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I have the following question :

$A$ is a $n \times n$ matrix, and this is the characteristic polynom $$p(x)=(x+3)^2(x-1)(x-5)$$

Then I can conclude that $n=4$ since the number of the roots is $4$, now this is my question :

Can I conclude for any $\lambda \neq -3,1,5$ that $\lambda I-A$ and $A- \lambda I$ are invertible matrixs?

I think this statement is true, but I'd like to be sure of that, since this is critical point.

Thank you!

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  • $\begingroup$ Now, you cannot conclude that $\lambda I-A$ and $A-\lambda I$ are inverses, their product is $-\lambda^2 I+2\lambda A-A^2$, which is not guaranteed to be the identity (try a few examples). All you know is that the matrices have inverses (but not what the inverse is). $\endgroup$ – Michael Burr Aug 18 '15 at 13:16
  • $\begingroup$ @MichaelBurr I think I wasn't clear enough I don't mean that $(A-\lambda I)$ and $(A-\lambda I)$ are inverse of one other meaning I don't mean that $(\lambda I-A)(A-\lambda I)=I$, I mean them both individually are inverse matrixs. $\endgroup$ – JaVaPG Aug 18 '15 at 13:19
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    $\begingroup$ No he interpreted correctly. You used the wrong term. "Invertible" is what you want. And "invertible" means "have inverses" which is what he already said. $\endgroup$ – user21820 Aug 18 '15 at 13:20
  • $\begingroup$ If you mean $A-\lambda I$ and $\lambda I-A$ are invertible, then, yes, they are invertible. $\endgroup$ – Michael Burr Aug 18 '15 at 13:20
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    $\begingroup$ And you can also accept Selva's answer since it is a correct and easy way to get the result you want. $\endgroup$ – user21820 Aug 18 '15 at 13:22
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$det(A-\lambda I)=(\lambda+3)^2(\lambda-1)(\lambda-5)$. Hence $det(A-\lambda $I) not zero except $\lambda \neq-3,1,5$

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