My name is Hanzhang Yin, and I have developed this website as a resource to facilitate the review of key concepts in abstract algebra, and it is not fully completed. Feel free to email me at hanyin@ku.edu if there are any errors or suggestions for improvement.
Problem 5. Suppose \( A \in \mathbb{C}^{m \times n} \) with \( m \geq n \).
(a) Show that matrix \( A^*A \) is always Hermitian positive semi-definite.
(b) Show that \( \|A^*A\|_2 = \|A\|_2^2 \).
(c) Show that if \( \text{rank}(A) = n \) then \( A^*A \) is invertible and \( \|A(A^*A)^{-1}A^*\|_2 = 1 \).
(d) Let \( A^\dagger \in \mathbb{C}^{n \times m} \) be the pseudoinverse (the Moore-Penrose inverse) of matrix \( A \) and suppose that \( \text{rank}(A) = n \). Show that \[ A^\dagger = (A^*A)^{-1}A^*. \]
Proof (a). Let \(x\in \mathbb{C}^n\) and we denote \(Ax = y\in \mathbb{C}^m\). Then, we have \[ x^*A^*Ax = (Ax)^*(Ax) = \|Ax\|_2^2 = \|y\|_2^2\geq 0, \] for any \(x\in \mathbb{R}^n\). Thus, we can know that \(A^*A\) is Hermitian positive semi-definite. \( \blacksquare \)
Proof (b). We denote the SVD of \(A\) as \(A = U\Sigma V^*\), where \(U\in \mathbb{C}^{m\times m}\) and \(V\in \mathbb{C}^{n\times n}\) are unitary matrices, and \(\Sigma\in \mathbb{R}^{m\times n}\) is a diagonal matrix with non-negative real numbers on the diagonal. Suppose that \(\Sigma_{1,1}\geq \Sigma_{2,2}\geq \cdots \geq \Sigma_{n,n} \geq 0\). Since the 2-norm of a matrix is the largest singular value of the matrix, we have \(\|A\|_2 = \Sigma_{1,1}\). Besides that, we can know that \(\Sigma_{1, 1}^2 \geq \Sigma_{2, 2}^2 \geq \cdots \geq \Sigma_{n, n}^2 \geq 0\). \[ A^*A = V\Sigma^*U^*U\Sigma V^* = V\Sigma^*\Sigma V^* = V\Sigma\Sigma V^* = V\Sigma^2 V^*. \] Since \(\Sigma_{1, 1}^2\) is the \((1, 1)\)-entry of \(\Sigma^2\), we can know that \(\|\Sigma^2\|_2 = \Sigma_{1, 1}^2\). Then, we can get \[ \|A^*A\|_2 = \|V\Sigma^2 V^*\|_2 = \|\Sigma^2\|_2 = \|\Sigma^2\|_2 = \Sigma_{1, 1}^2. \] Thus, we can know that \[ \|A^*A\|_2 = \|A\|_2^2. \tag*{\(\blacksquare\)} \]