Math Problem Statement
Find a basis for the kernel of T.
Solution
To find a basis for the kernel of the linear transformation , we need to determine the null space (or kernel) of the matrix given by:
1 & 2 & 1 \\ 0 & 1 & 3 \\ -1 & 3 & 4 \end{pmatrix}$$ The kernel of $$T$$ consists of all vectors $$\mathbf{x} \in \mathbb{R}^3$$ such that $$A\mathbf{x} = \mathbf{0}$$. Let's solve the equation $$A\mathbf{x} = \mathbf{0}$$, where $$\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$$. This leads to the following system of linear equations: $$\begin{aligned} 1. & \quad x_1 + 2x_2 + x_3 = 0, \\ 2. & \quad x_2 + 3x_3 = 0, \\ 3. & \quad -x_1 + 3x_2 + 4x_3 = 0. \end{aligned}$$ To find the solution set, we'll solve this system using Gaussian elimination or row reduction. ### Step 1: Write the augmented matrix $$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & 1 & 3 & 0 \\ -1 & 3 & 4 & 0 \\ \end{array}\right]$$ ### Step 2: Row reduction 1. Add row 1 to row 3 to eliminate the first entry in the third row: $$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 5 & 5 & 0 \\ \end{array}\right]$$ 2. Subtract 5 times row 2 from row 3 to eliminate the second entry in the third row: $$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & -10 & 0 \\ \end{array}\right]$$ 3. Divide the third row by -10: $$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 1 & 0 \\ \end{array}\right]$$ ### Step 3: Back-substitution From the reduced matrix, we have the following equations: $$\begin{aligned} x_3 & = 0, \\ x_2 + 3x_3 & = 0 \quad \Rightarrow \quad x_2 = 0, \\ x_1 + 2x_2 + x_3 & = 0 \quad \Rightarrow \quad x_1 = 0. \end{aligned}$$ This implies that the only solution to the system is the trivial solution $$\mathbf{x} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$. ### Conclusion Since the kernel consists only of the zero vector, the kernel is trivial, and therefore, the basis for the kernel of $$T$$ is the empty set. This indicates that the linear transformation $$T$$ is injective (one-to-one), meaning there are no non-trivial solutions to $$A\mathbf{x} = \mathbf{0}$$. Would you like any additional details or explanations? Here are five follow-up questions to explore this topic further: 1. What does it mean for a linear transformation to be injective? 2. How would the analysis change if there were non-trivial solutions in the kernel? 3. What is the rank-nullity theorem and how does it apply here? 4. Can you find the image (or range) of the transformation $$T$$? 5. How would you find a basis for the image of $$T$$? **Tip:** Understanding the relationship between the rank of a matrix and the dimension of its kernel is key in linear algebra.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Operations
Null Space
Linear Transformations
Formulas
Augmented matrix
Gaussian elimination
Theorems
Rank-nullity theorem
Suitable Grade Level
Advanced Undergraduate