91Ó°ÊÓ

Let \(A\) be a hermitian positive definite matrix. Since \(A\) is positive definite, by definition, its eigenvalues are all positive. We can use the spectral theorem for hermitian matrices, which states that a matrix \(A\) can be diagonalized as: \[ A = U \Lambda U^* \] where \(U\) is a unitary matrix and \(\Lambda\) is a diagonal matrix with the eigenvalues of \(A\) on its diagonal. Since all the eigenvalues are positive, we can take the square root of each eigenvalue and construct a new diagonal matrix: \[ \Lambda^{1/2} = \text{diag}(\sqrt{\lambda_1}, \sqrt{\lambda_2}, \dots, \sqrt{\lambda_n}) \] Now we can define the matrix \(N\) as: \[ N = U \Lambda^{1/2} \] Note that \(N\) is non-singular because both \(U\) and \(\Lambda^{1/2}\) are non-singular. Finally, we can multiply \(N^* N\): \[ N^* N = (\Lambda^{1/2})^* U^* U \Lambda^{1/2} = \Lambda^{*1/2} \Lambda^{1/2} = \Lambda \] Since \(A = U \Lambda U^*\) and \(N^* N = U \Lambda U^*\), we have shown that there exists a non-singular matrix \(N\) such that \(A = N^* N\).

Let \(A = N^* N\) for some non-singular complex matrix \(N\). First, we will show that \(A\) is hermitian. By taking the conjugate transpose of \(A\), we get: \[ A^* = (N^* N)^* = N^{**} (N^*)^* = N N^* \] Since \(A = N^* N\), we would like to show that \(A^* = N N^* = N^* N = A\). We know that \(N\) is non-singular, and so is \(N^*\). Since the product of non-singular matrices is non-singular, we can deduce that \(N^* N\) is also non-singular. Thus: \[ (N^* N) (N N^*) = N^* (NN^*) N = N^* N \] Now multiplying by the inverse of \(N^* N\), which exists because \(N^* N\) is non-singular, we get: \[ (N^* N)^{-1} (N^* N) (N N^*) = (N^* N)^{-1} N^* N \] \[ I (N N^*) = I \] \[ N N^* = N^* N = A \] So \(A\) is a hermitian matrix since \(A = A^*\). Now, we will show that \(A\) is positive definite. For a vector \(x \in \mathbb{C}^n\), let \(y = Nx\). Then: \[ x^* A x = x^* (N^* N) x = x^* N^* N x = (Nx)^* (Nx) = y^* y \] Since \(y = Nx\) and \(N\) is non-singular, we know that \(y \neq 0\) if \(x \neq 0\). Thus, \(y^* y > 0\) for \(x \neq 0\). Therefore, \(A\) is positive definite. In conclusion, if there exists a non-singular matrix \(N\) such that \(A = N^* N\), then \(A\) is hermitian positive definite. Combining both parts, we have shown that a non-singular complex matrix \(A\) is hermitian positive definite if and only if there exists a non-singular matrix \(N\) such that \(A = N^* N\).