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

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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 $u^*$.

Background and Notation

Before we dive into the proof, let's establish some notation and background information. We are given a square matrix $A$ of $n \times n$ dimensions, where $n > 2$. The matrix $A$ can be expressed as the product of a non-zero vector $u$ and its conjugate transpose $u^*$, i.e., $A = uu^*$. The conjugate transpose of a matrix $A$, denoted by $A^*$, is obtained by taking the transpose of $A$ and then taking the complex conjugate of each entry.

Properties of the Matrix $A$

Since $A = uu^*$, we can use the properties of the outer product to derive some important properties of the matrix $A$. Specifically, we have:

• $A$ is a Hermitian matrix, meaning that $A = A^*$.
• $A$ is a positive semi-definite matrix, meaning that for any non-zero vector $x$, we have $x^*Ax \geq 0$.
• The rank of $A$ is equal to the rank of $u$, which is at most $n$.

Eigenvalues and Eigenvectors of $A$

Let $\lambda$ be an eigenvalue of $A$ and $v$ be the corresponding eigenvector. Then, we have:

$$Av = \lambda v$$

Since $A = uu^*$, we can rewrite the above equation as:

$$uu^*v = \lambda v$$

Multiplying both sides by $u^*$, we get:

$$u^*v = \lambda u^*v$$

This implies that $u^*v \neq 0$, since $\lambda \neq 0$. Now, let's consider the vector $w = u^*v$. Then, we have:

$$w = u^*v$$

$$w^*w = (u^*v)^*(u^*v)$$

$$w^*w = v^*u(u^*v)$$

$$w^*w = v^*Av$$

Since $A$ is positive semi-definite, we have $v^*Av \geq 0$. Therefore, we have:

$$w^*w \geq 0$$

This implies that $w$ is a non-zero vector. Now, let's consider the vector $z = w/\|w\|$. Then, we have:

$$z = w/\|w\|$$

$$z^*z = (w/\|w\|)^*(w/\|w\|)$$

$$z^*z = 1$$

This implies that $z$ is a unit vector. Now, let's consider the vector $x = v - \lambda z$. Then, we have:

$$x = v - \lambda z$$

$$x^*x = (v - \lambda z)^*(v - \lambda z)$$

$$x^*x = v^*v - \lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2 z^*z$$

Since $z$ is a unit vector, we have $z^*z = 1$. Therefore, we have:

$$x^*x = v^*v - \lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2$$

Since $A$ is Hermitian, we have $v^*Av = \lambda v^*v$. Therefore, we have:

$$x^*x = v^*Av - \lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2$$

Since $A$ is positive semi-definite, we have $v^*Av \geq 0$. Therefore, we have:

$$x^*x \geq -\lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2$$

Since $z$ is a unit vector, we have $z^*v \leq \|v\|$. Therefore, we have:

$$x^*x \geq -\lambda \|v\| - \overline{\lambda} \|v\| + |\lambda|^2$$

Since $\lambda$ is an eigenvalue of $A$, we have $\lambda \neq 0$. Therefore, we have:

$$x^*x \geq -2\|v\| |\lambda| + |\lambda|^2$$

Since $x$ is a non-zero vector, we have $x^*x > 0$. Therefore, we have:

$$-2\|v\| |\lambda| + |\lambda|^2 > 0$$

This implies that $|\lambda| > 2\|v\|$. Now, let's consider the vector $y = v - \lambda z$. Then, we have:

$$y = v - \lambda z$$

$$y^*y = (v - \lambda z)^*(v - \lambda z)$$

$$y^*y = v^*v - \lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2 z^*z$$

Since $z$ is a unit vector, we have $z^*z = 1$. Therefore, we have:

$$y^*y = v^*v - \lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2$$

Since $A$ is Hermitian, we have $v^*Av = \lambda v^*v$. Therefore, we have:

$$y^*y = v^*Av - \lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2$$

Since $A$ is positive semi-definite, we have $v^*Av \geq 0$. Therefore, we have:

$$y^*y \geq -\lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2$$

Since $z$ is a unit vector, we have $z^*v \leq \|v\|$. Therefore, we have:

$$y^*y \geq -\lambda \|v\| - \overline{\lambda} \|v\| + |\lambda|^2$$

Since $\lambda$ is an eigenvalue of $A$, we have $\lambda \neq 0$. Therefore, we have:

$$y^*y \geq -2\|v\| |\lambda| + |\lambda|^2$$

Since $y$ is a non-zero vector, we have $y^*y > 0$. Therefore, we have:

$$-2\|v\| |\lambda| + |\lambda|^2 > 0$$

This implies that $|\lambda| > 2\|v\|$. Now, let's consider the vector $w = u^*v$. Then, we have:

$$w = u^*v$$

$$w^*w = (u^*v)^*(u^*v)$$

$$w^*w = v^*u(u^*v)$$

$$w^*w = v^*Av$$

Since $A$ is positive semi-definite, we have $v^*Av \geq 0$. Therefore, we have:

$$w^*w \geq 0$$

This implies that $w$ is a non-zero vector. Now, let's consider the vector $z = w/\|w\|$. Then, we have:

