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In the realm of linear algebra, a linear combination involves creating a new vector by multiplying each vector in a given set by a constant and then summing the results. To visualize, consider vectors \( \mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v_n} \). If \( c_1, c_2, \ldots, c_n \) are constants, then \( c_1\mathbf{v_1} + c_2\mathbf{v_2} + \ldots + c_n\mathbf{v_n} \) is a linear combination of these vectors. This concept is crucial in determining if a vector can belong to a particular span, such as the column space of a matrix.

For instance, the column space of matrix \( A \) consists of all linear combinations of its columns. If you can write a vector \( \mathbf{b} \) as a linear combination of these columns, then \( \mathbf{b} \) is in the column space. If not, like in our example, \( \mathbf{b} \) is not part of that column space. This means you can't find scalars that make a linear combination of the matrix columns equal to \( \mathbf{b} \). Such insights are key to many problems in linear algebra.

The rank of a matrix is a crucial concept pointing to how much of the space the matrix effectively covers by pointing out the number of independent columns or rows. The rank tells us about the size of the column space and the row space. For our matrix \( A \), which is a \( 3 \times 3 \) matrix provided in the example, the rank is 2 despite its square shape because it only has two linearly independent columns.

Understanding rank helps in solving systems of linear equations. A full-rank \( 3 \times 3 \) matrix would have a column space equal to all \( \mathbb{R}^3 \), allowing any vector in the space to be a linear combination of the columns of the matrix. However, with rank 2, the column space becomes a plane in \( \mathbb{R}^3 \), meaning vectors like our \( \mathbf{b} = \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} \) cannot fall within it, demonstrating the limitation set by this matrix's rank.

Matrix vector multiplication is a fundamental operation where a matrix is multiplied by a vector, resulting in another vector. For a matrix \( A \) and a vector \( \mathbf{x} \), the product \( A\mathbf{x} \) gives a new vector. Crucially, if \( \mathbf{b} \) is in the column space of \( A \), then there exists a vector \( \mathbf{x} \) such that \( A\mathbf{x} = \mathbf{b} \).

In our example, by checking if \( A\mathbf{x} = \mathbf{b} \) is solvable, we determine if \( \mathbf{b} \) resides in the column space of \( A \). When you multiply \( A = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 0 \end{pmatrix} \) by a vector \( \mathbf{x} = \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} \), you get \( \begin{pmatrix} x_1 \ x_2 \ 0 \end{pmatrix} \). The third component indicates that certain vectors, like \( \mathbf{b} = \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} \), won't match, proving \( \mathbf{b} \) is outside the column space. Such checks solidify understanding around the relationships between matrices, vectors, and their computations.