Math Problem Statement

q.7(a)

Let F: R^4 → R^3 be the linear mapping defined by F(x, y, z, t) = (-x + y + 2t, x + 2z - t, x + y + 3z - 3t). Find a basis and the dimension of the image of F. Find a basis and the dimension of the kernel of the map F.

Solution

For question 7(a), let's solve the problem step by step:

Given: The linear mapping F:R4→R3F: \mathbb{R}^4 \to \mathbb{R}^3 is defined as:

F(x,y,z,t)=(−x+y+2t,x+2z−t,x+y+3z−3t).F(x, y, z, t) = (-x + y + 2t, x + 2z - t, x + y + 3z - 3t).

Tasks:

  1. Find a basis and the dimension of the image of FF.
  2. Find a basis and the dimension of the kernel of FF.

Step 1: Write F(x,y,z,t)F(x, y, z, t) in matrix form

We express FF as a matrix transformation: F(x,y,z,t)=A[xyzt],F(x, y, z, t) = A \begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix}, where AA is a 3×43 \times 4 matrix. By reading the coefficients of x,y,z,tx, y, z, t in the components of FF:

-1 & 1 & 0 & 2 \\ 1 & 0 & 2 & -1 \\ 1 & 1 & 3 & -3 \end{bmatrix}.$$ --- ### Step 2: Row reduce $$A$$ to find the rank (image dimension) Perform row reduction on $$A$$: 1. Start with: $$A = \begin{bmatrix} -1 & 1 & 0 & 2 \\ 1 & 0 & 2 & -1 \\ 1 & 1 & 3 & -3 \end{bmatrix}.$$ 2. Add Row 1 to Row 2 and Row 3: $$\begin{bmatrix} -1 & 1 & 0 & 2 \\ 0 & 1 & 2 & 1 \\ 0 & 2 & 3 & -1 \end{bmatrix}.$$ 3. Replace Row 3 with Row 3 - 2 × Row 2: $$\begin{bmatrix} -1 & 1 & 0 & 2 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & -1 & -3 \end{bmatrix}.$$ 4. Multiply Row 3 by -1: $$\begin{bmatrix} -1 & 1 & 0 & 2 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 3 \end{bmatrix}.$$ 5. Eliminate the 2 in Row 2, Column 3 by subtracting 2 × Row 3 from Row 2: $$\begin{bmatrix} -1 & 1 & 0 & 2 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & 3 \end{bmatrix}.$$ 6. Eliminate the 1 in Row 1, Column 2 by subtracting Row 2 from Row 1: $$\begin{bmatrix} -1 & 0 & 0 & 7 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & 3 \end{bmatrix}.$$ Finally, multiply Row 1 by -1: $$\begin{bmatrix} 1 & 0 & 0 & -7 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & 3 \end{bmatrix}.$$ --- ### Step 3: Interpret the row-reduced form - The rank of $$A$$ (number of pivots) is 3. - Dimension of the image of $$F$$: $$\text{dim(Im } F) = 3$$. - A basis for the image of $$F$$ consists of the rows of the row-reduced matrix: $$\text{Basis for Im } F: \{ (1, 0, 0, -7), (0, 1, 0, -5), (0, 0, 1, 3) \}.$$ --- ### Step 4: Find the kernel of $$F$$ The kernel is the solution space to $$A \mathbf{x} = 0$$. Since the rank of $$A$$ is 3, the nullity of $$A$$ is: $$\text{nullity of } A = 4 - 3 = 1.$$ Thus, the kernel is 1-dimensional. From the row-reduced form, we write: $$\begin{bmatrix} 1 & 0 & 0 & -7 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix} = 0.$$ This gives: $$x - 7t = 0 \quad \Rightarrow \quad x = 7t,$$ $$y - 5t = 0 \quad \Rightarrow \quad y = 5t,$$ $$z + 3t = 0 \quad \Rightarrow \quad z = -3t.$$ Let $$t = t$$. The kernel is spanned by: $$\text{Basis for Ker } F: \{ (7, 5, -3, 1) \}.$$ --- ### Final Answer: 1. **Image:** - Dimension: 3. - Basis: $$\{ (1, 0, 0, -7), (0, 1, 0, -5), (0, 0, 1, 3) \}$$. 2. **Kernel:** - Dimension: 1. - Basis: $$\{ (7, 5, -3, 1) \}$$. Would you like a step-by-step diagram or further clarification? --- **Expanding Questions:** 1. How is the dimension of the kernel and image related to the rank-nullity theorem? 2. What is the geometric interpretation of the image and kernel of a linear map? 3. Can you construct a similar example with a 4×4 matrix and find its kernel and image? 4. How do eigenvalues relate to the kernel of a matrix? 5. What role does row reduction play in determining the properties of a matrix? **Tip:** Always verify row reduction steps carefully to avoid calculation errors!

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Transformations
Kernel and Image
Dimension Analysis

Formulas

Rank-Nullity Theorem: dim(Kernel) + dim(Image) = Total Dimension
Row reduction to echelon form

Theorems

Rank-Nullity Theorem
Properties of Linear Transformations

Suitable Grade Level

Undergraduate

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