Return to Answer

So you want to find a proof of this well-known fact, which avoids the usual "indirect" proofs? I've also pondered over this some time ago.

We have the following general assertion:

Let $M$ be a finite-dimensional $K$-algebra, and $a,b \in M$ such that $ab=1$, then $ba=1$. [For example, $M$ could be the algebra of $n \times n$ matrices]

Proof: The sequence of subspaces $\cdots \subseteq b^{k+1} M \subseteq b^k M \subseteq \cdots \subseteq M$ must be stationary, since $M$ is finite-dimensional. Thus there is some $k$ and some $c \in M$ such that $b^k = b^{k+1} c$. Now multiply with $a^k$ on the left to get $1=bc$. Then $ba=ba1 = babc=b1c=bc=1$. QED

No commutativity condition is needed. The proof shows more general that the claim holds in every left- or right-artinian ring $M$.

Remark that we needed, in a essential way, some finiteness condition. There is no purely algebraic manipulation with $a,b$, which shows $ab = 1 \Rightarrow ba=1$ (and shift operators provide a concrete counterexample). Every argument uses some argument of the type above. For example when you want to argue with linear maps, you have to know that every subspace of a finite-dimensional(!) vector space of the same dimension actually is the whole vector space, for which there is also no "direct" proof. I doubt that there is one.

PS. See here for a proof of $AB=1 \Rightarrow BA=1$ for square matrices over a commutative ring.

We have the following general assertion:

Lemma. Let $A$ be a finite-dimensional $K$-algebra, and $a,b \in A$. If $ab=1$, then $ba=1$.

For example, $A$ could be the algebra of $n \times n$ matrices over $K$.

Proof. The sequence of subspaces $\cdots \subseteq b^{k+1} A \subseteq b^k A \subseteq \cdots \subseteq A$ must be stationary, since $A$ is finite-dimensional. Thus there is some $k$ with $b^{k+1} A = b^k A$. So there is some $c \in A$ such that $b^k = b^{k+1} c$. Now multiply with $a^k$ on the left to get $1=bc$. Then $ba=ba1 = babc=b1c=bc=1$. $\square$

The proof also works in every left- or right-Artinian ring $A$. In particular, the statement is true in every finite ring.

Remark that we need in an essential way some finiteness condition. There is no purely algebraic manipulation with $a,b$ that shows $ab = 1 \Rightarrow ba=1$.

In fact, there is a $K$-algebra with two elements $a,b$ such that $ab=1$, but $ba \neq 1$. Consider the left shift $a : K^{\mathbb{N}} \to K^{\mathbb{N}}$, $a(x_0,x_1,\dotsc) := (x_1,x_2,\dotsc)$ and the right shift $b(x_0,x_1,\dotsc) := (0,x_0,x_1,\dotsc)$. Then $a \circ b = \mathrm{id} \neq b \circ a$ holds in the $K$-algebra $\mathrm{End}_K(K^{\mathbb{N}})$.

See SE/298791 for a proof of $AB=1 \Rightarrow BA=1$ for square matrices over a commutative ring.

So you want to find a proof of this well-known fact, which avoids the usual "indirect" proofs? I've also pondered over this some time ago.

We have the following general assertion:

Let $M$ be a finite-dimensional $K$-algebra, and $a,b \in M$ such that $ab=1$, then $ba=1$. [For example, $M$ could be the algebra of $n \times n$ matrices]

Proof: The sequence of subspaces $\cdots \subseteq b^{k+1} M \subseteq b^k M \subseteq \cdots \subseteq M$ must be stationary, since $M$ is finite-dimensional. Thus there is some $k$ and some $c \in M$ such that $b^k = b^{k+1} c$. Now multiply with $a^k$ on the left to get $1=bc$. Then $ba=ba1 = babc=b1c=bc=1$. QED

No commutativity condition is needed. The proof shows more general that the claim holds in every left- or right-artinian ring $M$.

Remark that we needed, in a essential way, some finiteness condition. There is no purely algebraic manipulation with $a,b$, which shows $ab = 1 \Rightarrow ba=1$ (and shift operators provide a concrete counterexample). Every argument uses some argument of the type above. For example when you want to argue with linear maps, you have to know that every subspace of a finite-dimensional(!) vector space of the same dimension actually is the whole vector space, for which there is also no "direct" proof. I doubt that there is one.

