Math Problem Statement

For the linear transformation L: M_3,3 → M_3,3 defined by L(A) = A − A^T, find a basis for ker(L) and a basis for range(L), and verify that dim(ker(L)) + dim(range(L)) = dim(V).

Solution

We are given a linear transformation L:M3,3→M3,3L : \mathcal{M}_{3,3} \to \mathcal{M}_{3,3} defined by

L(A)=A−AT,A∈M3,3L(A) = A - A^T, \quad A \in \mathcal{M}_{3,3}

where M3,3\mathcal{M}_{3,3} represents the space of all 3×33 \times 3 matrices. The problem asks to:

  1. Find a basis for ker⁡(L)\ker(L), the kernel of LL, which corresponds to the matrices that are mapped to the zero matrix.
  2. Find a basis for range(L)\text{range}(L), the range of LL, which corresponds to the set of matrices that can be expressed as A−ATA - A^T.
  3. Verify that dim⁡(ker⁡(L))+dim⁡(range(L))=dim⁡(M3,3)\dim(\ker(L)) + \dim(\text{range}(L)) = \dim(\mathcal{M}_{3,3}).

Step 1: Finding ker⁡(L)\ker(L)

The kernel of LL consists of matrices A∈M3,3A \in \mathcal{M}_{3,3} such that:

L(A)=A−AT=0.L(A) = A - A^T = 0.

This implies that A=ATA = A^T, meaning the kernel is the set of all symmetric 3×33 \times 3 matrices. A general symmetric matrix has the form:

a & d & e \\ d & b & f \\ e & f & c \end{pmatrix},$$ with 6 independent entries (since $$a, b, c, d, e, f$$ can be arbitrary). Therefore, the dimension of the kernel is 6. A basis for $$\ker(L)$$ can be given by the following 6 symmetric matrices: $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \quad \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \quad \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix},$$ $$\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \quad \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, \quad \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$ ### Step 2: Finding $$\text{range}(L)$$ The range of $$L$$ consists of matrices of the form $$A - A^T$$, which are **skew-symmetric** matrices, i.e., matrices where $$A^T = -A$$. A general skew-symmetric matrix has the form: $$A = \begin{pmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{pmatrix},$$ with 3 independent entries (since $$x, y, z$$ can be arbitrary). Therefore, the dimension of the range is 3. A basis for $$\text{range}(L)$$ can be given by the following 3 skew-symmetric matrices: \[ \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Transformations
Kernel
Range
Skew-Symmetric Matrices
Symmetric Matrices

Formulas

L(A) = A - A^T
Kernel of L: A = A^T
Range of L: A - A^T (Skew-Symmetric Matrices)
dim(ker(L)) + dim(range(L)) = dim(V)

Theorems

Rank-Nullity Theorem: dim(ker(L)) + dim(range(L)) = dim(V)

Suitable Grade Level

College/University Level

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