Linear Algebra Chap2.5

Challenge Problems


How does the identity ({\bf I}+{\bf AB})^{-1}{\bf A} = {\bf A}({\bf I}+{\bf BA})^{-1} connect the inverses of {\bf I}+{\bf BA} and {\bf I}+{\bf AB}?

Those are both invertible or both singular: not obvious.

Ans(Prof.Nagahara taught us):

{\bf I}+{\bf BA} is a singular matrix.

there exists {\bf x}\neq0 such that

({\bf I}+{\bf BA}){\bf x} &=& 0 \tag{1} \\
\therefore {\bf A}({\bf I}+{\bf BA}){\bf x} &=& 0 \\
\therefore ({\bf I}+{\bf AB}){\bf Ax} &=& 0 \tag{2}


ここで、{\bf I}+{\bf AB}が正則として仮定して矛盾を導く.

{\bf I}+{\bf AB}が正則より(2)より

({\bf I}+{\bf AB})^{-1}({\bf I}+{\bf AB}){\bf Ax} &=& 0 \\
\therefore {\bf Ax} &=& 0 \tag{3}

(1)より {\bf x}+{\bf BAx} = 0

(3)を代入すれば {\bf x} = 0

これは{\bf x}\neq 0に矛盾する.

ゆえに{\bf I}+{\bf AB}は特異.