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Prove that a $3\times3$


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 first –  Berci Oct 8 '12 at 21:53

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up vote 3 down vote accepted

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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Thank you marvis!! But we havent learned eigenvalues yet? What can I replace eigenvalues with to prove your hint? –  diimension Oct 8 '12 at 22:06
@diimension: Since $A$ is symmetric positive definite, you can find $B$ such that $A = B \,^t B$. So you get $\det(A) = \det(B) \det(^t B) = (\det(B))^2 \ge 0$. –  Joel Cohen Oct 8 '12 at 22:15
Beautiful thank you very much!! –  diimension Oct 8 '12 at 22:18
@JoelCohen how will I be able to do the first part?Do I take the quadratic function into account here? –  diimension Oct 8 '12 at 22:20
@diimension : The existence of $B$ is equivalent to saying the quadratic function can be written as a sum of squares : If $q(x) = (b_{1,1} x_1 + \ldots + b_{1,n} x_n)^2 + (b_{2,1} x_1 + \ldots + b_{2,n} x_n)^2 + \ldots + (b_{n,1} x_1 + \ldots + b_{n,n} x_n)^2$, then $A = \,^t B B$ with $B = (b_{i,j})$. –  Joel Cohen Oct 8 '12 at 22:28

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