Invertibility of AB+BA

**minimammut** · October 16th 2012, 12:50 PM

I have to show that AB+BA is invertible if A and B are invertible. I tried to work with determinants but I never manage to get rid of that nasty plus

. Maybe someone has some suggestions? Thanks!

**johnsomeone** · October 16th 2012, 01:24 PM

What you're attempting to prove isn't true.

$\text{Let } A = \left(\begin{matrix} 1& 2 \\ -2& -1 \end{matrix}\right), B = \left(\begin{matrix} 0& 1 \\ 1& 0 \end{matrix}\right).$

$\text{Then } det(A) = (1)(-1) - (-2)(2) = 3, \ det(B) = (0)(0) - (1)(1) = -1,$

$\text{ so both } A \text{ and } B \text{ are invertible in } M_{2, 2}(\mathbb{R}).$

$\text{Have } AB = \left(\begin{matrix} 1& 2 \\ -2& -1 \end{matrix}\right) \left(\begin{matrix} 0& 1 \\ 1& 0 \end{matrix}\right) = \left(\begin{matrix} 2& 1 \\ -1& -2 \end{matrix}\right)$

$\text{and } BA = \left(\begin{matrix} 0& 1 \\ 1& 0 \end{matrix}\right) \left(\begin{matrix} 1& 2 \\ -2& -1 \end{matrix}\right) = \left(\begin{matrix} -2& -1 \\ 1& 2 \end{matrix}\right),$

$\text{so that } AB + BA = \left(\begin{matrix} 0& 0 \\ 0& 0 \end{matrix}\right), \text{ which is clearly not invertible.}$

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