The image you've uploaded contains the proof of Theorem 5.1, which states that the system of normal equations $X^T X B = X^T Y$ is consistent. Here's a breakdown of the proof:

Summary of the Proof:

1. Setup:

   • Let $A = X^T X$ and $a = X^T Y$.
   • The goal is to show that the rank of $X^T X$ (denoted as $r(X^T X)$) is equal to the rank of the augmented matrix $[X^T X, X^T Y]$.

2. Rank Inequality:

   • We know that adding a column to a matrix can increase the rank or leave it unchanged. Thus: $r(X^T X) \leq r([X^T X, X^T Y])$

3. Second Rank Inequality:

   • The rank of $[X^T X, X^T Y]$ is bounded above by the minimum of the ranks of $X^T$ and $X^T Y$: $r(X^T X) \leq \min(r(X^T), r(X^T Y))$

4. Conclusion:

   • Since $r(X^T X) \leq r(X) = r(X^T X)$, the two ranks must be equal: $r(X^T X) = r([X^T X, X^T Y])$
   • This implies that the system is consistent, meaning the normal equations $X^T X B = X^T Y$ have solutions. However, these solutions may not be unique (there could be multiple solutions).

The core idea of this proof is based on properties of matrix rank and consistency in systems of linear equations.

Would you like further clarification on any part of this proof or more details about matrix ranks and consistency in linear systems?

Related Questions:

1. How is matrix rank related to the existence of solutions in linear systems?
2. Can you explain more about the uniqueness of solutions in the normal equations?
3. What happens when the matrix $X^T X$ is not invertible?
4. How do normal equations relate to least squares regression?
5. What are some methods to handle cases where normal equations do not have a unique solution?

Tip:

When working with normal equations, if the matrix $X^T X$ is not invertible (i.e., it doesn't have full rank), you can use methods like regularization (e.g., Ridge Regression) to ensure uniqueness.