Solution: Given that $A \in M_n(\mathbb{R} )$ such that $A^2 = A$.
-
Let us denote
\[
\mathcal{C} = \left\{ \mathbf{x} \in \mathbb{R} ^n: \mathbf{x} = A \mathbf{x} \right\}.
\]
It is clear that $\mathcal{S} \subseteq \mathbf{C} (A)$. For the other side, let $\mathbf{y} \in \mathbf{C} (A)$. Then, there exists $\mathbf{x} \in \mathbb{R} ^n$ such that $A \mathbf{x} = \mathbf{y} $. So we have
\begin{align*}
A \mathbf{x} = \mathbf{y} & \implies A A \mathbf{x} = A \mathbf{y} \\
& \implies A^2 \mathbf{x} = A \mathbf{y} \\
& \implies A \mathbf{x} = A \mathbf{y} \\
& \implies \mathbf{y} = A \mathbf{y} \\
& \implies \mathbf{y} \in \mathcal{S} .
\end{align*}
Thus, we proved that $\mathbf{C} (A) \subseteq \mathcal{C} $ and hence $\mathbf{C}(A) = \mathcal{C} $.
-
Let us denote
\[
\mathcal{N} = \left\{ \mathbf{x} \in \mathbb{R} ^n: \mathbf{x} = \mathbf{u} - A \mathbf{u} \text{ for some } \mathbf{u} \in \mathbb{R} ^n \right\}.
\]
Note that for any $\mathbf{x} \in \mathcal{N} $,
\[
A \mathbf{x} = A(\mathbf{u} - A \mathbf{u} ) = A \mathbf{u} - A^2 \mathbf{u} = A \mathbf{u} - A \mathbf{u} = \mathbf{0} .
\]
Thus, $\mathbf{x} \in \mathbf{N} (A)$ and hence, $\mathcal{N} \subseteq \mathbf{N} (A) $. On the other hand, let $\mathbf{x} \in \mathbf{N} (A)$. This means $A \mathbf{x} = \mathbf{0} $. Take
$\mathbf{u} = \mathbf{x}$. Then,
\[
\mathbf{x} = \mathbf{u} - A \mathbf{u} = \mathbf{x} - A \mathbf{x} = \mathbf{x} - \mathbf{0} = \mathbf{x}.
\]
Thus, $\mathbf{x} \in \mathcal{N}$ and hence, $\mathbf{N} (A) \subseteq \mathcal{N}$. Therefore, $\mathbf{N} (A) = \mathcal{N}$.
-
Let $\mathbf{x} \in \mathbf{C} (A) \cap \mathbf{N} (A)$. Since $x \in \mathbf{N} (A)$, $A \mathbf{x} = \mathbf{0} $. This implies, from the part (a),
\[
\mathbf{x} = A \mathbf{x} = \mathbf{0} .
\]
Thus, $\mathbf{C} (A) \cap \mathbf{N} (A) = \{ \mathbf{0} \} $.
-
By the definition of $\mathbf{C} (A)$ and $\mathbf{N} (A)$, we have $\mathbf{C} (A) + \mathbf{N} (A) \subseteq \mathbb{R} ^n$. Let $\mathbf{y} \in \mathbb{R}^n$. We need to show that $\mathbf{y} \in \mathbf{C}(A) + \mathbf{N}(A)$. Define $\mathbf{u} = \mathbf{y} - A\mathbf{y}$. Note that
\[
A\mathbf{u} = A(\mathbf{y} - A\mathbf{y}) = A\mathbf{y} - A^2\mathbf{y} = A\mathbf{y} - A\mathbf{y} = \mathbf{0}.
\]
Thus, $\mathbf{u} \in \mathbf{N}(A)$. Now, consider $\mathbf{v} = A\mathbf{y}$. We have
\[
A\mathbf{v} = A(A\mathbf{y}) = A^2\mathbf{y} = A\mathbf{y} = \mathbf{v}.
\]
Thus, $\mathbf{v} \in \mathbf{C}(A)$. Therefore,
\[
\mathbf{y} = \mathbf{v} + \mathbf{u} \in \mathbf{C}(A) + \mathbf{N}(A).
\]
Hence, $\mathbf{C}(A) + \mathbf{N}(A) = \mathbb{R}^n$.