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Let $A$ and $B$ be any $n \times n$ defined over the real numbers.

Assume that $A^2+AB+2I=0$.

  • Prove $AB=BA$

My solution (Not full)

I didn't managed to get so far.

$A(A+B)=-2I$

$-\frac{1}{2}(A(A+B)=I$

Therefore $A$ reversible and $A+B$ reversible.

I don't know how to get on from this point, What could I conclude about $A^2+AB+2I=0?$

Any ideas? Thanks.

JaVaPG
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2 Answers2

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If $-\frac{1}{2}A(A+B)=I$, then $(A+B)\cdot(-\frac{1}{2}A)$ is also $I$, since $-\frac{1}{2}A$ and $A+B$ are each other inverses. Then $A^2+AB=A^2+BA$, and so $AB=BA$.

Sasha Patotski
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  • Can you explain why since $-\frac{1}{2}A$ and $A+B$ inverses $A^2+AB=A^2+BA?$ – JaVaPG Oct 15 '14 at 20:23
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    Since $-\frac{1}{2}A$ and $A+B$ are each other inverses, you have $-\frac{1}{2}A\cdot (A+B)=(A+B)\cdot (-\frac{1}{2}A)$ (and both equal $I$). Now multiply both sides by $-2$. Get $A(A+B)=(A+B)A$, and then open the brackets. – Sasha Patotski Oct 15 '14 at 20:36
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From $$ A(A+B)=A^2+AB=-2I $$ we have that $$ A^{-1}=-\frac12(A+B) $$ then multiplying by $-2A$ on the right and adding $2I$ gives $$ A^2+BA+2I=0=A^2+AB+2I $$ Cancelling common terms yields $$ BA=AB $$

Another Approach

Using this answer (involving more work than the previous answer), which says that $$ AB=I\implies BA=I $$ we get $$ -\frac12A(A+B)=I\implies-\frac12(A+B)A=I $$ Cancelling common terms gives $AB=BA$.

Community
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robjohn
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  • What do you mean subtracting the original equation? $A^2+BA+2I=0$ is the original equation. – JaVaPG Oct 15 '14 at 20:26
  • Yes, subtract $A^2+AB+2I=0$ from $A^2+BA+2I=0$ gives $BA-AB=0$ which is the same as $BA=AB$. I've modified the last line in case that makes it clearer. – robjohn Oct 15 '14 at 20:32