Math Problem Statement
Skip to main content Homework due Feb 26, 2025 00:00 EST Completed 5.7. Repeated values of a linear transformation 0/12 points (graded) Let T:R4→R3 be a linear transformation given by T(v)=Av where A is the following matrix: a=[[1,0,-1,-1],[1,1,-1,1],[2,1,-2,0]]
Find vectors w, w′, z, z′ which satisfy the following conditions:
w≠w′ but T(w)=T(w′) z≠z′ but T(z)=T(z′) T(w)≠T(z)
Enter the vector w in the form [w1,w2,w3,w4]:
[1,0,0,0] incorrect ⎡⎣⎢⎢⎢⎢1000⎤⎦⎥⎥⎥⎥ Enter the vector w′ in the form [w′1,w′2,w′3,w′4]:
[2,-2,1,0] incorrect ⎡⎣⎢⎢⎢⎢2−210⎤⎦⎥⎥⎥⎥ Enter the vector z in the form [z1,z2,z3,z4]:
[0,1,0,0] incorrect ⎡⎣⎢⎢⎢⎢0100⎤⎦⎥⎥⎥⎥ Enter the vector z′ in the form [z′1,z′2,z′3,z′4]:
[1,1,0,1] incorrect ⎡⎣⎢⎢⎢⎢1101⎤⎦⎥⎥⎥⎥
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Solution
We need to find vectors and in such that:
- but .
- but .
- .
Given the transformation matrix:
Step 1: Find the Null Space of
Vectors in the null space of are mapped to zero, meaning if , then must be in the null space.
To find the null space, solve . That is, solve:
1 & 0 & -1 & -1 \\ 1 & 1 & -1 & 1 \\ 2 & 1 & -2 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$ This gives a system of equations: 1. $$x - z - w = 0$$ → $$x = z + w$$ 2. $$x + y - z + w = 0$$ → $$x + y = z - w$$ 3. $$2x + y - 2z = 0$$ → $$2x + y = 2z$$ Substituting $$x = z + w$$ in the third equation: $$2(z + w) + y = 2z$$ $$2z + 2w + y = 2z$$ $$2w + y = 0 \Rightarrow y = -2w$$ Substituting $$x = z + w$$ and $$y = -2w$$ into equation 2: $$z + w - 2w = z - w$$ which is always true. Thus, the null space is spanned by the vectors: $$\begin{bmatrix} 1 \\ -2 \\ 1 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$ ### Step 2: Choose Two Distinct Vectors in the Null Space Pick two different linear combinations: 1. $$w = (1, -2, 1, 0)$$ 2. $$w' = (1, 0, 0, 1)$$ Since both belong to the null space, we have $$T(w) = T(w')$$, satisfying the first condition. ### Step 3: Choose Two More Distinct Vectors Using other combinations: 1. $$z = (2, -4, 2, 0)$$ 2. $$z' = (3, -6, 3, 0)$$ These also satisfy $$T(z) = T(z')$$, while $$T(w) \neq T(z)$$, ensuring condition 3. ### Final Answer: \[ w = [1, -2, 1, 0], \quad w' = [1, 0, 0, 1] \] \[ z = [2, -4, 2, 0], \quad z' = [3, -6, 3, 0] \] Would you like further details or verification? #### Related Questions: 1. How do you determine the null space of a matrix? 2. What does it mean for a transformation to be linear? 3. How do you check if two vectors map to the same image under a transformation? 4. Why do we need two sets of vectors for this problem? 5. What does the rank-nullity theorem tell us about this transformation? **Tip:** The null space of a matrix is key to finding vectors that are mapped to the same image.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Transformations
Null Space
Vector Spaces
Systems of Linear Equations
Formulas
T(v) = A * v
A * v = 0 (Null Space)
Theorems
Rank-Nullity Theorem
Suitable Grade Level
Grades 11-12