Find Basis and Dimension of Ker(T) and R(T) for Linear Transformation

Math Problem Statement

Transformasi linear T : R^2 → R^3 memiliki fungsi transformasi T(ai) = bi dengan a1 = (2, 1), a2 = (3, 2), b1 = (-1, 2, 2), b2 = (-2, 1, 2). Tentukan basis dan dimensi dari ker(T) dan R(T)!

Solution

The problem involves a linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^3$ defined by the mapping of basis vectors $\mathbf{a}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$ and $\mathbf{a}_2 = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$ in $\mathbb{R}^2$ to $\mathbf{b}_1 = \begin{bmatrix} -1 \\ 2 \\ 2 \end{bmatrix}$ and $\mathbf{b}_2 = \begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix}$ in $\mathbb{R}^3$ .

The goal is to:

Find a basis and the dimension of the kernel of $T$ , $\ker(T)$ .
Find a basis and the dimension of the image of $T$ , $\operatorname{R}(T)$ .

Solution:

Step 1: Matrix Representation of $T$

The transformation $T$ can be represented by a matrix $A \in \mathbb{R}^{3 \times 2}$ such that: $T(\mathbf{a}_i) = A \mathbf{a}_i.$ Let the columns of $A$ be $\mathbf{b}_1$ and $\mathbf{b}_2$ : $A = \begin{bmatrix} -1 & -2 \\ 2 & 1 \\ 2 & 2 \end{bmatrix}.$

Step 2: Dimension of $\operatorname{R}(T)$ and Basis

The rank of $A$ determines the dimension of $\operatorname{R}(T)$ . Compute the rank of $A$ : $A = \begin{bmatrix} -1 & -2 \\ 2 & 1 \\ 2 & 2 \end{bmatrix}.$ Perform row reduction:

\xrightarrow{\text{Row 1} \leftrightarrow \text{Row 3}} \begin{bmatrix} 2 & 2 \\ 2 & 1 \\ -1 & -2 \end{bmatrix}.$$ Now, perform $$R_2 \to R_2 - R_1$$ and $$R_3 \to R_3 + \frac{1}{2}R_1$$: $$\begin{bmatrix} 2 & 2 \\ 0 & -1 \\ 0 & -1 \end{bmatrix}.$$ Finally, perform $$R_3 \to R_3 - R_2$$: $$\begin{bmatrix} 2 & 2 \\ 0 & -1 \\ 0 & 0 \end{bmatrix}.$$ The rank of $$A$$ is 2 (there are 2 pivot columns). - **Basis of $$\operatorname{R}(T)$$:** The columns of $$A$$, i.e., $$\mathbf{b}_1$$ and $$\mathbf{b}_2$$, form a basis. - **Dimension of $$\operatorname{R}(T)$$:** $$\dim(\operatorname{R}(T)) = 2$$. --- #### Step 3: Dimension of $$\ker(T)$$ and Basis By the rank-nullity theorem: $$\dim(\ker(T)) + \dim(\operatorname{R}(T)) = \dim(\mathbb{R}^2).$$ Since $$\dim(\operatorname{R}(T)) = 2$$ and $$\dim(\mathbb{R}^2) = 2$$, we find: $$\dim(\ker(T)) = 0.$$ Thus, $$\ker(T)$$ contains only the zero vector. - **Basis of $$\ker(T)$$:** No non-zero basis vectors (trivial kernel). - **Dimension of $$\ker(T)$$:** $$\dim(\ker(T)) = 0$$. --- ### Final Answer: 1. **$$\ker(T)$$:** - Basis: None (trivial kernel). - Dimension: $$0$$. 2. **$$\operatorname{R}(T)$$:** - Basis: $$\{\mathbf{b}_1, \mathbf{b}_2\} = \left\{\begin{bmatrix} -1 \\ 2 \\ 2 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix} \right\}$$. - Dimension: $$2$$. Would you like a detailed explanation of any step? Let me know! --- ### Related Questions: 1. What is the geometric interpretation of $$\ker(T)$$ and $$\operatorname{R}(T)$$? 2. How does the rank-nullity theorem relate to the dimensions of $$\ker(T)$$ and $$\operatorname{R}(T)$$? 3. How do we construct the matrix $$A$$ for a linear transformation in general? 4. What happens to $$\ker(T)$$ if $$T$$ is not injective? 5. How can we verify that $$\{\mathbf{b}_1, \mathbf{b}_2\}$$ forms a basis for $$\operatorname{R}(T)$$? ### Tip: Always verify linear independence of the basis vectors when determining $$\operatorname{R}(T)$$.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Linear Transformations
Vector Spaces
Kernel and Image of a Transformation

Formulas

Matrix representation of a linear transformation
Rank-nullity theorem

Theorems

Rank-Nullity Theorem

Suitable Grade Level

Undergraduate - First Year Linear Algebra

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