invertibility_remains_in_number_multiplication.md

若$A$可逆，数$\lambda\ne0$，则$\lambda A$可逆，且$(\lambda A)^{-1}=\frac{1}{\lambda}A^{-1}$

证明

由$AA^{-1}=A^{-1}A=E$，便有

$$ (\lambda A)(\frac1\lambda A^{-1})=(\frac1\lambda A^{-1})(\lambda A)=E $$

故知$A$可逆，并且

$$ (\lambda A)^{-1}=\frac1\lambda A^{-1} $$