Prooving That Eigenvalues Are In The Trace Of $A=uu^*$

$$

Introduction

In linear algebra, the concept of eigenvalues and eigenvectors plays a crucial role in understanding the properties of matrices. Given a square matrix $A$ of $n \times n$ dimensions, we are often interested in finding its eigenvalues and eigenvectors. In this article, we will focus on proving that the eigenvalues of a matrix $A$ are its trace ($Tr(A)$) and $0$, where $A$ is expressed as the product of a non-zero vector $u$ and its conjugate transpose $u^*$. This is a fundamental result in linear algebra, and we will provide a step-by-step proof to demonstrate its validity.

Preliminaries

Before we dive into the proof, let's establish some necessary background information. Given a square matrix $A$ of $n \times n$ dimensions, we can express it as the product of a non-zero vector $u$ and its conjugate transpose $u^*$, i.e., $A = uu^*$. This is a special case of a matrix decomposition, where the matrix $A$ is expressed as the outer product of a vector and its conjugate transpose.

Properties of the Matrix $A$

The matrix $A$ has several important properties that we will utilize in our proof. Specifically, we note that:

• $A$ is a square matrix of $n \times n$ dimensions.
• $A$ is expressed as the product of a non-zero vector $u$ and its conjugate transpose $u^*$, i.e., $A = uu^*$.
• $n > 2$, which implies that $A$ is a matrix of dimension greater than $2 \times 2$.

Proof of the Main Result

We are now ready to prove the main result, which states that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$. To do this, we will follow a step-by-step approach and utilize the properties of the matrix $A$.

Step 1: Show that $A$ is a Hermitian Matrix

First, we need to show that $A$ is a Hermitian matrix. A Hermitian matrix is a square matrix that is equal to its conjugate transpose, i.e., $A = A^*$. In this case, we have:

$$A = uu^* = (uu^*)^* = u^*u$$

Since $A = A^*$, we conclude that $A$ is a Hermitian matrix.

Step 2: Show that $A$ has Real Eigenvalues

Next, we need to show that $A$ has real eigenvalues. Since $A$ is a Hermitian matrix, we know that its eigenvalues are real. This is a fundamental property of Hermitian matrices, and it follows from the fact that the eigenvalues of a Hermitian matrix are the roots of the characteristic equation $|A - \lambda I| = 0$, where $I$ is the identity matrix.

Step 3: Show that the Eigenvalues of $A$ are its Trace ($Tr(A)$) and $0$

Now, we need to show that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$. To do this, we will utilize the fact that the trace of a matrix is equal to the sum of its eigenvalues. Specifically, we have:

$$Tr(A) = \sum_{i=1}^n \lambda_i$$

where $\lambda_i$ are the eigenvalues of $A$. Since $A$ is a Hermitian matrix, we know that its eigenvalues are real. Therefore, we can write:

$$Tr(A) = \sum_{i=1}^n \lambda_i = \lambda_1 + \lambda_2 + \cdots + \lambda_n$$

where $\lambda_i$ are the eigenvalues of $A$. Now, we need to show that $\lambda_i = 0$ for all $i \neq 1$. To do this, we will utilize the fact that $A$ is expressed as the product of a non-zero vector $u$ and its conjugate transpose $u^*$, i.e., $A = uu^*$. Specifically, we have:

$$A = uu^* = \begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_n \end{bmatrix} \begin{bmatrix} u_1^* & u_2^* & \cdots & u_n^* \end{bmatrix}$$

where $u_i$ are the components of the vector $u$. Now, we can write:

$$A = \begin{bmatrix} |u_1|^2 & u_1u_2^* & \cdots & u_1u_n^* \\ u_2u_1^* & |u_2|^2 & \cdots & u_2u_n^* \\ \vdots & \vdots & \ddots & \vdots \\ u_nu_1^* & u_nu_2^* & \cdots & |u_n|^2 \end{bmatrix}$$

where $|u_i|^2$ are the squared magnitudes of the components of the vector $u$. Now, we can see that the diagonal elements of $A$ are equal to the squared magnitudes of the components of the vector $u$, i.e., $|u_i|^2$. Therefore, we can write:

$$\lambda_i = |u_i|^2$$

for all $i \neq 1$. Now, we can see that $\lambda_i = 0$ for all $i \neq 1$, since the components of the vector $u$ are non-zero. Therefore, we conclude that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$.

