Problem 11
Question
Show that every square matrix \(A\) can be uniquely expressed as \(P+i Q\), where \(P\) and \(Q\) are Hermitian matrices.
Step-by-Step Solution
Verified Answer
Every square matrix \( A \) can be uniquely expressed as \( P + iQ \) with Hermitian matrices \( P \) and \( Q \).
1Step 1: Define Hermitian Matrices
A Hermitian matrix is defined as a square matrix that is equal to its own conjugate transpose. In mathematical terms, a matrix \( M \) is Hermitian if \( M = M^* \), where \( M^* \) denotes the conjugate transpose of the matrix.
2Step 2: Express Matrix A in Terms of Hermitian Matrices
To express a matrix \( A \) as \( P + iQ \), where \( P \) and \( Q \) are Hermitian, start by considering the decomposition of \( A \) using its real and imaginary parts. Define the matrices as follows:\[ P = \frac{A + A^*}{2} \]\[ Q = \frac{A - A^*}{2i} \]These decompositions ensure that \( P \) and \( Q \) are Hermitian.
3Step 3: Verify P is Hermitian
To show that \( P \) is Hermitian, we need to verify that \( P = P^* \). Calculate \( P^* \) and check:\[ P^* = \left(\frac{A + A^*}{2}\right)^* = \frac{A^* + (A^*)^*}{2} = \frac{A^* + A}{2} = P \]This shows that \( P \) is indeed Hermitian since \( P = P^* \).
4Step 4: Verify Q is Hermitian
Similar to \( P \), we verify that \( Q \) is Hermitian by checking that \( Q = Q^* \). Calculate \( Q^* \):\[ Q^* = \left(\frac{A - A^*}{2i}\right)^* = \frac{(A - A^*)^*}{2i} = \frac{A^* - A}{2(-i)} = \frac{A - A^*}{2i} = Q \]Since \( Q = Q^* \), \( Q \) is Hermitian.
5Step 5: Uniqueness of Expression
To prove uniqueness, assume there are two decompositions: \( A = P_1 + iQ_1 \) and \( A = P_2 + iQ_2 \), where both \( P_1, Q_1 \) and \( P_2, Q_2 \) are Hermitian. Then:\[ P_1 + iQ_1 = P_2 + iQ_2 \]Equating real and imaginary parts, we have \( P_1 = P_2 \) and \( Q_1 = Q_2 \). Hence, the decomposition is unique.
Key Concepts
Square Matrix DecompositionReal and Imaginary Parts of MatricesMatrix Conjugate Transpose
Square Matrix Decomposition
Decomposing a square matrix is an essential concept in linear algebra. A square matrix is a matrix with the same number of rows and columns. In this context, we're interested in expressing a square matrix \( A \) as the sum of two Hermitian matrices \( P \) and \( iQ \). To achieve this, the matrix \( A \) is split into real and imaginary components. This is known as the decomposition of the matrix into Hermitian components:
- \( P = \frac{A + A^*}{2} \)
- \( Q = \frac{A - A^*}{2i} \)
Real and Imaginary Parts of Matrices
In the realm of matrices, distinguishing between real and imaginary parts is crucial when handling complex matrices. Just as with complex numbers, matrices can have real and imaginary parts, allowing for further manipulation and understanding.The decomposition of a matrix \( A \) into its real and imaginary parts via Hermitian matrices \( P \) and \( Q \) comes from:
- Real part: \( P = \frac{A + A^*}{2} \)
- Imaginary part: \( Q = \frac{A - A^*}{2i} \)
Matrix Conjugate Transpose
The concept of the conjugate transpose is fundamental in understanding Hermitian matrices and, by extension, many matrix operations in linear algebra. The conjugate transpose of a matrix \( M \), denoted as \( M^* \), is obtained by taking the transpose of \( M \) and then taking the complex conjugate of each element.For any matrix \( M \), the conjugate transpose operation involves these steps:
- Transpose the matrix: switch rows with columns.
- Apply complex conjugation: replace each complex entry \( a + bi \) with \( a - bi \).
Other exercises in this chapter
Problem 10
If \(\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right]\) is to be the square root of two-rowed unit matrix, then \(\alpha, \beta\) and
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If \(A^{-1}=\left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 3 & 1 \\ 0 & 0 & -1 / 3\end{array}\right]\), then a. \(|A|=-1\) b. \(\operatorname{adj} A=\left[\begin{arr
View solution Problem 11
If \(A\) is a square matrix such that \(A^{2}=A\), then \((I+A)^{3}-7 A\) is a. \(3 I\) b. \(O\) c. \(I\) d. \(2 I\)
View solution Problem 11
If \(B\) is an idempotent matrix, and \(A=I-B\), then a. \(A^{2}=A\) b. \(A^{2}=I\) c. \(A B=O\) d. \(B A=O\)
View solution