(a) Show that the inverse of an orthogonal matrix is orthogonal. Hint: Let A = \(\mathrm{O}^{-1} ;\) from \((9.2),\) write the condition for \(\mathrm{O}\) to be orthogonal and show that \(\mathrm{A}\) satisfies it. (b) Show that the inverse of a unitary matrix is unitary. See hint in (a). (c) If \(\mathrm{H}\) is Hermitian and \(\mathrm{U}\) is unitary, show that \(\mathrm{U}^{-1} \mathrm{HU}\) is Hermitian.

Orthogonal matrices are square matrices with orthonormal columns. This means that each column is a unit vector and is orthogonal to the other columns. Mathematically, a matrix \(O\) is orthogonal if \(O^T O = I\), where \(O^T\) is the transpose of \(O\) and \(I\) is the identity matrix.

Important properties of orthogonal matrices include:

• Their determinant is either +1 or -1.
• Rows are also orthonormal vectors.
• The inverse of an orthogonal matrix is its transpose.

For example, let's imagine we have an orthogonal matrix \(O\). By definition, if we transpose it and then multiply it by itself, we get the identity matrix: \(O^T O = I\). Now, if we consider the inverse matrix \(O^{-1}\), it must also satisfy the condition of orthogonality to be considered orthogonal. By multiplying both sides of the orthogonality condition \(O^T O = I\) by \(O^{-1}\), we end up with \(O^T = O^{-1}\), confirming that the transpose of \(O\) is indeed its inverse, thus maintaining orthogonality.

Unitary matrices are a critical concept in linear algebra, particularly in quantum mechanics and signal processing. A matrix \(U\) is unitary if \(U^* U = I\), where \(U^*\) is the conjugate transpose of \(U\). The conjugate transpose (also called Hermitian transpose) means that you take the transpose of the matrix and then take the complex conjugate of each entry.

Key properties of unitary matrices include:

• Their columns form an orthonormal set in the complex inner product space.
• Unitary matrices preserve the inner product, meaning the dot product of two vectors remains the same after applying the unitary transformation.
• The inverse of a unitary matrix is its conjugate transpose, \(U^{-1} = U^*\).

Let's consider a unitary matrix \(U\). If \(U\) satisfies \(U^* U = I\), its inverse \(U^{-1}\) must also be unitary. To show this, we take the definition, \(U^* U = I\), and multiply both sides by \(U^{-1}\). This results in \(U^* = U^{-1}\). Hence, taking the conjugate transpose of \(U^{-1}\), we get \(U^* = U^{-1}\), confirming that \(U^{-1}\) satisfies the condition for unitarity.

Hermitian matrices are fundamental in various fields, including quantum mechanics and complex systems. A matrix \(H\) is Hermitian if it equals its own conjugate transpose, \(H = H^*\). This property implies that Hermitian matrices have real eigenvalues and their eigenvectors can be chosen to form an orthonormal basis.

Significant properties of Hermitian matrices include:

• They are always square matrices.
• Their eigenvalues are real numbers.
• Eigenvectors corresponding to different eigenvalues are orthogonal.

Now, consider a Hermitian matrix \(H\) and a unitary matrix \(U\). If we form the product \(U^{-1} H U\), we need to show that it remains Hermitian. By taking the conjugate transpose of this product: $$(U^{-1} H U)^* = (U^{-1})^* H^* U^*$$ Since \(U\) is unitary, we know that \((U^{-1})^* = U\) and \(U^* = U^{-1}\). Additionally, because \(H\) is Hermitian, we have \(H^* = H\). Thus, substituting these into our expression, we get $$(U^{-1} H U)^* = U H U^{-1}$$, which simplifies back to \(U^{-1} H U\). Therefore, \(U^{-1} H U\) is indeed Hermitian.