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Let $A,\,B,\,C$ be square matrices such that $A^2=A,\,B^2=B,\,C^2=C$ and $A+B+C=I.$ Prove that $AB=BC=CA=\mathbb{O}.$

Attempt. We have $A(B+C)=A(I-A)=A-A^2=\mathbb{O}$ and similarly we get $B(A+C)=C(A+B)=\mathbb{O}$, but this is where i get so far.

Thanks in advance for the help.

Nikolaos Skout
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2 Answers2

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Roland's answer gives a great and simple linear algebra argument.

Here is a purely algebraic development:

\begin{align*} (A+B+C)^2 & = A^2 + B^2 + C^2 + AB + AC + BC + BA + CA + CB \\ & = I + A(B+C) + BC + (B+C)A + CB \\ & = I + A(I - A) + (I-A)A + BC + CB \\ & = I + A - A^2 + A - A^2 + BC + CB \\ & = I + BC + CB \\ I & = I + BC + CB \\ 0 & = BC + CB \\ \end{align*} By permuting the roles of $A,B,C$, we get $BC + CB = AC + CA = AB + BA = 0$.

This would allow us to conclude if we knew that $AB = BA$. It is not one of the hypotheses, but at this point we can deduce it: \begin{align*} AB + BA & = 0 \\ AB & = - BA \\ A^2B & = - ABA \\ AB & = - ABA \\ AB & = BA^2 \\ AB & = BA \\ \end{align*}

Stef
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  • Doesn't it follow directly from $(A+B+C)^2=I+2AC$ that $AC=0$ because $(A+B+C)=I$ and thus $(A+B+C)^2=I$? – LegNaiB Sep 21 '22 at 09:24
  • Not to forget the great answer of Ewan Delanoy in the above duplicate, so not only Roland. – Dietrich Burde Sep 21 '22 at 09:24
  • @LegNaiB Indeed you're right! – Stef Sep 21 '22 at 09:27
  • @LegNaiB Also note that the step $AB + AC = AB + AC + A² - A² = A - A²$ is the same step that you did, $AB + AC = A(I - A) = A - A²$. – Stef Sep 21 '22 at 09:30
  • Yeah this is true. But I find it easier to deduce it directly from the $(A+B+C)^2=I$ fact - but that's probably just preference – LegNaiB Sep 21 '22 at 09:32
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    @Stef. You assumed AB=BA, AC=CA, BC=CB which maybe not true. So $(A+B+C)^2=A^2+B^2+C^2+AB+BA+AC+CA+BC+CA$ only. – Abel Wong Sep 21 '22 at 09:51
  • @AbelWong Thanks! I fixed it. – Stef Sep 21 '22 at 10:05
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    The conclusion that $AB + BA = 0$ can be achieved perhaps more easily in the following manner. Since $C$ is idempotent, it follows that $A+B = I - C$ is idempotent, so you can expand $(A+B)^2 = A+B$. Of, course, the same for other permutations of $A,B,C$. – Ennar Sep 21 '22 at 10:17
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    @Stef. Good ABA trick. – Abel Wong Sep 21 '22 at 11:01
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An idempotent matrix is a projection onto a subvectorspace. The assumption $A+B+C=I$ in particular says that the direct sum of the ranges of $A,B,C$ yields the whole space. To see this you have to know that $0,1$ are the only eigenvalues of a projection, since we have $\lambda v=Pv=P^2v=\lambda^2v$ for an eigenvector of a projection $P$. Let $v\in rg(A)$. Then we have $v=Iv=Av+Bv+Cv=v+Bv+Cv$, hence $0=Bv+Cv$ or equivalently $Bv=-Cv$. Set $w=Bv$ then $w\in \mathrm{rg}(B)$ and $w=C(-v)\in \mathrm{rg}(C)$. Then we have $w=Iw=Aw+Bw+Cw=Aw+w+w\Leftrightarrow Aw=-w$. Since the only eigenvalues of $A$ are $0$ and $1$ we necessarily obtain $w=0$, therefore $Bv=w=0$ and also $Cv=-Bv=0$.

