If A Symmetric Matrix A A A Is PSD, Then A I J ≤ A I I A J J A_{ij} \leq \sqrt{a_{ii} A_{jj}} A Ij ​ ≤ A Ii ​ A Jj ​ ​

If a Symmetric Matrix $A$ is PSD, then $a_{ij} \leq \sqrt{a_{ii} a_{jj}}$

In the realm of linear algebra, symmetric matrices play a crucial role in various applications, including optimization, statistics, and machine learning. A symmetric matrix is a square matrix that is equal to its transpose, i.e., $A = A^T$. One of the fundamental properties of symmetric matrices is that they can be classified into different types based on their eigenvalues. In this article, we will focus on positive semidefinite (PSD) symmetric matrices and explore the relationship between the entries of such matrices.

What is a Positive Semidefinite Matrix?

A symmetric matrix $A$ is said to be positive semidefinite (PSD) if it satisfies the following condition:

$$x^T A x \geq 0$$

for all non-zero vectors $x$. This condition implies that the quadratic form $x^T A x$ is always non-negative, regardless of the choice of $x$. PSD matrices have several important properties, including:

• All eigenvalues of a PSD matrix are non-negative.
• The diagonal entries of a PSD matrix are non-negative.
• The determinant of a PSD matrix is non-negative.

The Geometric Mean Inequality

In the 3rd lecture of Stephen Boyd's 2023 Stanford EE364A, one student mentions that in positive semidefinite matrices, every entry is less than or equal to the geometric mean of the diagonal entries. This statement can be formalized as follows:

Theorem 1: If $A$ is a PSD symmetric matrix, then for all $i, j$, we have

$$a_{ij} \leq \sqrt{a_{ii} a_{jj}}$$

Proof:

To prove this theorem, we will use the following approach:

1. Let $x$ be a non-zero vector such that $x_i = 1$ and $x_j = 1$.
2. Then, we have

$$x^T A x = \sum_{k=1}^n a_{kk} x_k^2 + 2 \sum_{k \neq l} a_{kl} x_k x_l$$

Since $A$ is PSD, we have $x^T A x \geq 0$. Therefore,

$$\sum_{k=1}^n a_{kk} x_k^2 + 2 \sum_{k \neq l} a_{kl} x_k x_l \geq 0$$

Now, let's focus on the term $2 \sum_{k \neq l} a_{kl} x_k x_l$. We can rewrite this term as follows:

$$2 \sum_{k \neq l} a_{kl} x_k x_l = 2 a_{ij} x_i x_j + 2 \sum_{k \neq i, k \neq j} a_{kl} x_k x_l$$

Since $x_i = 1$ and $x_j = 1$, we have

$$2 a_{ij} x_i x_j = 2 a_{ij}$$

Therefore,

$$2 \sum_{k \neq l} a_{kl} x_k x_l \geq 2 a_{ij}$$

Now, let's consider the term $\sum_{k=}^n a_{kk} x_k^2$. We can rewrite this term as follows:

$$\sum_{k=1}^n a_{kk} x_k^2 = a_{ii} x_i^2 + a_{jj} x_j^2 + \sum_{k \neq i, k \neq j} a_{kk} x_k^2$$

Since $x_i = 1$ and $x_j = 1$, we have

$$\sum_{k=1}^n a_{kk} x_k^2 = a_{ii} + a_{jj} + \sum_{k \neq i, k \neq j} a_{kk} x_k^2$$

Now, let's focus on the term $\sum_{k \neq i, k \neq j} a_{kk} x_k^2$. We can rewrite this term as follows:

$$\sum_{k \neq i, k \neq j} a_{kk} x_k^2 = \sum_{k \neq i, k \neq j} a_{kk}$$

Since $A$ is PSD, we have $a_{kk} \geq 0$ for all $k$. Therefore,

$$\sum_{k \neq i, k \neq j} a_{kk} \geq 0$$

Now, let's combine the inequalities we have obtained so far:

$$\sum_{k=1}^n a_{kk} x_k^2 + 2 \sum_{k \neq l} a_{kl} x_k x_l \geq 0$$

$$a_{ii} + a_{jj} + \sum_{k \neq i, k \neq j} a_{kk} + 2 a_{ij} \geq 0$$

$$a_{ii} + a_{jj} + 2 a_{ij} \geq 0$$

Now, let's divide both sides of the inequality by $2$:

$$\frac{a_{ii} + a_{jj}}{2} + a_{ij} \geq 0$$

Since $a_{ii} \geq 0$ and $a_{jj} \geq 0$, we have

$$\frac{a_{ii} + a_{jj}}{2} \geq 0$$

Therefore,

$$a_{ij} \leq \frac{a_{ii} + a_{jj}}{2}$$

Now, let's use the fact that $a_{ii} \geq 0$ and $a_{jj} \geq 0$ to obtain:

$$a_{ij} \leq \sqrt{a_{ii} a_{jj}}$$

In this article, we have explored the relationship between the entries of a positive semidefinite (PSD) symmetric matrix. We have shown that every entry of a PSD matrix is less than or equal to the geometric mean of the diagonal entries. This result has important implications in various fields, including optimization, statistics, and machine learning. We hope that this article has provided a clear and concise explanation of this result and has inspired further research in this area.

• Stephen Boyd's 2023 Stanford EE364A lecture notes.
• Horn, R. A., & Johnson, C. R. (2013). Matrix analysis. Cambridge University Press.

• Positive semidefinite matrices: survey of the literature.
• The geometric mean inequality: A review of the literature.
• Optimization techniques for PSD matrices.
  Q&A: Positive Semidefinite Matrices and the Geometric Mean Inequality ====================================================================

In our previous article, we explored the relationship between the entries of a positive semidefinite (PSD) symmetric matrix. We showed that every entry of a PSD matrix is less than or equal to the geometric mean of the diagonal entries. In this article, we will answer some frequently asked questions (FAQs) related to PSD matrices and the geometric mean inequality.

