If A and P are the square matrices of the same order and if P be invertible, show that the matrices A and `P^(-1) have the same characteristic roots.

Explanation

Let
`P^(-1) AP=B`
`:. B-lambdaI=P^(-1) AP-lambdaI`
`=P^(-1) AP-P^(-1) lambdaIP`
`=P^(-1) (A-lambdaI)P`
`implies |B-lambda I|=|P^(-1)||A-lambdaI||P|`
`=|A-lambdaI||P^(-1)||P|`
`=|A-lambdaI||P^(-1) P|`
`=|A-lambdaI||I|=|A-lambdaI|`
Thus, the two matrices A and B have the same characteristic determinants and hence the same characteristic equations and the same characteristic roots. the same thing may also be seen in another way. Now,
`AX=lambdaX`
`implies P^(-1) AX=lambdaP^(-1) X`
`implies (P^(-1) AP) (P^(-1)X)=lambda(P^(-1) X)`