Chapter 5: Problem 29
When you multiply a Hermitian matrix by a real number \(c\), is \(c A\) still Hermitian? If \(c=i\), show that \(i A\) is skew-Hermitian. The 3 by 3 Hermitian matrices are a subspace, provided that the "scalars" are real numbers.
Short Answer
Expert verified
Yes, \(cA\) is Hermitian if \(c\) is real. If \(c=i\), \(iA\) is skew-Hermitian.
Step by step solution
01
Definitions and Properties
A Hermitian matrix, also called self-adjoint, is a complex square matrix that is equal to its own conjugate transpose. If \(A\) is Hermitian, then \(A^* = A\), where \(A^*\) denotes the conjugate transpose.
02
Multiplying a Hermitian Matrix by a Real Number
To determine if \(cA\) is Hermitian when \(c\) is a real number, we consider the properties of a Hermitian matrix. When we multiply a Hermitian matrix \(A\) by a real number \(c\), the conjugate transpose of \(cA\) is \((cA)^* = cA^* = cA\) because the conjugate of a real scalar is itself. This satisfies the condition for Hermitian matrices.
03
Conclusion for Real Scalars
Thus, if \(c\) is a real number and \(A\) is Hermitian, then \(cA\) remains Hermitian since \((cA)^* = cA\).
04
Multiplying by the Imaginary Unit
Now, consider \(c = i\), the imaginary unit. The matrix \(iA\) is formed by multiplying each entry of \(A\) by \(i\). The conjugate transpose of \(iA\) is \((iA)^* = -iA^* = -iA\) (since \(A^* = A\) and conjugating \(i\) gives \(-i\)).
05
Conclusion for \(c = i\)
Because \((iA)^* = -iA\), the matrix \(iA\) is skew-Hermitian, which means it is equal to the negative of its own conjugate transpose.
06
Subspace Under Real Scalars
The set of 3x3 Hermitian matrices forms a subspace when the scalars are real numbers. This is because they are closed under addition and scalar multiplication by real numbers, as shown in Step 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conjugate Transpose
The conjugate transpose is a critical concept when dealing with Hermitian matrices. It is denoted by the symbol \(A^*\) for any given matrix \(A\). Essentially, the conjugate transpose involves two actions: taking the transpose of the matrix and replacing each element with its complex conjugate.
If a matrix is represented by \(A = [a_{ij}]\), then the conjugate transpose \(A^*\) is obtained by:
If a matrix is represented by \(A = [a_{ij}]\), then the conjugate transpose \(A^*\) is obtained by:
- Switching the rows and columns, i.e., exchanging the position of \(a_{ij}\) with \(a_{ji}\).
- Taking the complex conjugate of each element, which for any complex number \(z = x + yi\) (where \(x\) and \(y\) are real numbers) is given by \(\overline{z} = x - yi\).
Skew-Hermitian
A skew-Hermitian matrix is a complex square matrix that is equal to the negative of its own conjugate transpose. Mathematically, for a matrix \(A\), it is skew-Hermitian if \((A^*) = -A\). These matrices occur naturally when you consider what happens with imaginary coefficients in matrices.
For example, if you multiply a Hermitian matrix \(A\) by the imaginary unit \(i\), you obtain \(iA\). The conjugate transpose of \(iA\) is \((iA)^* = -iA\), showing that \(iA\) is skew-Hermitian.
Here's a simple way to think about the differences:
For example, if you multiply a Hermitian matrix \(A\) by the imaginary unit \(i\), you obtain \(iA\). The conjugate transpose of \(iA\) is \((iA)^* = -iA\), showing that \(iA\) is skew-Hermitian.
Here's a simple way to think about the differences:
- Hermitian matrices have purely real diagonal elements.
- Skew-Hermitian matrices have purely imaginary or zero diagonal elements.
Subspace
The term "subspace" is a key concept in linear algebra. It represents a special kind of set that exists within a larger vector space, following certain rules. A subspace must satisfy three properties:
- It must include the zero vector.
- It must be closed under addition, meaning that adding any two vectors in the subspace yields another vector in the subspace.
- It must be closed under scalar multiplication, such that multiplying any vector in the subspace by a real number still produces a vector in the subspace.
- The sum of any two Hermitian matrices is Hermitian.
- When you multiply a Hermitian matrix by a real number, it remains Hermitian, as shown in the exercise.