Math Problem Statement
Dada la transformación lineal T: P_≤3 → R^2; T(x_1b_1 + x_2b_2 + x_3b_3) = (2x_1 − 4x_2 + 5x_3, −x_2 + 3x_3), calcule ρ(T), ν(T), Im T ∧ Nu(T).
Solution
Let’s analyze and solve the given linear transformation problem step by step.
Problem Setup:
We are given a linear transformation , where:
\begin{pmatrix} 2x_1 - 4x_2 + 5x_3 \\ -x_2 + 3x_3 \end{pmatrix}.$$ Here, $$P_{\leq 3}$$ represents the space of polynomials of degree up to 3, and the basis $$\{b_1, b_2, b_3\}$$ corresponds to coefficients $$x_1, x_2, x_3$$. The task is to calculate: 1. $$\rho(T)$$: Rank of $$T$$, 2. $$\nu(T)$$: Nullity of $$T$$, 3. $$\text{Im} \, T$$: The image (range) of $$T$$, 4. $$\text{Nu} \, T$$: The null space (kernel) of $$T$$. --- ### 1. Analyze the Linear Transformation The transformation maps $$P_{\leq 3}$$ (dimension 3) into $$\mathbb{R}^2$$. To compute $$\rho(T)$$ and $$\nu(T)$$, we use the **Rank-Nullity Theorem**: $$\dim(P_{\leq 3}) = \rho(T) + \nu(T).$$ The matrix associated with $$T$$ is: $$A = \begin{pmatrix} 2 & -4 & 5 \\ 0 & -1 & 3 \end{pmatrix}.$$ --- ### 2. Compute the Rank $$\rho(T)$$ The rank $$\rho(T)$$ is the number of linearly independent rows or columns of $$A$$. We perform row reduction to find the rank. #### Row Reduction: \[ \begin{pmatrix} 2 & -4 & 5 \\ 0 & -1 & 3 \end{pmatrix} \xrightarrow[]{R_2 \to R_2/(1)} \begin{pmatrix} 2 & -4 & 5 \\ 0 & -1 & 3 \end as sp # Define the matrix A A = sp.Matrix([[2, -4, 5], [0, -1, 3]]) # Perform row reduction rref_A, pivots = A.rref() # Calculate rank and nullity rank_T = A.rank() nullity_T = 3 - rank_T # Dimension of domain - rank rref_A,The row-reduced form of the matrix $$A$$ is: $$\begin{pmatrix} 1 & 0 & -\frac{7}{2} \\ 0 & 1 & -3 \end{pmatrix}.$$ From this, we determine: - The rank of $$T$$, $$\rho(T)$$, is $$2$$ (number of pivot columns). - The nullity of $$T$$, $$\nu(T)$$, is $$1$$ (since $$\dim(P_{\leq 3}) - \rho(T) = 3 - 2 = 1$$). --- ### 3. Image and Null Space #### Image of $$T$$ ($$\text{Im} \, T$$): The image of $$T$$ is the span of the rows of the original matrix: $$\text{Im} \, T = \text{span} \left\{ \begin{pmatrix} 2 \\ 0 \end{pmatrix}, \begin{pmatrix} -4 \\ -1 \end{pmatrix}, \begin{pmatrix} 5 \\ 3 \end{pmatrix} \right\}.$$ Row reduction confirms this spans $$\mathbb{R}^2$$. #### Null Space of $$T$$ ($$\text{Nu} \, T$$): The null space is determined by solving: $$A \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. for the null space null_space_TThe null space ($$\text{Nu} \, T$$) is spanned by: \[ \text{Nu} \, T = \text{span} \left\{ \begin{pmatrix} \frac{7}{2} \\ 3 \\ 1 \end{pmatrix} \right\}.$$ --- ### Final Results: 1. **Rank ($$\rho(T)$$)**: $$2$$. 2. **Nullity ($$\nu(T)$$)**: $$1$$. 3. **Image ($$\text{Im} \, T$$)**: The entire $$\mathbb{R}^2$$ (spanned by the rows of $$A$$). 4. **Null Space ($$\text{Nu} \, T$$)**: Spanned by $$\begin{pmatrix} \frac{7}{2} \\ 3 \\ 1 \end{pmatrix}$$. Would you like a detailed explanation of any step? Here are related questions to consider: 1. How is the rank-nullity theorem applied to transformations? 2. Can you explain more about row reduction and its significance? 3. What does it mean for the image to span $$\mathbb{R}^2$$? 4. How can the null space be visualized in higher-dimensional spaces? 5. How does this transformation map polynomials into $$\mathbb{R}^2$$? **Tip:** Always check the consistency of the domain and codomain dimensions when analyzing linearAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Linear Transformations
Rank-Nullity Theorem
Kernel
Image
Formulas
Rank-Nullity Theorem: dim(V) = rank(T) + nullity(T)
Row Reduction for Matrix Rank
Theorems
Rank-Nullity Theorem
Suitable Grade Level
Undergraduate Mathematics
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