# If $A,B\in M(2,\mathbb{F})$ and $AB=I$, then $BA=I$

This is Exercise 7, page 21, from Hoffman and Kunze's book.

Let $A$ and $B$ be $2\times 2$ matrices such that $AB=I$. Prove that $BA=I.$

I wrote $BA=C$ and I tried to prove that $C=I$, but I got stuck on that. I am supposed to use only elementary matrices to solve this question.

I know that there is this question, but in those answers they use more than I am allowed to use here.

I would appreciate your help.

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You can prove this in a low-tech way using row reduction. –  Qiaochu Yuan Feb 21 '12 at 19:32
I agree with your answers, but I cannot use $\det$ here, just elementary matrices. –  spohreis Feb 21 '12 at 19:38
–  JavaMan Feb 21 '12 at 19:43
@JavaMan: None of those answers can help me. –  spohreis Feb 21 '12 at 21:26
When you say "use elementary matrices"... what is it you know about the connection between elementary matrices and invertible matrices? What do you know about invertible matrices? –  Arturo Magidin Feb 21 '12 at 21:52
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I will give a sketch of a proof. Let $A= \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$ and $B= \left( \begin{array}{cc} x & x \\ z & w \end{array} \right)$ such that $AB=I.$ Then we get $\left\{\begin{array}{c} ax + bz = 1 \\ cx + dz = 0 \\ \end{array}\right.$ and $\left\{\begin{array}{c} ay + bw = 1 \\ cy + dw = 0 \\ \end{array}\right.$

I will assume that $a\neq 0$ (since there is no $B$ such that BO=I.) Then we have $x=\frac{1}{a}-\frac{bz}{a}$ and we get $(ad-bc)z=-c$. Let suppose that $ad=bc$. If $b=0$ or $c=0$ then $d=0$ and we would have $A= \left( \begin{array}{cc} a & b \\ 0 & 0 \end{array} \right)$, or $A= \left( \begin{array}{cc} a & 0 \\ c & 0 \end{array} \right)$, or $A= \left( \begin{array}{cc} a & 0 \\ 0 & 0 \end{array} \right)$ but in any case there is no $B$ such that $BA=I$ (It is easy to prove that). So we have $(a,b,c,d)\neq (0,0,0,0)$. Then we have $a=\frac{bc}{d}$, but in this case the systems above do not have solution. Then $ad-bc\neq 0$ and we get $z=\frac{-c}{ad-bc}$. In the end we will find that $B= \frac{1}{ad-bc}\left( \begin{array}{cc} d & -b \\ -c & a \end{array} \right).$ It is easy to check that $BA=I.$ Now if $a=0$ then we have $b\neq0$ and $\dots$

I don't know how to solve the exercise in a different way. This is my best effort.

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I know this is old, but I think I have found the answer that was intended. I also struggled with this one for a while because, as spohreis mentioned, you don't have much to go on at the time this is asked (no determinants, no transposes, no inverses even).

That being said, in problem 3 of section 1.4 you prove that all $2\times 2$ row-reduced echelon matrices are of the following form:

$$\left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right]\quad,\quad \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]\quad,\quad \left[ \begin{array}{cc} 1 & c \\ 0 & 0 \end{array} \right]\quad,\quad \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right] \,.$$

Now assume that $A$ and $B$ are $2 \times 2$ matrices such that $AB=I$. By theorem 5 (pg 12) we have that $B$ is row-equivalent to a row-reduced echelon matrix $R$, and by the corollary to theorem 9 (pg 20) this implies that $B=PR$ where $P$ is a product of elementary matrices. Similarly we have that $A=QT$ (where $Q$ is a product of elementary matrices and $T$ is in row-reduced echelon form).

Now we have that $AB=I \implies QTPR=I$, but now clearly $T=I$ because if the bottom row of $T$ were all zeros then the bottom row of $TPR$ would be zero, and this implies that the product $QTPR$ would have the form $$\left[ \begin{array}{cc} aQ_{11} & bQ_{11} \\ aQ_{21} & bQ_{21} \end{array} \right]$$ for some $a,b\in F$ and thus clearly could not be $I$. A similar argument shows that $R=I$. Thus $A$ and $B$ are both actually products of elementary matrices. By theorem 2 (pg 7) each elementary row operation has an inverse and using theorem 9 (pg 20) each elementary matrix therefore has an inverse, now we can write

\begin{align} AB=QP=E_{q_1}E_{q_2} \cdots E_{q_t}E_{p_1} \cdots E_{p_s}&=I\\ E_{q_1}^{-1}E_{q_1}E_{q_2} \cdots E_{q_t}E_{p_1} \cdots E_{p_s}E_{q_1}&=E_{q_1}^{-1}E_{q_1}=I\\ &\vdots\\ E_{p_1} \cdots E_{p_s}E_{q_1} \cdots E_{q_t}&=I\\ PQ&=I\\ BA&=I\,. \end{align}

(Note that at the end here, although I chose to use the standard inverse notation, it really is enough that such a matrix exists, which follows from theorem 2 and theorem 9 alone - no need to really "know" about inverse matrices yet. You could, if you so chose, just use theorem 9 to rewrite $QP=I$ in terms of elementary row operations, and then just use theorem 2 directly without ever mentioning inverse matrices.)

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$AB= I$, $Det(AB) = Det (A) . Det(B) = 1$. Hence $Det(B)\neq 0$ Hence $B$ is invertible.

Now let $BA= C$ then we have $BAB= CB$ which gives $B= CB$ that is $B. B^{-1} = C$ this gives $C= I$

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$AB = I$ implies that $ABAB = I$ and $AABB = A(I)B = AB = I$, hence $ABAB = AABB$. Since $\det(AB) = \det(A)\det(B) = 1$, the determinant of $A$ and the determinant of $B$ are units, so that $A$ and $B$ have inverses (using the adjoint matrix thing), hence $AABB = ABAB$ implies $BA=AB = I$.

Hope that helps,

Note : This proof works when $\mathbb F$ is an arbitrary commutative ring with unity. You didn't specify what $\mathbb F$ was so that I'm stating in which generality this proof holds.

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I don't know why you specify $2 \times 2$ matrix, but perhaps it is for explicitly using the adjoint formula for the inverse. –  Patrick Da Silva Feb 21 '12 at 19:33
Because I cannot use $\det$ here. I can use very little indeed. Determinants will be defined in the chapter $5$. I am trying to solve the exercises by using what I have. Thanks! –  spohreis Feb 21 '12 at 19:44
Hmm. Then perhaps you should've mentioned that, it would have made my answer different. –  Patrick Da Silva Feb 21 '12 at 19:44
I did! I wrote: I am supposed to use only elementary matrices to solve this question. –  spohreis Feb 21 '12 at 19:46
Then perhaps I should've read better. =P –  Patrick Da Silva Feb 21 '12 at 19:54
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