Sunday, 27 August 2017

linear algebra - AinMn(mathbbR) symmetric matrix , AlambdaI is Positive-definite matrix- prove: detAgeqan



Let a>1 and AMn(R) symmetric matrix such that AλI is Positive-definite matrix (All eigenvalues >0) for every $\lambda


First, I'm not sure what does it mean that AλI is positive definite for every $\lambda 0andalleigenvaluesarebiggerthan0$ or it's not.



Then, If it's symmetric I can diagonalize it, I'm not sure what to do...



Thanks!


Answer



AλI is positive definite for every $\lambda0,A-(a-\epsilon) Iispositivedefinite.Thatmeans,eacheigenvalueofAislargerthana-\epsilon,thustheirproduct\det A\ge \prod (a-\epsilon)...let\epsilon$ goes to zero, you get what you want.


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