Solution: We begin by decomposing the general vector in the set as a linear combination of two fixed vectors, separating the contributions from \(a\) and \(b\):

\[\begin{pmatrix} 8a + 22b \\ -3b \\ a + 4b \end{pmatrix} = a\begin{pmatrix} 8 \\ 0 \\ 1 \end{pmatrix} + b\begin{pmatrix} 22 \\ -3 \\ 4 \end{pmatrix}.\]

Therefore, the set equals

\[\left\{ a\begin{pmatrix} 8 \\ 0 \\ 1 \end{pmatrix} + b\begin{pmatrix} 22 \\ -3 \\ 4 \end{pmatrix} \;\middle|\; a, b \in \mathbb{R} \right\} = \operatorname{Span}\left\{\begin{pmatrix}8\\0\\1\end{pmatrix},\begin{pmatrix}22\\-3\\4\end{pmatrix}\right\}.\]

Since \(\operatorname{Col} A\) is defined as all linear combinations of the columns of \(A\), we use the two spanning vectors as columns. One valid choice is

\[A = \begin{pmatrix} 8 & 22 \\ 0 & -3 \\ 1 & 4 \end{pmatrix}.\]

Note that the column order does not affect the span, so \(A = \begin{pmatrix} 22 & 8 \\ -3 & 0 \\ 4 & 1 \end{pmatrix}\) is equally valid.

Solution:

(a) Because \(A\) has 4 columns, vectors \(\mathbf{x}\) in \(A\mathbf{x} = \mathbf{0}\) belong to \(\mathbb{R}^4\), so \(k = 4\).

(b) Because \(A\) has 2 rows, the columns of \(A\) are vectors in \(\mathbb{R}^2\), so \(k = 2\).

Solution: (a) To test whether \(\mathbf{v} \in \operatorname{Col} A\), we check whether \(A\mathbf{x} = \mathbf{v}\) is consistent by row-reducing the augmented matrix \([A \mid \mathbf{v}]\):

\begin{align*}\begin{pmatrix} 2 & 4 & 6 & 2 \\ 1 & 3 & 2 & 1 \end{pmatrix} &\xrightarrow{r_1 \to r_1-r_2} \begin{pmatrix} 1 & 1 & 4 & 1 \\ 1 & 3 & 2 & 1 \end{pmatrix}\\ &\xrightarrow{r_2 \to r_2-r_1} \begin{pmatrix} 1 & 1 & 4 & 1 \\ 0 & 2 & -2 & 0 \end{pmatrix}\\ &\xrightarrow{r_2\to\frac{r_2}{2}} \begin{pmatrix} 1 & 1 & 4 & 1 \\ 0 & 1 & -1 & 0 \end{pmatrix}\\ &\xrightarrow{r_1\to r_1-r_2} \begin{pmatrix} 1 & 0 & 5 & 1 \\ 0 & 1 & -1 & 0 \end{pmatrix}.\end{align*}

The system is consistent, so \(\mathbf{v} \in \operatorname{Col} A\). However, \(\mathbf{v}\) cannot belong to \(\operatorname{Nul} A\). The null space is a subspace of \(\mathbb{R}^3\) (since \(A\) has 3 columns), and \(\mathbf{v}\) has only 2 entries — a dimension mismatch makes membership impossible.

(b) To test whether \(\mathbf{u} \in \operatorname{Nul} A\), we compute \(A\mathbf{u}\):

\[A\mathbf{u} = \begin{pmatrix} 2\cdot 1 + 4\cdot (-2) + 6\cdot 1 \\ 1\cdot 1 + 3\cdot (-2) + 2\cdot 1 \end{pmatrix} = \begin{pmatrix} 2 - 8 + 6 \\ 1 - 6 + 2 \end{pmatrix} = \begin{pmatrix} 0 \\ -3 \end{pmatrix} \neq \mathbf{0}.\]

Since \(A\mathbf{u} \neq \mathbf{0}\), we have \(\mathbf{u} \notin \operatorname{Nul} A\). Furthermore, \(\mathbf{u}\) cannot belong to \(\operatorname{Col} A\). The column space is a subspace of \(\mathbb{R}^2\) (since \(A\) has 2 rows), and \(\mathbf{u}\) has 3 entries — again a dimension mismatch.

Let's first verify that \(T\) is a linear transformation. For additivity, the sum rule of differentiation gives \[T(f + g) = (f + g)' = f' + g' = T(f) + T(g)\] for all \(f, g \in V\). For homogeneity, the constant multiple rule gives \[T(cf) = (cf)' = cf' = cT(f)\] for all \(f \in V\) and scalars \(c\). Both properties hold, so \(T\) is indeed a linear transformation.

The kernel of \(T\) consists of all differentiable functions \(f\) on \([a, b]\) satisfying \(f' = 0\). A function with zero derivative everywhere on an interval must be constant. Therefore,

\[\ker T = \{f \in V \mid f \text{ is constant on } [a, b]\}.\]

By the Fundamental Theorem of Calculus, every continuous function on \([a, b]\) is the derivative of some differentiable function (namely, any of its antiderivatives). Therefore \(\operatorname{im} T = W\); the differentiation operator is surjective onto the continuous functions.