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QUESTION: If A and P be $2$ non-singular $n\times n$ matrices and $\lambda$ is the eigen-value of $A$, then show that $\lambda$ is also the eigen-values of a matrix $P^{-1}AP$.

I could simply show that $\lambda$ being the eigen-value of $A$, we have that $$det (A-\lambda I_n)=0$$

But I could not proceed further and make any comment on the question asked.

NOTE: I am unaware of diagonalisation of matrices, if at all it is playing any part in this problem. And also I require a method which does not utilise this principle of diagonalisation to solve this.

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  • $\begingroup$ @Moo I know that. But the problem was given in a mathematics paper where the syllabus did not include diagonlisation. So surely, the examiner wants a solution not based on diagonlisation. $\endgroup$ – SchrodingersCat May 26 '16 at 12:40
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Actually $A$ and $P^{-1}AP$ share the same characteristic polynomial, hence they have the same eigenvalues. Note that

$$\begin{align*} \det(P^{-1}AP-\lambda I_n) & = \det(P^{-1}(A-\lambda I_n)P))\\ & = \det(P^{-1})\det(A-\lambda I_n)\det(P)\\ & = \det(A-\lambda I_n). \end{align*} $$

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  • $\begingroup$ How do you make that transformation $P^{-1}AP-\lambda I=P^{-1}(A-\lambda I)P$? I mean, $\lambda I$ contains no $P$, then how is it that you can factor out the $P$? $\endgroup$ – SchrodingersCat May 26 '16 at 12:35
  • $\begingroup$ Do the distribution to get $P^{-1}AP - P^{-1}\lambda I P$. But the second term is clearly equal to $\lambda I$. $\endgroup$ – C. Dubussy May 26 '16 at 12:37
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It's just that $$P^{-1}AP-\lambda I=P^{-1}(A-\lambda I)P. $$

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  • $\begingroup$ How do you make that transformation $P^{-1}AP-\lambda I=P^{-1}(A-\lambda I)P$? I mean, $\lambda I$ contains no $P$, then how is it that you can factor out the $P$? $\endgroup$ – SchrodingersCat May 26 '16 at 12:32
  • $\begingroup$ $I=P^{-1}P. \ \ $ $\endgroup$ – Martin Argerami May 26 '16 at 13:01

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