I am looking for a rather simple proof that if the matrix A has eigenvalues>0, then (x^t)Ax>0 for any vector x not 0.(adsbygoogle = window.adsbygoogle || []).push({});

My first tought was if the eigenvalues are bigger than 0, then (x^t)Ax=(x^t)"eigenvaulue"x="eigenvalue"(x^t)x>0, if x is nonzero and eigenvalues is bigger than 0.

Is this proof good enough? I am a little unsure, becuase I think I have not proven it for all x, just the eigenvectors.

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# Mathematical proof

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