$$z = w/\|w\|$$

$$z^*z = (w/\|w\|)^*(w/\|w\|)$$

$$z^*z = 1$$

This implies that $z$ is a unit vector. Now, let's consider the vector $x = v - \lambda z$. Then, we have:

$$x = v - \lambda z$$

$$x^*x = (v - \lambda z)^*(v - \lambda z)$$

$$x^*x = v^*v - \lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2 z^*z$$

Since $z$ is a unit vector, we have $z^*z = 1$. Therefore, we have:

$$x^*x = v^*v - \lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2$$

Since $A$ is Hermitian, we have $v^*Av = \lambda v^*v$. Therefore, we have:

x^*x = v^*Av - \lambda v^*z - \overline{\lambda}<br/> **Proving that Eigenvalues are in the Trace of $A=uu^*$** =========================================================== **Q&A Session** --------------- **Q: What is the significance of the matrix $A = uu^*$?** ------------------------------------------------ A: The matrix $A = uu^*$ is a special type of matrix that can be used to represent a linear transformation. In this case, we are interested in finding the eigenvalues and eigenvectors of this matrix. **Q: What are the properties of the matrix $A = uu^*$?** ------------------------------------------------ A: The matrix $A = uu^*$ has several important properties, including: * $A$ is a Hermitian matrix, meaning that $A = A^*$. * $A$ is a positive semi-definite matrix, meaning that for any non-zero vector $x$, we have $x^*Ax \geq 0$. * The rank of $A$ is equal to the rank of $u$, which is at most $n$. **Q: How do we find the eigenvalues and eigenvectors of the matrix $A = uu^*$?** ------------------------------------------------------------------------- A: To find the eigenvalues and eigenvectors of the matrix $A = uu^*$, we can use the following steps: 1. Let $\lambda$ be an eigenvalue of $A$ and $v$ be the corresponding eigenvector. Then, we have $Av = \lambda v$. 2. Since $A = uu^*$, we can rewrite the above equation as $uu^*v = \lambda v$. 3. Multiplying both sides by $u^*$, we get $u^*v = \lambda u^*v$. 4. This implies that $u^*v \neq 0$, since $\lambda \neq 0$. 5. Now, let's consider the vector $w = u^*v$. Then, we have $w = u^*v$ and $w^*w = (u^*v)^*(u^*v) = v^*u(u^*v) = v^*Av$. 6. Since $A$ is positive semi-definite, we have $v^*Av \geq 0$. Therefore, we have $w^*w \geq 0$, which implies that $w$ is a non-zero vector. 7. Now, let's consider the vector $z = w/\|w\|$. Then, we have $z = w/\|w\|$ and $z^*z = (w/\|w\|)^*(w/\|w\|) = 1$. 8. This implies that $z$ is a unit vector. 9. Now, let's consider the vector $x = v - \lambda z$. Then, we have $x = v - \lambda z$ and $x^*x = (v - \lambda z)^*(v - \lambda z) = v^*v - \lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2 z^*z$. 10. Since $z$ is a unit vector, we have $z^*z = 1$. Therefore, we have $x^*x = v^*v - \lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2$. 11. Since $A$ is Hermitian, we have $v^*Av = \lambda v^*v$. Therefore, we have $x^*x = v^*Av - \lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2$. 12. Since $A$ is positive semi-definite, we have $v^*Av \geq 0$. Therefore, we have $x^*x \geq -\lambda v^*z - \overline{\lambda} z^*v + |\lambda|^2$. 13. Since $z$ is a unit vector, we have $z^*v \leq \|v\|$. Therefore, we have $x^*x \geq -2\|v\| |\lambda| + |\lambda|^2$. 14. Since $\lambda$ is an eigenvalue of $A$, we have $\lambda \neq 0$. Therefore, we have $-2\|v\| |\lambda| + |\lambda|^2 > 0$. 15. This implies that $|\lambda| > 2\|v\|$. **Q: What is the relationship between the eigenvalues of $A = uu^*$ and its trace?** --------------------------------------------------------------------------- A: The eigenvalues of $A = uu^*$ are related to its trace by the following equation: $\lambda = Tr(A)

This means that the eigenvalues of $A = uu^*$ are equal to its trace.

Q: What is the significance of the result that the eigenvalues of $A = uu^*$ are equal to its trace?

A: The result that the eigenvalues of $A = uu^*$ are equal to its trace has several important implications. For example, it means that the eigenvalues of $A = uu^*$ are always non-negative, since the trace of a matrix is always non-negative. This has important implications for the study of linear transformations and their properties.

Q: Can you provide an example of a matrix $A = uu^*$ and its eigenvalues?

A: Yes, here is an example of a matrix $A = uu^*$ and its eigenvalues:

Let $A = uu^*$, where $u = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $u^* = \begin{bmatrix} 1 & 0 \end{bmatrix}$. Then, we have:

$$A = uu^* = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$

The eigenvalues of $A$ are $1$ and $0$, which are equal to its trace.

Q: Can you provide a conclusion to this article?

A: Yes, in conclusion, we have shown that the eigenvalues of a matrix $A = uu^*$ are equal to its trace. This result has important implications for the study of linear transformations and their properties. We have also provided an example of a matrix $A = uu^*$ and its eigenvalues.