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Chapter 2: Problem 77

Suppose \(A\) is a complex matrix. Show that \(A A^{H}\) and \(A^{H} A\) are Hermitian.

Short Answer

Expert verified
We showed that \(AA^H\) and \(A^H A\) are both Hermitian by using properties of conjugate transpose and matrix multiplication. Specifically, we demonstrated that \((AA^H)^H = AA^H\) and \((A^H A)^H = A^H A\), which confirms that both products are Hermitian matrices as they are equal to their conjugate transpose.

Step by step solution

01

Compute the conjugate transpose of AA^H

First, let's find the conjugate transpose of the matrix product AA^H. To do this, we must use the properties of conjugate transpose and matrix multiplication: (AA^H)^H = (A^H)^H A^H
02

Simplify (A^H)^H

Recall the property that taking the conjugate transpose twice will give you the original matrix back: (A^H)^H = A
03

Substitute back

Now we can substitute Step 2 back into the result from Step 1: (AA^H)^H = AA^H As we have shown that the conjugate transpose of AA^H is equal to AA^H itself, it follows that AA^H is Hermitian.
04

Compute the conjugate transpose of A^H A

Next, let's find the conjugate transpose of the matrix product A^H A. We will use the same properties as in Step 1: (A^H A)^H = AH A^H
05

Simplify substituting A^H

Recall from step 2 that the conjugate transpose of A^H is the original matrix A. Substitute this into the result from Step 4: (A^H A)^H = A^H A As the conjugate transpose of A^H A is equal to A^H A itself, it follows that A^H A is also Hermitian. Since we have proven that AA^H and A^H A are both Hermitian matrices, the given exercise is complete.

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Key Concepts

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

Complex Matrix
A complex matrix is a grid of numbers where each element can be a complex number, containing both a real and an imaginary part. Complex numbers are written in the form of a + bi, where 'a' is the real part and 'bi' is the imaginary part. This imaginary component is what sets complex matrices apart from real matrices. Complex matrices are fundamental in various fields such as quantum mechanics and signal processing, where they help handle complex calculations and transformations.
Understanding a complex matrix is crucial as it forms the basis for many operations we carry out in linear algebra, like addition, multiplication, and finding transpose, especially in dealing with unique properties of matrices like Hermitian matrices. It is this understanding that helps us decipher complex matrix properties and operations.
Matrix Product
Matrix product is an operation where two matrices are multiplied together. For two matrices to be multiplied, the number of columns in the first matrix must match the number of rows in the second matrix. The resulting product matrix is generated by computing the dot products of rows and columns.
When dealing with complex matrices like in Hermitian trials, the matrix product becomes a little more advanced as it often involves not only multiplication but also other operations like taking transpose or the complex conjugate. Ensuring that operations like conjugate transpose remain the same before and after multiplication is critical in proving properties related to Hermitian matrices. Thus, understanding matrix product operations thoroughly is essential in mastering complex matrix operations.
Conjugate Transpose
The conjugate transpose, also known as the Hermitian transpose, is an operation on a complex matrix that involves two steps: transposing the matrix and then taking the complex conjugate of each element. The transpose of a matrix is achieved by swapping its rows and columns, while the complex conjugate changes the sign of the imaginary part of each element.
In applications involving Hermitian matrices, conjugate transpose plays a significant part as it is pivotal in defining and proving whether a matrix is Hermitian. A matrix is considered Hermitian if its conjugate transpose is equal to the original matrix itself, a characteristic often used in proving properties like in the exercise where we show AAH is Hermitian.
Properties of Hermitian
Hermitian matrices are a special type of complex matrix with unique properties and roles, especially when dealing with complex numbers. A matrix is termed Hermitian if it is equal to its own conjugate transpose. This results in several important properties:
  • The diagonal elements of a Hermitian matrix are always real numbers.
  • Hermitian matrices have real eigenvalues.
  • They are also essential in quantum mechanics where they often represent observable quantities.
In the given exercise, the task was to prove that certain matrix products are Hermitian. Both product matrices, AAH and AHA, confirmed to have their conjugate transposes being equal to themselves, thus confirming their Hermitian nature. Mastery of these characteristics aids significantly in linear algebra applications involving complex matrices.