Conclusion

In this article, we have proved that the eigenvalues of a matrix $A$ are its trace ($Tr(A)$) and $0$, where $A$ is expressed as the product of a non-zero vector $u$ and its conjugate transpose $u^*$. This is a fundamental result in linear algebra, and it has important implications for the study of matrices and their properties. We hope that this article has provided a clear and concise proof of this result, and that it will be useful to students and researchers in the field of linear algebra.

References

• [1] Horn, R. A., & Johnson, C. R. (2013). Matrix analysis. Cambridge University Press.
• [2] Strang, G. (2016). Linear algebra and its applications. Thomson Learning.
• [3] Trefethen, L. N., & Bau, D. (1997). Numerical linear algebra. SIAM.

Further Reading

For further reading on this topic, we recommend the following resources:

• [1] Linear Algebra and Its Applications by Gilbert Strang
• [2] Matrix Analysis by Roger A. Horn and Charles R. Johnson
• [3] Numerical Linear Algebra by Lloyd N. Trefethen and David Bau

Q: What is the significance of the result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$?

A: The result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$ has significant implications for the study of matrices and their properties. It provides a fundamental understanding of the relationship between the eigenvalues and the trace of a matrix, and has important applications in various fields such as linear algebra, differential equations, and quantum mechanics.

Q: What is the relationship between the eigenvalues and the trace of a matrix?

A: The trace of a matrix is equal to the sum of its eigenvalues. This is a fundamental property of matrices, and it has important implications for the study of matrix properties and applications.

Q: How does the result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$ relate to the concept of Hermitian matrices?

A: The result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$ is closely related to the concept of Hermitian matrices. A Hermitian matrix is a square matrix that is equal to its conjugate transpose, and it has real eigenvalues. The result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$ is a consequence of the fact that $A$ is a Hermitian matrix.

Q: What are some of the applications of the result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$?

A: The result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$ has important applications in various fields such as linear algebra, differential equations, and quantum mechanics. Some of the applications include:

• Linear Algebra: The result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$ provides a fundamental understanding of the relationship between the eigenvalues and the trace of a matrix, and has important implications for the study of matrix properties and applications.
• Differential Equations: The result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$ has important implications for the study of differential equations, particularly in the context of linear systems.
• Quantum Mechanics: The result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$ has important implications for the study of quantum mechanics, particularly in the context of quantum systems and their properties.

Q: How can I apply the result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$ in my research or studies?

A: The result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$ can be applied in various ways, depending on your research or studies. Some possible applications include:

• Linear Algebra: Use the result to study the properties of matrices and their eigenvalues, and to develop new algorithms and techniques for matrix analysis.
• Differential Equations: Use the result to study the properties of linear systems and their solutions, and to develop new techniques for solving differential equations.
• Quantum Mechanics: Use the result to study the properties of quantum systems and their eigenvalues, and to develop new techniques for quantum computing and simulation.

Q: What are some of the limitations of the result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$?

A: The result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$ has some limitations, including:

• Assumptions: The result assumes that $A$ is a Hermitian matrix, which may not be the case in all situations.
• Dimensionality: The result assumes that $A$ is a square matrix of dimension $n \times n$, which may not be the case in all situations.
• Eigenvalue Distribution: The result assumes that the eigenvalues of $A$ are distributed in a specific way, which may not be the case in all situations.

Q: How can I extend the result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$ to more general cases?

A: The result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$ can be extended to more general cases by relaxing the assumptions and considering more general matrix properties. Some possible extensions include:

• Non-Hermitian Matrices: Consider the case where $A$ is a non-Hermitian matrix, and study the properties of its eigenvalues and trace.
• Non-Square Matrices: Consider the case where $A$ is a non-square matrix, and study the properties of its eigenvalues and trace.
• Eigenvalue Distribution: Consider the case where the eigenvalues of $A$ are distributed in a specific way, and study the properties of its eigenvalues and trace.

Conclusion

In this article, we have provided a comprehensive overview of the result that the eigenvalues of $A$ are its trace ($Tr(A)$) and $0$, and have discussed its significance, applications, and limitations. We have also provided some possible extensions and generalizations of the result, and have highlighted some of the key concepts and techniques involved. We hope that this article has been helpful in providing a clear and concise understanding of this important result in linear algebra.