't know $A$ and $B$ are invertible. You can set $(AB)^{-1}=C$, and think about what the product $(AB)C$ would equal, to attempt to find an inverse to $A$)

user64742
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Lil
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4 Answers4

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Hint: $AB$ is invertible implies $AB$ is surjective, we suppose it is an endomorphism of $E$ for every $y\in E$ there exists $x\in E$ such that $AB(x)=y$ thus $A(B(x))=y$ and $A$ represents a surjective linear map, so $A$ is invertible, since it is a square matrix.

Tsemo Aristide
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$AB$ is invertible right? So assume there is a matrix $C$ where $ABC = I$ (identity). Then $(AB)C = A(BC) ,A(BC)= I.

Take the inverse now. ,

$A^{-1}A(BC) = I A^{-1}$

$BC = A^{-1}$

Since$ BC = A^{-1}$ then $A$ is invertible.

user64742
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user321151
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  • I'm a litte confused by your formatting for taking the inverse... – Lil Mar 09 '16 at 14:39
  • should taking the inverse of the equation be done on the same side for both sides of the equation? – Lil Mar 09 '16 at 14:41
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Multiplication of matrices is associative. $$I=(AB)(AB)^{-1}\Rightarrow I=A(B(AB)^{-1})\Rightarrow A^{-1}=B(AB)^{-1}$$ Reason similarly with $I=(AB)^{-1}AB$ to find an inverse for $B$.

Matt Samuel
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By the invertability of $AB$ we deduce that $det(AB)\neq0$ which implies that both $detA\neq 0$ and $det B\neq 0$. Therefore, both $A$ and $B$ must be invertable.

user64742
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Ali
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  • This is rather overkill as it assumes properties of the determinant, while the original question is easily answered in arbitrary unital rings that don’t admit a determinant function. – Erick Wong Mar 10 '21 at 17:47