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Unformatted text preview: positive denite, contradicting the Fact that 1 > 0 and 2 > 0. Thus 2 and 3 are both positive , and hence Q is positive denite on all oF R 3 . What about deciding iF Q is negative denite? The easiest way to get at this is to note that Q is negative denite precisely iF its negative ( Q )( x ) := Q ( x ) is positive denite, and that Q = [ Q ]. Now, the determinant oF a k k matrix M is related to the determinant oF its negative by det( M ) = ( 1) k det M so we see that For k = 1 , 2 , 3 k ( Q ) = ( 1) k k ( Q )...
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 Spring '08
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 Calculus

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