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Why must a symmetric positive definite matrix must be invertible? I'm reading a proof of the Levi-Civita theorem in differential geometry but the author states this without proof and I haven't been able to prove it.

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  • $\begingroup$ Because its null space is trivial. $\endgroup$ – Arin Chaudhuri Apr 25 '13 at 17:16
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Suppose a positive definite matrix $P$ were not invertible. Then there would be a nonzero vector $x$ such that $Px = 0$. Multiplying both sides of this equation on the left by $x^T$ gives $x^TPx = x^T0 = 0$ for $x\neq 0$, contradicting positive definiteness.

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  • $\begingroup$ Short and sweet. $\endgroup$ – Alex Provost Apr 25 '13 at 18:36
  • $\begingroup$ Precisely. This is a great solution. Thank you very much. $\endgroup$ – Sak Apr 26 '13 at 16:39
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Symmetric matrices are diagonalizable: $$ S = U D U^T$$ for $U^T = U^{-1}$ and $D$ a diagonal matrix whose entries are the eigenvalues. When $S$ is positive definite, all the eigenvalues must be strictly positive. Hence if $D = \text{diag}(d_1,\dotsc,d_n)$ then $D$ is invertible with $D^{-1} = \text{diag}(d_1^{-1},\dotsc,d_n^{-1})$. Then one may check that $S^{-1} = U D^{-1} U^T$, and in particular $S$ is invertible.

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Symmetric matrices are diagonalizable,and diagonal entries are real. if it's positive definite,then all the eigenvalues must be strictly positive.And similar matrices have the same eigenvalues. so,it's invertible.

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