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Given a square matrix $A$ that has full row rank we know that the matrix is invertible. So there is a matrix $B$ such that

$$ AB=1 $$

writing this in component notation,

$$ A_{ij}B_{jk}=\delta_{ik} $$

Now, we tend to write $A^{-1}$ instead of $B$ but let's leave it like that for now.

My question is how can we show that $BA=1$? We mechanically jump to the conclusion that if the inverse exists, $AA^{-1}=A^{-1}A=1$ but how to show that? Equivalently why is the left inverse equal to the right inverse? It seems intuitively obvious!

Thanks a bunch, I appreciate.

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marked as duplicate by rschwieb, Davide Giraudo, Namaste linear-algebra Jan 13 '15 at 13:33

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  • $\begingroup$ This is not a stupid question at all. It is not that trivial (although not super hard either) to show that if $AB=1$ then $BA=1$. $\endgroup$ – mickep Jan 13 '15 at 11:55
  • $\begingroup$ It is easier to prove it in the abstract, like in group theory. Also, for the conclusion, you need the condition that the matrix be invertible. Some non-invertible (for instance, rectangular) matrives can have right inverse, but no left inverse (or the oposite). $\endgroup$ – kjetil b halvorsen Jan 13 '15 at 12:04
  • $\begingroup$ In general, this is not true, and $BA$ is not defined. You need to have that $A,B$ are square matrices. $\endgroup$ – Ofir Schnabel Jan 13 '15 at 12:09
  • $\begingroup$ @ kjetil b halvorsen: I assumed that $A$ has full row rank in line 1 $\endgroup$ – Georgy Jan 13 '15 at 12:26
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So you have a right inverse, and you know there is a left inverse too, let's say $C$. Then you have: $$1 = AB,$$ and multiplying both sides for C in the left, you get $$ C = C(AB) = (CA)B = B,$$ that is, if both exist, they must be equal.

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Let the endomorphism

$$\Phi:\mathcal M_n(\Bbb R)\to\mathcal M_n(\Bbb R),\; X\mapsto XA$$ then $\Phi(X)=0\iff XA=0\iff XAB=X=0$ hence $\Phi$ is injective and by the rank-nullity theorem it's bijective hence surjective so there's $B'$ such that $\Phi(B')=I_n\iff B'A=I_n$ which means that $A$ has a left inverse. It's routine to prove that $B=B'$.

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Given $AB=I$ then $B=BI=B(AB)=(BA)B$ Since $B=(BA)B$ then $BA=I$.

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  • $\begingroup$ You don't know $B$ is invertible. $\endgroup$ – Pedro Tamaroff Jan 13 '15 at 12:19
  • $\begingroup$ In fact you need $B$ to have a right inverse. $\endgroup$ – Georgy Jan 13 '15 at 12:30
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I just had a thought, using the commutator matrix $C=AB-BA=1-BA$. Post-multiply by $B$ we get

$$ CB=B-BAB=B-B=0 $$

using $AB=1$. Now $B$ has full row rank and therefore $C=0$ implying that $A$ and $B$ commute. Is that a way to go or did I miss something?

EDIT:

Better yet, just consider the matrix $S=ABA$. $$ S=(AB)A=A $$ on the other hand $$ S=A(BA) $$ Therefore $$ A=A(BA) \Rightarrow A(BA-1)=0 \Rightarrow BA=1 $$

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Lemma: Suppose there is a $v_1\ne0$ so that $Av_1=0$, then the column space of $A$ has dimension at most $n-1$.

Proof: Suppose there is a $v_1\ne0$ so that $Av_1=0$. Create a basis $\{v_k\}_{k=1}^n$ for $\mathbb{R}^n$ including $v_1$. We can write all vectors in $\mathbb{R}^n$ as $$ \sum_{k=1}^nc_kv_k $$ which means we can write all vectors in the column space of $A$ as $$ \sum_{k=2}^nc_kAv_k $$ Therefore, the column space of $A$ has dimension at most $n-1$. $$\square$$

Since $AB=I$, the column space of $A$ has dimension $n$. Therefore, $Ax=0\implies x=0$.

If $AB=I$, then $ABA=A$. Therefore, $A(BA-I)=0$. This means that each column of $BA-I$ is $0$. That is, $$ BA=I $$


Note that this depend on the vector space having finite dimension. Consider the two linear operators on sequences, $A$ and $B$, where $A$ shifts left: $$ (Av)_k=v_{k+1} $$ and $B$ shifts right, filling in with $0$: $$ (Bv)_k=\left\{\begin{array}{} 0&\text{if }k=0\\ v_{k-1}&\text{if }k\ge1 \end{array}\right. $$ We have $AB=I$, but $BA\ne I$ since $BA$ sets the first element of any sequence to $0$.

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  • $\begingroup$ Depending on the level of the reader, I thought a very basic approach might be useful. $\endgroup$ – robjohn Jan 13 '15 at 14:36

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