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vish_maths

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Q. Given that A is invertible , check whether A + A^(-1) is invertible or not ?

A. A is invertible . => A^-1 exists

( sorry about not using the powers directly. i am not able to click on the required icon ).

Suppose a non trivial solution exists to ( A + A^(-1) ) X = 0

( which means let's assume A + A^(-1) is non - invertible ).

=> a non trivial solution exists to AX = A^(-1) ( -X)

or to AX = A A^(-2) ( -X)

or to A [ A^(-2) + I ] X = 0

clearly , [ A^(-2) + I ] X = Y is equal to 0 as A is invertible

i.e. [ A^(-2) + I ] X =0

We also know that a non trivial solution for X exists

=> [ A^(-2) + I ] is non - invertible.

=> A^(-2) must have an eigen value = -1

=> A also must have an eigen value = -1 ( by the theorem of eigen value decomp.)

=> A + A^(-1) is non-invertible only when the matrix A has an eigen value = -1.

Hence, generally speaking, it is not invertible.

Thanks a lot.

A. A is invertible . => A^-1 exists

( sorry about not using the powers directly. i am not able to click on the required icon ).

Suppose a non trivial solution exists to ( A + A^(-1) ) X = 0

( which means let's assume A + A^(-1) is non - invertible ).

=> a non trivial solution exists to AX = A^(-1) ( -X)

or to AX = A A^(-2) ( -X)

or to A [ A^(-2) + I ] X = 0

clearly , [ A^(-2) + I ] X = Y is equal to 0 as A is invertible

i.e. [ A^(-2) + I ] X =0

We also know that a non trivial solution for X exists

=> [ A^(-2) + I ] is non - invertible.

=> A^(-2) must have an eigen value = -1

=> A also must have an eigen value = -1 ( by the theorem of eigen value decomp.)

=> A + A^(-1) is non-invertible only when the matrix A has an eigen value = -1.

Hence, generally speaking, it is not invertible.

Thanks a lot.

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