91Ó°ÊÓ

Recall the definition of a symmetric positive definite matrix: A matrix A is symmetric positive definite if it is symmetric (i.e., \(A = A^T\)) and positive definite (i.e., for all non-zero vectors x, \(x^T A x > 0\)).

To show that the determinant of A is positive, we can use the Cholesky decomposition. A symmetric positive definite matrix A can be factorized into a lower triangular matrix L and its transpose, such that \(A = LL^T\). Now, let's compute the determinant of A using this factorization. Recall that the determinant of a product of matrices is the product of their determinants, i.e., \(\operatorname{det}(AB) = \operatorname{det}(A) \operatorname{det}(B)\). Therefore, we have: \(\operatorname{det}(A) = \operatorname{det}(LL^T) = \operatorname{det}(L) \operatorname{det}(L^T)\). Since L is a lower triangular matrix, its determinant is equal to the product of its diagonal elements. If \(L = \begin{pmatrix} l_1 & 0 & \cdots & 0 \\ \vdots & \ddots & \ddots & \vdots \\ 0 & \cdots & l_n & 0 \end{pmatrix}\), then \(\operatorname{det}(L) = l_1 l_2 \cdots l_n\). Also, \(L^T\) is an upper triangular matrix, so its determinant is also the product of its diagonal elements, which are the same as the diagonal elements of L. Thus, \(\operatorname{det}(L^T) = l_1 l_2 \cdots l_n\). So, we have \(\operatorname{det}(A) = (l_1 l_2 \cdots l_n)^2\). Since the square of any non-zero scalar is positive, the determinant of a symmetric positive definite matrix A is positive (i.e., \(\operatorname{det}(A) > 0\)).

Now we should find a 2x2 matrix that has a positive determinant but is not positive definite. A matrix is not positive definite if there exists a non-zero vector x such that \(x^T A x \leq 0\). Consider the matrix \(B = \begin{pmatrix} 2 & -3 \\ -3 & 5\end{pmatrix}\). First, let's check its determinant: \(\operatorname{det}(B) = (2)(5) - (-3)(-3) = 10 - 9 = 1 > 0\). Now, let's check if it's positive definite. We need to find at least one non-zero vector x for which \(x^T B x \leq 0\). Let \(x = \begin{pmatrix} 1 \\ -1\end{pmatrix}\). Then, calculating \(x^T B x\): \(x^T B x = \begin{pmatrix} 1 & -1\end{pmatrix} \begin{pmatrix} 2 & -3 \\ -3 & 5\end{pmatrix} \begin{pmatrix} 1 \\ -1\end{pmatrix} = \begin{pmatrix} 5 & -2\end{pmatrix} \begin{pmatrix} 1 \\ -1\end{pmatrix} = 5(1) - 2(-1) = 5 + 2 = 7\). Since \(x^T B x = 7 > 0\), and we can't find any other non-zero vector x that satisfies \(x^T B x \leq 0\), the matrix B is positive definite. However, we need a matrix that is not positive definite. Let's try another matrix: \(C = \begin{pmatrix} 2 & 4 \\ 4 & 8\end{pmatrix}\). First, let's check its determinant: \(\operatorname{det}(C) = (2)(8) - (4)(4) = 16 - 16 = 0\). Now, for a matrix to be positive definite, its determinant must be positive, but the determinant of matrix C is 0. Thus, matrix C is a 2x2 matrix with positive (non-negative) determinant that is not positive definite.