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Understanding the column space of a matrix is crucial in linear algebra. It represents the set of all possible linear combinations of the column vectors of a matrix. For the given matrix \(A\), which is \[A = \begin{bmatrix} 1 & 5 \ 3 & 1 \ -2 & 4 \end{bmatrix}\]the column space, denoted as \( \text{Col } A \), is the span of its column vectors: \( \mathbf{a}_1 = \begin{bmatrix} 1 \ 3 \ -2 \end{bmatrix} \) and \( \mathbf{a}_2 = \begin{bmatrix} 5 \ 1 \ 4 \end{bmatrix} \).

• The column space gives us insight into the dimension of a vector space formed by the columns of the matrix.
• It is important for understanding linear transformations and systems of linear equations.

The span of these vectors indicates that any vector within \( \text{Col } A \) can be expressed as a combination of \( \mathbf{a}_1 \) and \( \mathbf{a}_2 \). This is a key concept for determining solutions to equations involving matrices, such as finding orthogonal projections and least-squares solutions.

The Gram-Schmidt process is essential for obtaining an orthonormal basis from a set of vectors. It involves orthogonalizing a set of vectors and then normalizing them to create an orthonormal set. Here's how it works for our matrix \(A\):

• Start with \(\mathbf{a}_1\) and \(\mathbf{a}_2\).
• Set \(\mathbf{u}_1 = \mathbf{a}_1\). Normalize it to get \(\mathbf{e}_1 = \frac{\mathbf{u}_1}{\|\mathbf{u}_1\|}\).
• Project \(\mathbf{a}_2\) onto \(\mathbf{u}_1\) to make it orthogonal: \(\mathbf{u}_2 = \mathbf{a}_2 - \text{proj}_{\mathbf{u}_1} \mathbf{a}_2\).
• Normalize \(\mathbf{u}_2\) to get \(\mathbf{e}_2\).

This process gives

• \(\mathbf{e}_1 = \frac{1}{\sqrt{14}}\begin{bmatrix} 1 \ 3 \ -2 \end{bmatrix}\)
• An orthogonal vector \(\mathbf{u}_2\) that can then be normalized.

The result is an orthonormal basis for the column space of \(A\). These orthonormal vectors simplify many calculations, especially projections.

A least-squares solution deals with minimizing the discrepancy between an observed vector and its approximation in the column space. For the given system \(A \mathbf{x} = \mathbf{b}\), where \[A = \begin{bmatrix} 1 & 5 \ 3 & 1 \ -2 & 4 \end{bmatrix}\] and \[\mathbf{b} = \begin{bmatrix} 4 \ -2 \ -3 \end{bmatrix}\],the least-squares solution is found as:

• Compute \(A^T A\) and \(A^T \mathbf{b}\).
• Solve \(A^T A \mathbf{x} = A^T \mathbf{b}\), also known as the normal equation, to find \(\mathbf{x}\).

This approach is useful when the system does not have an exact solution, allowing us to find the closest possible approximation. The least-squares method ensures that the sum of the square of the residuals is minimized, providing the best fit to the data.

Normal equations are derived from the least-squares method and are used to find the best approximation for a solution. For the linear system \(A \mathbf{x} = \mathbf{b}\), the normal equations are represented as:\[A^T A \mathbf{x} = A^T \mathbf{b}\]Here's how you can solve this:

• First, compute the transpose of \(A\), which is \(A^T\).
• Calculate \(A^T A\), a symmetric matrix.
• Calculate \(A^T \mathbf{b}\).
• Solve the resulting linear system using matrix methods such as Gaussian elimination or matrix inversion.

In our specific example, the computation of\[A^T A = \begin{bmatrix} 14 & 12 \ 12 & 42 \end{bmatrix}\]and \[A^T \mathbf{b} = \begin{bmatrix} 1 \ -2 \end{bmatrix}\]gives us a system we can solve for \(\mathbf{x}\) to find the least-squares solution. Normal equations are fundamental in linear regression and data fitting scenarios, providing a straightforward way to approximate solutions.