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Chapter 8: Problem 1

Let \(A\) be an invertible hermitian matrix. Show that \(A^{-1}\) is hermitian.

Short Answer

Expert verified
To prove that the inverse of a Hermitian matrix is also Hermitian, we start from the inverse property \(AA^{-1} = I\), take the conjugate transpose of both sides, and apply properties of Hermitian matrices and conjugate transposes. We ultimately obtain the equation \((A^{-1})^H A = I\), and by comparing it to the original inverse property equation, we deduce that \((A^{-1})^H = A^{-1}\), which means \(A^{-1}\) is Hermitian.

Step by step solution

01

Write the inverse property

Given that \(A\) is an invertible matrix, we have the property \(AA^{-1} = I\), where \(I\) is the identity matrix.
02

Take the conjugate transpose of both sides

To find the conjugate transpose of the given equation, we can apply the fact that the conjugate transpose of the product of two matrices is equal to the product of the conjugate transposes of the individual matrices in reverse order: \((AB)^H = B^H A^H\). Thus, taking the conjugate transpose of both sides, we get: \((AA^{-1})^H = I^H\) By applying the rule mentioned above, this simplifies to: \((A^{-1})^H A^H = I^H\)
03

Use the Hermitian property of A

Since \(A\) is a Hermitian matrix, we have \(A = A^H\). Also, we know that the conjugate transpose of the identity matrix is equal to itself: \(I^H = I\). Substituting these into the expression obtained in step 2, we get: \((A^{-1})^H A = I\)
04

Compare the two expressions

Compare the expressions from Step 1 and Step 3: \(AA^{-1} = I\) \((A^{-1})^H A = I\) Notice that both equations have the form of matrices multiplying its inverse to form an identity matrix. Since \(A\) and \((A^{-1})^H\) both multiply with \(A^{-1}\) and \(A\), respectively, to obtain the identity matrix, it implies that: \((A^{-1})^H = A^{-1}\)
05

Conclusion

Since we have shown that \((A^{-1})^H = A^{-1}\), we can conclude that the inverse of a Hermitian matrix is also Hermitian.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse of a Matrix
Understanding the inverse of a matrix is fundamental in linear algebra and is particularly relevant when dealing with Hermitian matrices. An inverse of a matrix, denoted as A^{-1}, is a matrix that, when multiplied with the original matrix A, yields the identity matrix I. The equation AA^{-1} = I represents this fundamental relationship.

It's important to note that not all matrices have an inverse; a matrix must be square (same number of rows and columns) and non-singular (having a non-zero determinant) to possess an inverse. When we're given an invertible Hermitian matrix, as in the exercise, we can employ several properties of matrix operations to demonstrate that its inverse will also retain the Hermitian quality, which leads to deeper insights in quantum mechanics, signal processing, and other fields.
Conjugate Transpose
The conjugate transpose of a matrix, often denoted by A^H or A^*, is a critical concept when working with complex matrices such as Hermitian matrices. It is achieved by taking the transpose of the matrix 鈥� swapping rows with columns 鈥� followed by taking the complex conjugate of each element. This operation is essential in the context of Hermitian matrices because for any Hermitian matrix A, it holds that A = A^H.

In the provided exercise, we saw that taking the conjugate transpose of the product of two matrices results in the reverse order of their conjugate transposes, meaning (AB)^H = B^H A^H. This property is crucial for proving that the inverse of a Hermitian matrix is also Hermitian, as seen in the later steps of the solution.
Identity Matrix
An identity matrix, denoted as I, plays the role of 1 in matrix multiplication. Just as multiplying any number by 1 yields the same number, multiplying any square (n x n) matrix by the corresponding identity matrix (of the same dimension) will yield the original matrix. For every n-dimensional square matrix A, we have AI = IA = A.

The identity matrix contains 1s on the diagonal and 0s elsewhere. Due to its simplicity and unique properties, the identity matrix is a pivotal element in the discussion of matrix inverses. It is also invariant under conjugate transpose, which means I^H = I, a property used in the exercise to show that the inverse of a Hermitian matrix is itself Hermitian.

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Most popular questions from this chapter

Let \(A\) be a hermitian matrix. Show that \(^{t} A\) and \(\bar{A}\) are hermitian. If \(A\) is invertible, show that \(A^{-1}\) is hermitian.

Let \(V\) be a finite dimensional space over the field \(K\), with a non- degenerate scalar product. Let \(A: V \rightarrow V\) be a linear map. Show that the image of \(^{t} A\) is the orthogonal space to the kernel of \(A\).

Let \(A\) be a diagonal complex unitary matrix. Show that each diagonal element has absolute value 1 , and hence is of type \(e^{i \theta}\), with real \(\theta\). The following exercises describe various properties of real unitary maps of the plane \(\mathbf{R}^{2}\).

If \(A\) is a complex square matrix, show that \(\operatorname{Det}(\bar{A})=\operatorname{Det}(A)\). Conelude that the absolute value of the determinant of a complex unitary matrix is equal to \(1 .\)

Show that a triangular hermitian matrix is diagonal.
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