Chapter 6: Problem 1
Show that (a) \((\widetilde{A B})=\tilde{B} A\) (b) \((A B)^{\dagger}=B^{\dagger} A^{\dagger}\) (c) \(A(B C)=(A B) C\) \((d) \operatorname{Tr} A B C=\operatorname{Tr} B C A\) where \(A, B, C\) are matrices.
Short Answer
Expert verified
(a) \(\tilde{A B} = \tilde{B} \tilde{A}\) (b) \( (A B)^{\dagger} = B^{\dagger} A^{\dagger}\) (c) \( A(B C) = (A B) C\) (d) \(\operatorname{Tr}(A B C) = \operatorname{Tr}(B C A)\).
Step by step solution
01
Show that \(\tilde{A B}=\tilde{B} A\)
Recall that \( \tilde{A} \) is the transpose of matrix A. By definition, if \( C = AB \), then the transpose \( \tilde{C} = (AB)^T \). Using the property of transposes, \( (AB)^T = B^T A^T \), we get: \[ \tilde{A B} = (A B)^T = B^T A^T = \tilde{B} \tilde{A}. \] Hence, \( \tilde{A B} = \tilde{B} \tilde{A} \).
02
Show that \((A B)^\dagger = B^\dagger A^\dagger\)
Recall that \( A^{\dagger} = (A^*)^T \), where \( A^* \) is the complex conjugate of A. For any matrices A and B, we need to show that: \[ (A B)^{\dagger} = (A B)^*^T = (B^*) (A^*)^T = B^{\dagger} A^{\dagger}. \] Thus, \( (A B)^{\dagger} = B^{\dagger} A^{\dagger} \).
03
Show that \((A(B C) = (A B) C \)
The associative property of matrix multiplication states that for any matrices \(A, B, \) and \(C \), the following holds: \[ A (BC) = (AB) C. \] This is due to the way matrix multiplication is defined, making sure that the order in which the matrices are multiplied does not change the product.
04
Show that \(\operatorname{Tr}(A B C) = \operatorname{Tr}(B C A)\)
The cyclic property of the trace of a matrix ensures that the trace remains invariant under cyclic permutations. For matrices \(A, B,\) and \(C\), \[ \operatorname{Tr}(A B C) = \operatorname{Tr}(B C A) = \operatorname{Tr}(C A B). \] Since the trace operates on the sum of diagonal elements, the order in terms of a cyclic rotation doesn't affect the value.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Transpose of a Matrix
The transpose of a matrix, denoted as \(\tilde{A}\), is an operation that flips a matrix over its diagonal. This means that the row and column indices of each element are swapped. For example, if matrix \(\text{A}\) has elements \(\text{A}_{ij}\), its transpose \(\tilde{A}\) will have elements \(\text{A}_{ji}\). The transpose is useful in many applications, including solving linear equations and in transformations. Here's how you can compute the transpose of a simple 2x2 matrix:
If \(\text{A} = \begin{bmatrix}a & b \ c & d\end{bmatrix}\), then \(\tilde{A} = \begin{bmatrix}a & c \ b & d\end{bmatrix}\). It’s important to note that the transpose of the product of two matrices follows this rule: \((A B)^T = B^T A^T\). This property shows that the transpose operation is distributive over matrix multiplication, helping simplify complex expressions.
If \(\text{A} = \begin{bmatrix}a & b \ c & d\end{bmatrix}\), then \(\tilde{A} = \begin{bmatrix}a & c \ b & d\end{bmatrix}\). It’s important to note that the transpose of the product of two matrices follows this rule: \((A B)^T = B^T A^T\). This property shows that the transpose operation is distributive over matrix multiplication, helping simplify complex expressions.
Complex Conjugate
The complex conjugate of a matrix involves taking the complex conjugate of each individual element. If an element in the matrix is represented by \(\text{a} + \text{bi}\), where \(\text{a}\) and \(\text{b}\) are real numbers and \(\text{i}\) is the imaginary unit, its complex conjugate is \(\text{a} - \text{bi}\). This is usually denoted by \(\text{A}^*\). For example, if \(\text{A} = \begin{bmatrix}1 + i & 2 - i \ 3 + 2i & 4 - i\end{bmatrix}\), then the complex conjugate \(\text{A}^*\) is \(\text{A}^* = \begin{bmatrix}1 - i & 2 + i \ 3 - 2i & 4 + i\end{bmatrix}\). This concept is particularly important in quantum mechanics and various fields of engineering. When combined with the transpose, we get the Hermitian adjoint or conjugate transpose, denoted by \(\text{A}^{\text{†}} \) which is calculated as \((A^T)^* \).
Matrix Multiplication
Matrix multiplication is a binary operation that produces a matrix from two matrices. It’s a bit more complex than simple element-wise multiplication. For two matrices \(\text{A}_{m×n}\) and \(\text{B}_{n×p}\), their product \(\text{C}_{m×p}\) is computed as \(\text{C}_{ij} = \text{A}_{i1}\text{B}_{1j} + \text{A}_{i2}\text{B}_{2j} + ... + \text{A}_{in}\text{B}_{nj} \). This means each element of the resulting matrix is a dot product of a row from the first matrix and a column from the second. One key point to remember is that matrix multiplication is not commutative, meaning \(\text{A}\text{B} e \text{B}\text{A} \) in general. However, it is associative: \(\text{A}(\text{B}\text{C}) = (\text{A}\text{B})\text{C} \), which simplifies the multiplication of multiple matrices.
Trace of a Matrix
The trace of a matrix, denoted as \(\text{Tr} A \), is the sum of the elements on the main diagonal of a square matrix \(\text{A} \). If \(\text{A} = [a_{ij}] \) is an \(\text{n} \) x \(\text{n} \) matrix, then \(\text{Tr} A = a_{11} + a_{22} + ... + a_{nn} \). For example, if \(\text{A} = \begin{bmatrix}1 & 2 \ 3 & 4\end{bmatrix}\), then \(\text{Tr} A = 1 + 4 = 5 \). The trace has some important properties: it is invariant under cyclic permutations, meaning that \(\text{Tr}(ABC) = \text{Tr}(BCA) = \text{Tr}(CAB) \). This property is particularly useful in simplifying the trace of a product of matrices. Furthermore, the trace is linear, meaning that for matrices \(\text{A}\) and \(\text{B}\), \(\text{Tr}(A + B) = \text{Tr} A + \text{Tr} B \).