PS. See here for a proof of $AB=1 \Rightarrow BA=1$ for square matrices over a commutative ring.

So you want to find a proof of this well-known fact, which avoids the usual "indirect" proofs? I've also pondered over this some time ago.

We have the following general assertion:

Let $M$ be a finite-dimensional $K$-algebra, and $a,b \in M$ such that $ab=1$, then $ba=1$. [For example, $M$ could be the algebra of $n \times n$ matrices]

Proof: The sequence of subspaces $\cdots \subseteq b^{k+1} M \subseteq b^k M \subseteq \cdots \subseteq M$ must be stationary, since $M$ is finite-dimensional. Thus there is some $k$ and some $c \in M$ such that $b^k = b^{k+1} c$. Now multiply with $a^k$ on the left to get $1=bc$. Then $ba=ba1 = babc=b1c=bc=1$. QED

No commutativity condition is needed. The proof shows more general that the claim holds in every left- or right-artinian ring $M$.

Remark that we needed, in a essential way, some finiteness condition. There is no purely algebraic manipulation with $a,b$, which shows $ab = 1 \Rightarrow ba=1$ (and shift operators provide a concrete counterexample). Every argument uses some argument of the type above. For example when you want to argue with linear maps, you have to know that every subspace of a finite-dimensional(!) vector space of the same dimension actually is the whole vector space, for which there is also no "direct" proof. I doubt that there is one.

PS. See here for a proof of $AB=1 \Rightarrow BA=1$ for square matrices over a commutative ring.

So you want to find a proof of this well-known fact, which avoids the usual "indirect" proofs? I've also pondered over this some time ago.

We have the following general assertion:

Let $M$ be a finite-dimensional $K$-algebra, and $a,b \in M$ such that $ab=1$, then $ba=1$. [For example, $M$ could be the algebra of $n \times n$ matrices]

Proof: The sequence of subspaces $... \subseteq b^{k+1} M \subseteq b^k M \subseteq ... \subseteq M$ must be stationary, since $M$ is finite-dimensional. Thus there is some $k$ and some $c \in M$ such that $b^k = b^{k+1} c$. Now multiply with $a^k$ on the left to get $1=bc$. Then $ba=ba1 = babc=b1c=bc=1$. QED

No commutativity condition is needed. The proof shows more general that the claim holds in every left- or right-artinian ring $M$.

Remark that we needed, in a essential way, some finiteness condition. There is no purely algebraic manipulation with $a,b$, which shows $ab = 1 \Rightarrow ba=1$ (and shift operators provide a concrete counterexample). Every argument uses some argument of the type above. For example when you want to argue with linear maps, you have to know that every subspace of a finite-dimensional(!) vector space of the same dimension actually is the whole vector space, for which there is also no "direct" proof. I doubt that there is one.

PS. See here for a proof of $AB=1 \Rightarrow BA=1$ for square matrices over a commutative ring.

So you want to find a proof of this well-known fact, which avoids the usual "indirect" proofs? I've also pondered over this some time ago.

We have the following general assertion:

Let $M$ be a finite-dimensional $K$-algebra, and $a,b \in M$ such that $ab=1$, then $ba=1$. [For example, $M$ could be the algebra of $n \times n$ matrices]

Proof: The sequence of subspaces $\cdots \subseteq b^{k+1} M \subseteq b^k M \subseteq \cdots \subseteq M$ must be stationary, since $M$ is finite-dimensional. Thus there is some $k$ and some $c \in M$ such that $b^k = b^{k+1} c$. Now multiply with $a^k$ on the left to get $1=bc$. Then $ba=ba1 = babc=b1c=bc=1$. QED

No commutativity condition is needed. The proof shows more general that the claim holds in every left- or right-artinian ring $M$.

Remark that we needed, in a essential way, some finiteness condition. There is no purely algebraic manipulation with $a,b$, which shows $ab = 1 \Rightarrow ba=1$ (and shift operators provide a concrete counterexample). Every argument uses some argument of the type above. For example when you want to argue with linear maps, you have to know that every subspace of a finite-dimensional(!) vector space of the same dimension actually is the whole vector space, for which there is also no "direct" proof. I doubt that there is one.

PS. See here for a proof of $AB=1 \Rightarrow BA=1$ for square matrices over a commutative ring.