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Prove that a $3\times3$

$K=\pmatrix{a&b&c\\b&d&e\\c&e&f}$

is positive definite if and only if $a > 0$, $ad - b^2 > 0 $ , $\det K > 0$. Also prove that an $n\times n$ matrix $K > 0$ is positive definite if and only if all the upper left square $k \times k$ subdeterminants are positive for $k = 1, \ldots, n$.

For the first part I know that in order it to be a positive definite then the quadratic function $q(x)$ can be transformed by completing the squares but I don't know how to prove $a > 0$, $ad - b^2 > 0 $ , $\det K > 0$. For the second part I have no idea how to do.

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    Second part implies the first2012-10-08

1 Answers 1

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HINT Note that if $A = \begin{pmatrix}A_{11} & A_{12}\\ A_{21} & A_{22} \end{pmatrix}$, where $A_{ij} \in \mathbb{R}^{n_i \times n_j}$ is positive definite, then choosing $x = \begin{pmatrix}x_1 \\ 0 \end{pmatrix}$, where $x_1 \in \mathbb{R}^{n_1 \times 1}$, we get $x^TAx = x_1^T A_{11} x_1$Hence, if $A$ is positive definite so is $A_{11}$ or for that matter any submatrix along the diagonal.

Hence, all you need to prove is that the determinant of a positive definite matrix is positive.

To prove the above fact, first prove that the eigenvalues of a positive definite matrix are positive (Why? Look at the definition of eigenvalue). Then prove that the determinant of a matrix is nothing but the product of all its eigenvalues (Why?).

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    Got it. Thank $y$ou ver$y$ much!2012-10-08