Question

Prove that if a is invertible, then a + b and i + ba1 are both invertible or both not invertible

Answer 1

Answer:

We can prove this with the help of the determinant.

As we are given that:

a is invertible this means that:

$\det(a)\neq0$ where $\det (a)$represents the determinant of the matrix a.

Also:

Let us consider the quantity a+b

$\det (a+b)=\det (a(i+ba^{-1}))$

( where i denotes the identity operator)

⇒ $\det (a+b)=\det(a)\times det(i+ba^{-1})$

( As

$\det(a.b)=\det(a).\det(b)$  ).

Hence,

the $\det(a+b)\neq0$ iff $\det (i+ba^{-1})$.

Hence we could say that:

if a is invertible, then (a+b)and (i +ba^-1) are both invertible or both not invertible.