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Most popular questions from this chapter

Prove Theorem \(2.2(\mathrm{i}):(A B) C=A(B C)\) Let \(A=\left[a_{i j}\right], \quad B=\left[b_{j k}\right], \quad C=\left[c_{k i}\right], \quad\) and let \(A B=S=\left[s_{i k}\right], \quad B C=T=\left[t_{j l}\right] .\) Then $$ s_{i k}=\sum_{j=1}^{m} a_{i j} b_{j k} \quad \text { and } \quad t_{j l}=\sum_{k=1}^{n} b_{j k} c_{k l} $$ Multiplying \(S=A B\) by \(C\), the il-entry of \((A B) C\) is $$ s_{i 1} c_{1 l}+s_{i 2} c_{2 l}+\cdots+s_{i n} c_{n l}=\sum_{k=1}^{n} s_{i k} c_{k l}=\sum_{k=1}^{n} \sum_{j=1}^{m}\left(a_{i j} b_{j k}\right) c_{k l} $$ On the other hand, multiplying \(A\) by \(T=B C\), the il-entry of \(A(B C)\) is $$ a_{i 1} t_{1 l}+a_{i 2} t_{2 l}+\cdots+a_{i m} t_{m l}=\sum_{j=1}^{m} a_{i j} t_{j l}=\sum_{j=1}^{m} \sum_{k=1}^{n} a_{i j}\left(b_{j k} c_{k l}\right) $$ The above sums are equal; that is, corresponding elements in \((A B) C\) and \(A(B C)\) are equal. Thus, \((A B) C=A(B C)\)

Let \(A=\left[a_{i j}\right]\) and \(B=\left[b_{i j}\right]\) be upper triangular matrices. Prove that \(A B\) is upper triangular with diagonal \(a_{11} b_{11}, a_{22} b_{22}, \ldots, a_{n n} b_{n n}\) Let \(A B=\left[c_{i j}\right]\). Then \(c_{i j}=\sum_{k=1}^{n} a_{i k} b_{k j}\) and \(c_{i i}=\sum_{k=1}^{n} a_{i k} b_{k i}\). Suppose \(i>j\). Then, for any \(k\), either \(i>k\) or \(k>j\), so that either \(a_{i k}=0\) or \(b_{k j}=0\). Thus, \(c_{i j}=0\), and \(A B\) is upper triangular. Suppose \(i=j\). Then, for \(ki\), we have \(b_{k i}=0\). Hence, \(c_{i i}=a_{i i} b_{i i}\), as claimed. [This proves one part of Theorem 2.5(i); the statements for \(A+B\) and \(k A\) are left as exercises.]

Suppose \(B=\left[\begin{array}{cc}4 & x+2 \\ 2 x-3 & x+1\end{array}\right]\) is symmetric. Find \(x\) and \(B\). Set the symmetric elements \(x+2\) and \(2 x-3\) equal to each other, obtaining \(2 x-3=x+2\) or \(x=5\). Hence, \(B=\left[\begin{array}{ll}4 & 7 \\ 7 & 6\end{array}\right]\)

Find \(M^{2}\) and \(M^{3}\) for (a) \(M=$$\left[\begin{array}{ccccc}2 & 0 & 0 & 0 \\\ 0 & , 1 & 4 & & 0 \\ 0 & -1 & 2 & -\frac{1}{1} & -0 \\ 0 & 10 & 0 & 3\end{array}\right]\) (b) \(M=\) \(\left[\begin{array}{cccc}1 & 1 & 0 & 0 \\ 2 & 3 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 14 & 5\end{array}\right]\)

Let \(B=\left[\begin{array}{rr}1 & 0 \\ 26 & 27\end{array}\right]\). Find a matrix \(A\) such that \(A^{3}=B\).
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