91Ó°ÊÓ

In linear algebra, the concept of a column space, or \( \text{Col} \ A \), plays a significant role in understanding matrix solutions. It refers to the set of all possible linear combinations of the column vectors of a matrix. Imagine \( A \) as an array of columns. Each column is like a vector in a coordinate space.
The column space is essentially the span of these vectors; it tells us all the possible vectors we can get by scaling and adding these columns together.

• This can also be interpreted as the image of the matrix when treated as a transformation.
• When we say vector \( \mathbf{w} \) is in \( \text{Col} \ A \), we're looking to see if \( \mathbf{w} \) can be expressed as some combination of the columns of \( A \).

In the exercise above, this means solving the equation \( A\mathbf{x} = \mathbf{w} \) using the augmented matrix form and row operations. If a solution exists, then \( \mathbf{w} \) belongs to the column space of \( A \). Here, the last row of the reduced matrix indicated an inconsistency, confirming that \( \mathbf{w} \) is not in \( \text{Col} \ A \).

The null space, denoted as \( \text{Nul} \ A \), of a matrix \( A \) is another fundamental concept. It includes all vectors \( \mathbf{x} \) that satisfy the equation \( A \mathbf{x} = \mathbf{0} \). In simple terms, it's the set of all solutions to the homogeneous equation.

Understanding the null space is crucial because:

• It reveals the linear dependencies among the columns of \( A \).
• The dimension of \( \text{Nul} \ A \), known as the nullity, tells us about the number of free variables in the system.

For the vector \( \mathbf{w} \) to be in \( \text{Nul} \ A \), multiplying \( A \) by \( \mathbf{w} \) should yield the zero vector. In the exercise, computing \( A \mathbf{w} \) indeed results in \( \mathbf{0} \), which confirms that \( \mathbf{w} \) lies within the null space of \( A \). This hints at underlying relationships among \( A \)'s columns.

Gaussian elimination is a powerful procedure used to solve linear systems and find matrix properties. It involves performing row operations to reduce a matrix to a simpler form, often aiming for the row-echelon form.
Here’s a step-by-step look at why it's handy:

• It systematically eliminates variables from equations, simplifying the system one row at a time.
• By achieving row-echelon form, it becomes easier to back-substitute and find solutions, if they exist.
• It helps identify inconsistencies, as seen in the step where we ended up with a row indicating no solution for \( A \mathbf{x} = \mathbf{w} \).

In the exercise, the row operations led to a matrix row showing \[ 0x_1 + 0x_2 + 0x_3 = 3 \].
This inconsistency (a non-zero entry in the augmented part but zeros in the coefficients) meant there was no solution for that system, and hence, \( \mathbf{w} \) wasn't in the column space.