matrices - Linear Algebra - Prove $AB=BA$

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.

Answer

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
$$BA=AB$$

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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 $AB=BA$.