# Assuming $AB=I$ prove $BA=I$ [duplicate]

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If $AB = I$ then $BA = I$

Most introductory linear algebra texts define the inverse of a square matrix $A$ as such:

Inverse of $A$, if it exists, is a matrix $B$ such that $AB=BA=I$.

That definition, in my opinion, is problematic. A few books (in my sample less than 20%) give a different definition:

Inverse of $A$, if it exists, is a matrix $B$ such that $AB=I$. Then they go and prove that $BA=I$.

Do you know of a proof other than defining inverse through determinants or through using rref?

Is there a general setting in algebra under which $ab=e$ leads to $ba=e$ where $e$ is the identity?

## marked as duplicate by Thomas, Emily, rschwieb, Noah Snyder, Hagen von EitzenOct 19 '12 at 21:17

• You might want to take a look at this question. – EuYu Oct 18 '12 at 22:15
• What's "problematic" about the first definition? In general there's no relationship between having a left inverse and having a right inverse, and two-sided inverse is the correct generalization of the naive notion of inverse. – Qiaochu Yuan Oct 18 '12 at 22:19
• There’s nothing at all problematic about the definition. There is no compelling reason to incorporate into the definition the fact that a one-sided inverse of a square matrix can be proved to be a two-sided inverse. – Brian M. Scott Oct 18 '12 at 22:21
• Maybe it's of some interest to notice that if $A$ is a non-square matrix, with more rows that columns, and its columns are independent, then there is a matrix $B$ such that $BA$ is a (small) identity matrix, and $AB$ is the projection matrix onto the column space of $A$. The matrix $B$ is $(A^T A)^{-1} A^T$. It is a left-inverse of $A$. Multiplying boths sides of an equality on the left by $B$ has the effect of canceling the $A$ when $A$ is the leftmost factor on one side of the equality. – Michael Hardy Oct 18 '12 at 22:21
• @Maesumi: they are both required, in general. For matrices it happens to be the case that a left inverse is also a right inverse and vice versa, but in more general contexts this is false and the correct generalization of "inverse" is "two-sided inverse." – Qiaochu Yuan Oct 18 '12 at 22:31

Multiply both sides of $AB-I=0$ on the left by $B$ to get $$(BA-I)B=0\tag{1}$$ Let $\{e_j\}$ be the standard basis for $\mathbb{R}^n$. Note that $\{Be_j\}$ are linearly independent: suppose that $$\sum_{j=1}^n a_jBe_j=0\tag{2}$$ then, multiplying $(2)$ on the left by $A$ gives $$\sum_{j=1}^n a_je_j=0\tag{3}$$ which implies that $a_j=0$ since $\{e_j\}$ is a basis. Thus, $\{Be_j\}$ is also a basis for $\mathbb{R}^n$.

Multiplying $(1)$ on the right by $e_j$ yields $$(BA-I)Be_j=0\tag{4}$$ for each basis vector $Be_j$. Therefore, $BA=I$.

Failure in an Infinite Dimension

Let $A$ and $B$ be operators on infinite sequences. $B$ shifts the sequence right by one, filling in the first element with $0$. $A$ shifts the sequence left, dropping the first element.

$AB=I$, but $BA$ sets the first element to $0$.

Arguments that assume $A^{-1}$ or $B^{-1}$ exist and make no reference to the finite dimensionality of the vector space, usually fail to this counterexample.

• For the infinite-dimensional operators, we could take the shifts to be $A=\tfrac{d}{dx}$, and $B=\int_0^x$ (acting on polynomials, for example). – mr_e_man Dec 26 '18 at 2:15

Without the assumption of $A$ and $B$ being square matrices, we can find counterexamples. For example: $$\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array}\right) \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{array}\right) = \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}\right)$$ and $$\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{array}\right) \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array}\right) = \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{array}\right).$$

For square matrices, it was proved in several ways for square matrices in the question:

If $AB = I$ then $BA = I$

Is there a general setting in algebra under which ab=e leads to ba=e where e is the identity?

Finiteness or finite-dimensionality or rigidities that follow from those, such as:

• a Dedekind-finite set is not infinite
• the double-dual $V^{**}$ being naturally isomorphic to $V$,
• antipode^2=identity, and other fancier analogues (think here of phrases like rigid tensor categories with duals).

There is a duality between injective and surjective, or left and right, and you need some setting in which the transpose from one to the other is its own inverse. The linear algebra result for finite matrices ultimately rests on the same principle for functions on finite sets, and on the dimension of a finite-dimensional vector space being well-defined (which is closely related to the cardinality of a finite set being well-defined).

For $A$ and $B$ square, $AB=I$ implies $(AB)^{-1} = B^{-1}A^{-1}=I$. Then multiply $B$ from the left side, $A$ from the right side. $BB^{-1}A^{-1}A=BIA$ implies $BA=I$.

• And how exactly do you show that $(AB)^{-1} = B^{-1}A^{-1}$ in this case? – EuYu Oct 18 '12 at 23:23
• Multiply $B$ from the left side, $A$ from the right side. – Stefan Oct 18 '12 at 23:23
• As the counter-examples for my question yesterday show, it is not possible to prove this using only "associativity" and "existence of identity" (as the same proof would work for all monoids). – Douglas S. Stones Oct 18 '12 at 23:27
• @Stefan: $A^{-1}$ and $B^{-1}$ do not exist yet. – wj32 Oct 18 '12 at 23:27
• @ EuYu, $(AB)(B^{-1}A^{-1})=I$ whenever the inverses exist [and they generally do as far as $(AB)^{-1}$ is meaningful]. Clearly, in our case, $AB=I$ – basta Oct 19 '12 at 5:24