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A thread I saw recently has led me to believe that this is not a valid proof of the fact that for matrices $A$ and $B$, $AB=I\implies BA=I$.

Suppose $AB=I$. Then

$$\begin{align}A^{-1}AB&=A^{-1}I\\\implies B&=A^{-1}\\\implies BA&=A^{-1}A\\\implies BA&=I\end{align}$$

what step is wrong in this? I assume $A$ has an inverse because $\det A\det B=\det AB=\det I=1$, so $\det A\neq 0$.

Integreek
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TY Mathers
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  • What leads you to believe this is invalid? I've seen exactly this proof in Schaum's Linear Algebra text. – Dion Bridger Jun 27 '15 at 23:43
  • http://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i

    It is correct since the existence of a left inverse of a square matrix follows from the existence of a right inverse.

     – Hasan Saad Jun 27 '15 at 23:43
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    How do you know that $A$ has an inverse? Once you do, then of course $B$ is it. But the problem is to start with $AB=I$, not knowing that $A$ has an inverse, and conclude $BA=I$. And from this it will follow that $B$ is the inverse of $A$. – GEdgar Jun 27 '15 at 23:43
  • Are you assuming that $A$, $B$, and $I$ are all $n\times n$ matrices? If that is not assumed, your statement is false and the proof is invalid since some of the products will be undefined. Is that what you mean? – Rory Daulton Jun 27 '15 at 23:43
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    Pretty sure he's assuming they're square. Else, this would have been obviously faulty to anyone. – Hasan Saad Jun 27 '15 at 23:44
  • The question I think you might be thinking about (this) asks for proofs where there are some restrictions imposed (i.e. if you don't know about inverses). – Winther Jun 27 '15 at 23:45
  • @HasanSaad that's what's being proved here though? And to the others, yes I am assuming they're square and same dimension. – TY Mathers Jun 28 '15 at 00:31
  • Also, thanks @Winther that helps – TY Mathers Jun 28 '15 at 00:32
  • The first equality may be incorrect. There is no assumption that $A^{-1}$ exists – sleeve chen Jun 28 '15 at 00:33
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    @sleevechen For a square matrix, the existence of a right inverse implies the existence of a left one.

    As for OP, the proof here is that if $AB=I$ then $BA=I$ and this proof is completely valid. However, this "proof" misses the gist of the subject which is all about proving the existence of $A^{-1}$ first. If it were proved before, then this proof is perfectly valid.

     – Hasan Saad Jun 28 '15 at 00:47

2 Answers2

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If $A$ and $B$ are both square matrices of the same dimension, then your proof is certainly correct. However, $AB=I\not\Longrightarrow BA=I$ when $B$ and $A$ are not square, and your step of $\det(AB)=\det(A)\det(B)$ is wrong since $\det(A)$ and $\det(B)$ are not defined for non-square matrices. In general for a non-square matrix there are so-called left- and right-inverses, which may not be identical for a given matrix.

For example, let $$A=\begin{pmatrix}1&1&0\\0&0&1\end{pmatrix}, B=\begin{pmatrix}1&0\\0&0\\0&1\end{pmatrix}$$ You can check $AB=I_\text{2x2}$ but $BA\neq I_\text{3x3}$

Leo
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I think the bigger problem is where the original poster is coming from. The biggest question is

In the first step while pre-multiplying by $A^{-1}$, are you sure that you know such a matrix $A^{-1}$ exists such that $A^{-1}A = I$? The answer is obviously true, but have you gone through enough Linear Algebra?

There are two possibilities, you know enough linear algebra to show that such a matrix $A^{-1}$ exists and is unique, then the proof is sound, there is no problem with it.

But if you're unsure with it, then I'll answer your question. The first step is to show that the matrix $A$ represents an invertible linear map. Partition $B = [C_1: C_2: \ldots : C_n]$ where $C_i$'s are the columns of the matrix $B$. Since $AC_i = e_i$ where $e_i$ is the column matrix of order $n \times 1$ with $i$th entry as $1$ and others $0$, we can conclude that $A$ spans entire $\mathbb C^n$. Applying the rank nullity theorem, we can conclude that $A$ is a bijective linear map or an isomorphism.

Therefore, there exists a unique matrix $B$ such that $AB = I$. Since $A$ is an isomorphism, the left and right inverse of $A$ will coincide.

Therefore $BA = I$ as well, and now the proof is sound. Multiplying by $A^{-1}$ has no problems now.

If you aren't convinced yet, then I think its better to specify what stuff you're familiar with. If you don't know the Rank-Nullity Theorem, then there other possible proofs which can simplify all of this into linear combinations. Giving your own background while doing basics of any subject is very important while asking the question.

Pragyan Pranay
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