This shows that any vector in the range of $A$ is sent to zero by $B$ and $C$, hence $BA=0$ and $CA=0$. The rest follows by intertwining the roles of $A,B$ and $C$.

The above also shows that a basechange to the basis given by the union of the bases of the ranges simplyfies the problem to $A,B,C$ being matrices which are zero everywhere but the diagonal and one of the three is zero on the diagonal if and only if one of the other two is not on the diagonal, e.g.

$A=\begin{pmatrix}1\\ & \ddots\\ & & 1\\ & & & & 0\\ & & & & & \ddots\end{pmatrix}$

$B=\begin{pmatrix}0\\ & \ddots\\ & & 0\\ & & & & 1\\ & & & & & \ddots\\ &&&&&&1\\&&&&&&&0\\&&&&&&&&\ddots\end{pmatrix}$

$C=\begin{pmatrix}0\\ & \ddots\\ & & 0\\ & & & & 1\\ & & & & & \ddots\end{pmatrix}$

Roland
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  • How do you get $AB = BC = CA$ all of a sudden? Also the kernel of a projection need not be the orthogonal complement of the range. – Adayah Sep 21 '22 at 10:16
  • Since all of them are zero – Roland Sep 21 '22 at 12:02
  • Why are they zero? (I don't know why I forgot to rewrite $=0$ in the first question) – Adayah Sep 21 '22 at 12:55
  • If you have two projections $A,B$ with disjoint ranges, then for any vector $x$, such that $Bx\neq 0$ you have $Bx$ in the range of $B$ and in particular $Bx$ not in the range of $A$ (by assumption). Since $A$ is the projection onto its range, we obtain $ABx=0$. – Roland Sep 29 '22 at 10:21
  • How do you get $ABx = 0$? It's not true that $Av = 0$ for every $v$ not in the range of $A$. – Adayah Sep 29 '22 at 12:26
  • By a change of basis you may assume that all entries other than the diagonal of $A$ and $B$ are zero. Now $A_{ii}\neq 0$ implies $B_{ii}=0$, since their ranges are disjoint. Then $AB$ clearly is the zero matrix. – Roland Oct 02 '22 at 14:32
  • How do you change the basis so that all entries other than on the diagonal of $A$ and $B$ are zero? – Adayah Oct 02 '22 at 21:07
  • We choose for example the first $n$ basis vectors as the basis of the range of $A$, the next $m$ as the basis of the range of $B$. – Roland Oct 03 '22 at 13:52
  • That doesn't work since you don't know if vectors from the range of $A$ are eigenvectors of $B$. At any rate, this conversation indicates that your answer is hardly a satisfactory solution. To be honest, I think it's none of a solution at all. – Adayah Oct 03 '22 at 22:32
  • Can you give a counterexample? – Roland Oct 05 '22 at 07:32
  • A counterexample to what? – Adayah Oct 05 '22 at 12:04
  • Such that a vector from the range of $A$ is not a eigenvector of $B$. I think this is not possible since $A$ and $B$ are projections and their ranges are disjoint. Assume $v$ is in the range of $A$, then since $A$ and $B$ have disjoint range and are projections onto their range, we have $Bv=0$, hence $v$ is an eigenvector of $B$. – Roland Oct 06 '22 at 09:43
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    $$\begin{pmatrix} 1 & 0 \ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \ 0 \end{pmatrix} \neq \begin{pmatrix} 0 \ 0 \end{pmatrix}$$ – Adayah Oct 06 '22 at 13:49
  • But there is no third projection such that they sum up to the identity. There is even no example in dimension $\leq 3$ (assuming that all projections are non-zero). You have to consider that the only eigenvalues of a projection are $1$ and $0$, since these are the only one to satisfy $\lambda^2=\lambda$. I think $A+B+C=I$ is even equivalent to saying that the ranges are disjoint and their direct sum is the whole space. – Roland Oct 06 '22 at 16:43
  • Of course I was not providing a counterexample to the original question, because it does not exist. I showed a counterexample to the implication you were using in your proof, which was not correct. – Adayah Oct 06 '22 at 17:49
  • From your example I think I know the confusion: Instead of disjoint range, i should rather write that the direct some of them yields the whole space. – Roland Oct 07 '22 at 12:28