Problem 31 - Differential Equations and Linear Algebra

Question

Let \(V\) denote the vector space of all \(3 \times 3\) skewsymmetric matrices over \(\mathbb{R}\). For what positive integer \(n\) is \(V\) isomorphic to \(\mathbb{R}^{n} .\) Construct an isomorphism.

Step-by-Step Solution

Answer

The dimension of the vector space \(V\) of all \(3 \times 3\) skew-symmetric matrices over \(\mathbb{R}\) is 3, so \(n = 3\). A basis for \(V\) is given by the set of matrices \(\begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\), \(\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}\), and \(\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}\). The isomorphism between \(V\) and \(\mathbb{R}^3\) can be constructed as a linear transformation \(T: V \rightarrow \mathbb{R}^3\) given by \(T \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix} = (a, b, c)\).

Step 1: Structure of a skew-symmetric matrix

A skew-symmetric (or antisymmetric) matrix is a square matrix A with the property that its transpose is equal to its negative, i.e., \(A^T = -A\). In our case, a \(3 \times 3\) skew-symmetric matrix A looks like this: \[A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}\] with \(a, b, c \in \mathbb{R}\).

Step 2: Determine the dimension of vector space V

For a \(3 \times 3\) skew-symmetric matrix, there are 3 independent components, namely \(a, b, c\). Therefore, the dimension of the vector space \(V\) is 3, which means that \(V\) can be isomorphic to \(\mathbb{R}^3\). So, \(n = 3\) in our case.

Step 3: Find a basis for V

A basis for the vector space \(V\) consists of linearly independent vectors that span the entire vector space. In this case, we are working with skew-symmetric matrices, so the basis matrices will also be skew-symmetric. The following set of matrices forms a basis for \(V\): \[B = \left\{ \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} , \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix} , \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix} \right\}\]

Step 4: Construct the isomorphism between V and \(\mathbb{R}^3\)

To construct an isomorphism, we need to find a linear transformation \(T: V \rightarrow \mathbb{R}^3\) that is both injective (one-to-one) and surjective (onto). Define the linear transformation \(T\) by \[T \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix} = (a, b, c).\] It is straightforward to verify that this linear transformation is an isomorphism, as it is a bijection between the elements of \(V\) and \(\mathbb{R}^3\). Therefore, taking n as 3, we have shown that the vector space of all \(3 \times 3\) skew-symmetric matrices over \(\mathbb{R}\) is isomorphic to \(\mathbb{R}^3\), and we have constructed the isomorphism \(T\).

Key Concepts

Skew-Symmetric Matrices

Skew-symmetric matrices are a fascinating concept in linear algebra. These matrices are square, meaning they have the same number of rows and columns. The defining property of a skew-symmetric matrix is that its transpose is the negative of itself. Mathematically, if you have a skew-symmetric matrix \(A\), it satisfies the equation \(A^T = -A\). This means that the diagonal elements must all be zero because they would otherwise contradict this condition.

For a \(3 \times 3\) skew-symmetric matrix, its general form is:

• The diagonal elements are all zero.
• The element in the \(i^{th}\) row and \(j^{th}\) column is the negative of the element in the \(j^{th}\) row and \(i^{th}\) column.

This gives the matrix a specific structure: \[ A = \begin{pmatrix} 0 & a & b \ -a & 0 & c \ -b & -c & 0 \end{pmatrix} \] where \(a, b,\) and \(c\) can be any real numbers.

Matrix Transformation

Matrix transformations are essential tools in linear algebra, particularly when working with vector spaces. A transformation involves converting one set of data or vectors into another, maintaining certain properties. In this context, we're looking at a transformation that maps skew-symmetric matrices to vectors in \(\mathbb{R}^3\).

This transformation, denoted as \(T\), takes the skew-symmetric matrix's variable components \(a, b,\) and \(c\) and maps them to a 3D vector. Specifically:\[T \begin{pmatrix}0 & a & b \-a & 0 & c \-b & -c & 0\end{pmatrix} = (a, b, c).\]

• This transformation is linear because it preserves vector addition and scalar multiplication.
• Each skew-symmetric matrix uniquely maps to a vector, ensuring the transformation is one-to-one and onto, or bijective.

By constructing this linear transformation, we demonstrate an isomorphism - a special kind of transformation that preserves the structure between vector spaces.

Linear Algebra Concepts

Linear algebra is a branch of mathematics that deals with vectors, matrices, and their transformations. Within this field, understanding vector space isomorphism is crucial. An isomorphism is a kind of mapping between two structures (in our case, vector spaces) that show they are essentially the same in terms of structure but not necessarily in the physical representation.

In our example, the vector space \(V\) of all \(3 \times 3\) skew-symmetric matrices is isomorphic to \(\mathbb{R}^3\). This is because both have a dimension of 3, meaning they can contain the same number of independent vectors:

• A basis for \(\mathbb{R}^3\) is the standard basis vectors \((1, 0, 0), (0, 1, 0), (0, 0, 1)\).
• A basis for \(V\) is given by the skew-symmetric matrices discussed earlier.

By using a linear transformation, we can map matrices in \(V\) to the standard vectors in \(\mathbb{R}^3\). This mapping is called isomorphic because it maintains the properties and operations of the vector spaces between \(V\) and \(\mathbb{R}^3\). Isomorphisms are powerful because they allow us to simplify problems by working in a more convenient vector space without losing any of the original's properties.