To analyze the given 2 x 2 matrix \( M = \begin{pmatrix} 0 & a \\ a & b \end{pmatrix} \), with \( a, b > 0 \), we need to check each of the statements provided in the options:
M is a real symmetric matrix: A matrix is symmetric if it is equal to its transpose. Let's compute the transpose of matrix \( M \): \(\begin{pmatrix} 0 & a \\ a & b \end{pmatrix}^T = \begin{pmatrix} 0 & a \\ a & b \end{pmatrix}\). Since the transpose of \( M \) is equal to \( M \) itself, matrix \( M \) is indeed a real symmetric matrix.
One of the eigenvalues of M is greater than b: To find the eigenvalues of the matrix \( M \), we solve the characteristic equation: \(\det(M - \lambda I) = 0\), where \( I \) is the identity matrix.
\[\begin{vmatrix} -\lambda & a \\ a & b - \lambda \end{vmatrix} = 0\]
. The determinant is calculated as: \(-\lambda(b - \lambda) - a^2 = \lambda^2 - b\lambda - a^2 = 0\). Solving this quadratic equation: \(\lambda = \frac{b \pm \sqrt{b^2 + 4a^2}}{2}\). The larger root: \(\lambda_1 = \frac{b + \sqrt{b^2 + 4a^2}}{2}\) is clearly greater than \( b \), since \( a, b > 0 \).
One of the eigenvalues of M is negative: Using the solutions from before, the smaller eigenvalue is: \(\lambda_2 = \frac{b - \sqrt{b^2 + 4a^2}}{2}\). Since \( a > 0 \), the second term \(\sqrt{b^2 + 4a^2}\) is greater than \( b \), hence \(\lambda_2\) is negative.
Product of eigenvalues of M is b: The product of the eigenvalues of a matrix is equal to its determinant. Calculating the determinant of \( M \): \(\det M = \begin{vmatrix} 0 & a \\ a & b \end{vmatrix} = 0 \times b - a \times a = -a^2\). Therefore, the product of the eigenvalues is \(-a^2\), not \( b \).
Therefore, the correct statements about the matrix \( M \) are: