So my book says that $\lambda$ is an eigenvalue of $A$ iff $det (A- \lambda I) = 0$ but khan academy says that $\lambda $ is an eigenvalue iff $det (\lambda I - A ) = 0$.
Who is right? Both?
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$\begingroup$ Both are right :) $\endgroup$Jika– Jika2014-05-27 14:26:22 +00:00Commented May 27, 2014 at 14:26
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4 Answers
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$\lambda I-A=(-I)\cdot(A-\lambda I)$ and $\det$ is multiplicative so $$\det(\lambda I-A)=\det((-I)\cdot(A-\lambda I))=\det(-I)\cdot \det(A-\lambda I)=(-1)^n\cdot \det(A-\lambda I).$$
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Both of them are right since $$\det(A-\lambda I)=(-1)^n\det(\lambda I-A)$$.
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Both. Remember this equality $Av=\lambda v$. Here $\lambda$ is the Eigen value, $v$ the Eigen vector.