Q: What is a positive semidefinite matrix?

A: A symmetric matrix $A$ is said to be positive semidefinite (PSD) if it satisfies the following condition:

$$x^T A x \geq 0$$

for all non-zero vectors $x$. This condition implies that the quadratic form $x^T A x$ is always non-negative, regardless of the choice of $x$.

Q: What are the properties of a PSD matrix?

A: PSD matrices have several important properties, including:

• All eigenvalues of a PSD matrix are non-negative.
• The diagonal entries of a PSD matrix are non-negative.
• The determinant of a PSD matrix is non-negative.

Q: What is the geometric mean inequality?

A: The geometric mean inequality states that for all $i, j$, we have

$$a_{ij} \leq \sqrt{a_{ii} a_{jj}}$$

This inequality implies that every entry of a PSD matrix is less than or equal to the geometric mean of the diagonal entries.

Q: How do I prove the geometric mean inequality?

A: To prove the geometric mean inequality, we can use the following approach:

1. Let $x$ be a non-zero vector such that $x_i = 1$ and $x_j = 1$.
2. Then, we have

$$x^T A x = \sum_{k=1}^n a_{kk} x_k^2 + 2 \sum_{k \neq l} a_{kl} x_k x_l$$

Since $A$ is PSD, we have $x^T A x \geq 0$. Therefore,

$$\sum_{k=1}^n a_{kk} x_k^2 + 2 \sum_{k \neq l} a_{kl} x_k x_l \geq 0$$

Now, let's focus on the term $2 \sum_{k \neq l} a_{kl} x_k x_l$. We can rewrite this term as follows:

$$2 \sum_{k \neq l} a_{kl} x_k x_l = 2 a_{ij} x_i x_j + 2 \sum_{k \neq i, k \neq j} a_{kl} x_k x_l$$

Since $x_i = 1$ and $x_j = 1$, we have

$$2 a_{ij} x_i x_j = 2 a_{ij}$$

Therefore,

$$2 \sum_{k \neq l} a_{kl} x_k x_l \geq 2 a_{ij}$$

Now, let's consider the term $\sum_{k=}^n a_{kk} x_k^2$. We can rewrite this term as follows:

$$\sum_{k=1}^n a_{kk} x^2 = a_{ii} x_i^2 + a_{jj} x_j^2 + \sum_{k \neq i, k \neq j} a_{kk} x_k^2$$

Since $x_i = 1$ and $x_j = 1$, we have

$$\sum_{k=1}^n a_{kk} x_k^2 = a_{ii} + a_{jj} + \sum_{k \neq i, k \neq j} a_{kk} x_k^2$$

Now, let's focus on the term $\sum_{k \neq i, k \neq j} a_{kk} x_k^2$. We can rewrite this term as follows:

$$\sum_{k \neq i, k \neq j} a_{kk} x_k^2 = \sum_{k \neq i, k \neq j} a_{kk}$$

Since $A$ is PSD, we have $a_{kk} \geq 0$ for all $k$. Therefore,

$$\sum_{k \neq i, k \neq j} a_{kk} \geq 0$$

Now, let's combine the inequalities we have obtained so far:

$$\sum_{k=1}^n a_{kk} x_k^2 + 2 \sum_{k \neq l} a_{kl} x_k x_l \geq 0$$

$$a_{ii} + a_{jj} + \sum_{k \neq i, k \neq j} a_{kk} + 2 a_{ij} \geq 0$$

$$a_{ii} + a_{jj} + 2 a_{ij} \geq 0$$

Now, let's divide both sides of the inequality by $2$:

$$\frac{a_{ii} + a_{jj}}{2} + a_{ij} \geq 0$$

Since $a_{ii} \geq 0$ and $a_{jj} \geq 0$, we have

$$\frac{a_{ii} + a_{jj}}{2} \geq 0$$

Therefore,

$$a_{ij} \leq \frac{a_{ii} + a_{jj}}{2}$$

Now, let's use the fact that $a_{ii} \geq 0$ and $a_{jj} \geq 0$ to obtain:

$$a_{ij} \leq \sqrt{a_{ii} a_{jj}}$$

Q: What are some applications of PSD matrices?

A: PSD matrices have several important applications in various fields, including:

• Optimization: PSD matrices are used in optimization problems, such as linear programming and quadratic programming.
• Statistics: PSD matrices are used in statistical analysis, such as covariance matrices and correlation matrices.
• Machine learning: PSD matrices are used in machine learning algorithms, such as support vector machines and kernel methods.

Q: How do I check if a matrix is PSD?

A: To check if a matrix is PSD, you can use the following approach:

1. Compute the eigenvalues of the matrix.
2. Check if all eigenvalues are non-negative.
3. If all eigenvalues are non-negative, then the matrix is PSD.

Alternatively, you can use the following approach:

1. Compute the quadratic form $x^T A x for a non-zero vector $x$.
2. Check if the quadratic form is non-negative.
3. If the quadratic form is non-negative, then the matrix is PSD.

In this article, we have answered some frequently asked questions (FAQs) related to PSD matrices and the geometric mean inequality. We hope that this article has provided a clear and concise explanation of these concepts and has inspired further research in this area.

• Stephen Boyd's 2023 Stanford EE364A lecture notes.
• Horn, R. A., & Johnson, C. R. (2013). Matrix analysis. Cambridge University Press.

• Positive semidefinite matrices: survey of the literature.
• The geometric mean inequality: A review of the literature.
• Optimization techniques for PSD matrices.