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Definite matrix
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== Quadratic forms == {{Main|Definite quadratic form}} The (purely) [[quadratic form]] associated with a real <math>n \times n</math> matrix <math>M</math> is the function <math>Q : \mathbb{R}^n \to \mathbb{R}</math> such that <math>Q(\mathbf{x}) = \mathbf{x}^\mathsf{T} M \mathbf{x}</math> for all <math>\mathbf{x}.</math> <math>M</math> can be assumed symmetric by replacing it with <math>\tfrac{1}{2} \left(M + M^\mathsf{T} \right),</math> since any asymmetric part will be zeroed-out in the double-sided product. A symmetric matrix <math>M</math> is positive definite if and only if its quadratic form is a [[strictly convex function]]. More generally, any [[quadratic function]] from <math>\mathbb{R}^n</math> to <math>\mathbb{R}</math> can be written as <math>\mathbf{x}^\mathsf{T} M \mathbf{x} + \mathbf{b}^\mathsf{T} \mathbf{x} + c</math> where <math>M</math> is a symmetric <math>n \times n</math> matrix, <math>\mathbf{b}</math> is a real {{mvar|n}} vector, and <math>c</math> a real constant. In the <math>n = 1</math> case, this is a parabola, and just like in the <math>n = 1</math> case, we have '''Theorem:''' This quadratic function is strictly convex, and hence has a unique finite global minimum, if and only if <math>M</math> is positive definite. '''Proof:''' If <math>M</math> is positive definite, then the function is strictly convex. Its gradient is zero at the unique point of <math>M^{-1} \mathbf{b},</math> which must be the global minimum since the function is strictly convex. If <math>M</math> is not positive definite, then there exists some vector <math>\mathbf{v}</math> such that <math>\mathbf{v}^\mathsf{T} M \mathbf{v} \leq 0,</math> so the function <math>f(t) \equiv ( t \mathbf{v} )^\mathsf{T} M ( t\mathbf{v} ) + b^\mathsf{T} (t \mathbf{v}) + c</math> is a line or a downward parabola, thus not strictly convex and not having a global minimum. For this reason, positive definite matrices play an important role in [[optimization (mathematics)|optimization]] problems.
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