Math Problem Statement

Seien V = ⟨(1 0) (0 1)⟩ und W = ⟨(1 0), (0 1)⟩ Unterräume von M_22(ℝ). Sei f : V → W definiert durch f((x y) (0 z)) = ((x + y) z (z x + y)) für alle (x y) (0 z) ∈ M_22(ℝ). 1. Beweisen Sie, dass f linear ist. 2. Bestimmen Sie eine Basis von V + W und von V ∩ W. 3. Bestimmen Sie eine Basis von Kern(f) und von Bild(f).

Solution

Let's analyze and address each part of this problem step-by-step.

Given:

  • Subspaces VV and WW of M22(R)M_{22}(\mathbb{R}): V=span{(1000),(0100),(0001)}V = \text{span} \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\} W=span{(1001),(0110)}W = \text{span} \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right\}
  • A linear map f:VWf: V \to W defined by f((xy0z))=(x+yzzx+y)f\left( \begin{pmatrix} x & y \\ 0 & z \end{pmatrix} \right) = \begin{pmatrix} x + y & z \\ z & x + y \end{pmatrix} for any (xy0z)M22(R)\begin{pmatrix} x & y \\ 0 & z \end{pmatrix} \in M_{22}(\mathbb{R}).

Problem Breakdown

  1. Prove that ff is linear.

    To show that ff is linear, we need to check two properties:

    • Additivity: f(A+B)=f(A)+f(B)f(A + B) = f(A) + f(B) for all A,BVA, B \in V.
    • Scalar Multiplication: f(αA)=αf(A)f(\alpha A) = \alpha f(A) for any scalar α\alpha and AVA \in V.
  2. Find a basis for V+WV + W and VWV \cap W.

    • To find V+WV + W, we look for the smallest set that spans all elements of both VV and WW.
    • To find VWV \cap W, we identify elements common to both VV and WW.
  3. Find a basis for the kernel ker(f)\ker(f) and the image Im(f)\text{Im}(f).

    • ker(f)\ker(f) is the set of elements in VV that ff maps to the zero matrix in WW.
    • Im(f)\text{Im}(f) is the span of all matrices of the form f(A)f(A) for AVA \in V.

Let's proceed with each part in more detail.


Solution Outline

  1. Proof of Linearity of ff: We need to show that ff preserves addition and scalar multiplication. That is:

    • For A=(xy0z)A = \begin{pmatrix} x & y \\ 0 & z \end{pmatrix} and B=(xy0z)B = \begin{pmatrix} x' & y' \\ 0 & z' \end{pmatrix}: f(A+B)=f((x+xy+y0z+z))=((x+x)+(y+y)z+zz+z(x+x)+(y+y))f(A + B) = f\left( \begin{pmatrix} x + x' & y + y' \\ 0 & z + z' \end{pmatrix} \right) = \begin{pmatrix} (x + x') + (y + y') & z + z' \\ z + z' & (x + x') + (y + y') \end{pmatrix} and separately f(A)+f(B)=(x+yzzx+y)+(x+yzzx+y)=((x+x)+(y+y)z+zz+z(x+x)+(y+y)).f(A) + f(B) = \begin{pmatrix} x + y & z \\ z & x + y \end{pmatrix} + \begin{pmatrix} x' + y' & z' \\ z' & x' + y' \end{pmatrix} = \begin{pmatrix} (x + x') + (y + y') & z + z' \\ z + z' & (x + x') + (y + y') \end{pmatrix}. Therefore, f(A+B)=f(A)+f(B)f(A + B) = f(A) + f(B).

    • For scalar α\alpha: f(αA)=f((αxαy0αz))=(α(x+y)αzαzα(x+y))=αf(A).f(\alpha A) = f\left( \begin{pmatrix} \alpha x & \alpha y \\ 0 & \alpha z \end{pmatrix} \right) = \begin{pmatrix} \alpha (x + y) & \alpha z \\ \alpha z & \alpha (x + y) \end{pmatrix} = \alpha f(A).

    Thus, ff is linear.

  2. Finding a Basis for V+WV + W and VWV \cap W:

    • Compute spanning sets for V+WV + W and common elements in VWV \cap W.
  3. Finding Bases for ker(f)\ker(f) and Im(f)\text{Im}(f):

    • Solve f(A)=0f(A) = 0 for elements in ker(f)\ker(f).
    • Determine the span of f(A)f(A) for AVA \in V to find Im(f)\text{Im}(f).

Would you like a detailed, step-by-step solution for each of these parts?

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Spaces
Linear Maps
Subspaces
Basis
Kernel and Image of a Map

Formulas

Linear map properties: f(A + B) = f(A) + f(B), f(αA) = αf(A)
Definition of Kernel: ker(f) = {v ∈ V | f(v) = 0}
Definition of Image: im(f) = {f(v) | v ∈ V}

Theorems

Subspace Theorem for sum and intersection of vector spaces
Basis theorem for vector spaces
Rank-Nullity Theorem

Suitable Grade Level

Undergraduate Linear Algebra