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Show that the two matrices A, `P^(-1) AP` have the same characteristic roots. |
Answer» 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)` |
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