To construct an example of a linear operator $A$ such that $\dim(\ker(A)) = 3$ and $\dim(\text{range}(A)) = 2$, we use the Rank-Nullity Theorem:

$\dim(\ker(A)) + \dim(\text{range}(A)) = \dim(V),$ where $V$ is the vector space on which $A$ acts.

Step 1: Choose the dimension of $V$

Let $V$ be a 5-dimensional vector space ($\dim(V) = 5$).

Step 2: Set the kernel and range dimensions

From the Rank-Nullity Theorem: $\dim(\ker(A)) + \dim(\text{range}(A)) = \dim(V),$ $3 + 2 = 5.$

Thus, we need to construct $A$ such that $\dim(\ker(A)) = 3$ and $\dim(\text{range}(A)) = 2$.

Step 3: Define $A$

Let $V = \mathbb{R}^5$. Represent $A$ as a $5 \times 5$ matrix. For the desired properties, $A$ should have rank 2 (dimension of the range is the rank of $A$). We construct $A$ such that it maps $\mathbb{R}^5$ to a 2-dimensional subspace.

Here is one such matrix:

\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}.$$ ### Verification 1. **Kernel (null space):** Solve $$A\mathbf{x} = 0$$. $$A\mathbf{x} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_2 \\ 0 \\ 0 \\ 0 \end{bmatrix}.$$ For $$A\mathbf{x} = 0$$, we must have $$x_1 = 0$$ and $$x_2 = 0$$. Thus, $$\ker(A)$$ consists of all vectors of the form: $$\ker(A) = \text{span}\{\mathbf{e}_3, \mathbf{e}_4, \mathbf{e}_5\},$$ where $$\mathbf{e}_3, \mathbf{e}_4, \mathbf{e}_5$$ are the standard basis vectors in $$\mathbb{R}^5$$. Therefore, $$\dim(\ker(A)) = 3$$. 2. **Range:** The range of $$A$$ consists of all linear combinations of the first two columns of $$A$$. These are: $$\text{range}(A) = \text{span}\{\mathbf{e}_1, \mathbf{e}_2\}.$$ Hence, $$\dim(\text{range}(A)) = 2$$. ### Conclusion The matrix $$A = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ satisfies $$\dim(\ker(A)) = 3$$ and $$\dim(\text{range}(A)) = 2$$. --- Would you like further clarification or examples? Here are related questions you might find interesting: 1. What are the properties of the kernel and range of a linear operator? 2. How does the Rank-Nullity Theorem apply to infinite-dimensional vector spaces? 3. Can you provide a geometric interpretation of the kernel and range? 4. How can such a matrix $$A$$ be transformed while maintaining its kernel and range dimensions? 5. What is the significance of the null space in practical applications? **Tip:** Always verify the dimensions of the kernel and range when constructing examples of linear operators to ensure consistency with the Rank-Nullity Theorem.