91Ó°ÊÓ

Matrix invertibility is a crucial property in linear algebra. When a matrix is invertible, it essentially means that there exists another matrix, called the inverse, which reverses its effect. For a square matrix \( B \), it is invertible if there exists a matrix \( B^{-1} \) such that \( BB^{-1} = B^{-1}B = I \), where \( I \) is the identity matrix of the same dimensions.
However, in the context of the problem, we're not dealing with a square matrix \( A \) directly, but rather the product \( A^T A \). When \( A^T A \) is invertible, it implies a few key things:

• The determinant of \( A^T A \) is non-zero.
• It has full rank, which means that its columns are linearly independent.
• No vector except the zero vector is mapped to the zero vector by \( A^T A \).

In this problem, the invertibility of \( A^T A \) leads us to conclude that the columns of \( A \) itself must be linearly independent.

Matrix rank is a measure of a matrix's dimensionality, specifically how many independent columns (or rows) it has. For a matrix \( A \), the rank can be thought of as the number of linearly independent column vectors in the matrix.
When we consider \( A^T A \), its rank is particularly informative because:

• If \( A^T A \) has full rank, it means that the columns of \( A \) must also have enough freedom to interact in such a way that makes \( A^T A \) invertible.
• In this case, the rank of \( A^T A \) is the same as the number of columns in \( A \), since \( A^T A \) is invertible.

Thus, the rank directly affects the solution to the equation, ensuring there is no nontrivial solution to \( Ax = 0 \), which confirms our claim of linear independence of the columns of \( A \).

Understanding transposes is vital for working with matrices like \( A^T A \). The transpose of a matrix is an operation where you "flip" a matrix over its diagonal, which turns rows into columns and vice versa.
In the context of our matrix \( A \), its transpose \( A^T \) has some helpful properties:

• For matrices, \( (A^T)^T = A \), which means applying the transpose operation twice returns the original matrix.
• The transpose of a product of matrices is the product of their transposes in reverse order: \( (AB)^T = B^T A^T \). This result is important for understanding why \( A^T A \) forms a square matrix even when \( A \) is not.

These properties are often used to simplify expressions in linear algebra. For example, in our exercise, applying \( A^T \) to both sides of an equation helps determine invertibility and solve for potential solutions.

Determinants provide key insights into the characteristics of matrices. For a square matrix \( C \), the determinant \( \det(C) \) is a scalar that informs us about several properties:

• If \( \det(C) eq 0 \), the matrix is invertible, meaning there is a unique solution for the matrix equation \( Cx = b \) for every vector \( b \).
• If the determinant is zero, it indicates that the matrix does not have full rank and its columns are linearly dependent.

In the problem, the non-zero determinant of \( A^T A \) is pivotal. It ensures that \( A^T A \) is invertible, which in turn implies that the columns of \( A \) are linearly independent. This usage of the determinant is a classic linear algebra tool, linking matrix properties and linear independence closely.