https://file.aiquickdraw.com/mathbot/file/17264098283183gvay73e.jpg

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

\[
L(A) = A - A^T, \quad A \in \mathcal{M}_{3,3}
\]

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

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

### Step 1: Finding \( \ker(L) \)
The kernel of \( L \) consists of matrices \( A \in \mathcal{M}_{3,3} \) such that:

\[
L(A) = A - A^T = 0.
\]

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

\[
A = \begin{pmatrix} 
a & d & e \\
d & b & f \\
e & f & c 
\end{pmatrix},
\]
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

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).

Learn how to find the kernel and range of the linear transformation L(A) = A - A^T, where A belongs to M_3,3. This problem explores matrix transformations, symmetric and skew-symmetric matrices, and the Rank-Nullity Theorem. A detailed step-by-step solution is provided.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation