Symmetric and Skew-Symmetric Matrices

Symmetric Matrix

A symmetric matrix is a square matrix that is equal to its transpose. For a matrix \( \mathbf{A} \) to be symmetric, it must satisfy the condition:

\[ \mathbf{A}^\top = \mathbf{A} \]

This means that for all elements \( a_{ij} \) of the matrix, the following condition holds:

\[ a_{ij} = a_{ji} \]

This property indicates that the elements of a symmetric matrix are symmetric about the main diagonal (from the top left to the bottom right).

Example

Consider the following 3x3 matrix:

\[ \mathbf{A} = \begin{bmatrix} 2 & -1 & 3 \\ -1 & 5 & 4 \\ 3 & 4 & 6 \end{bmatrix} \]

To check if this matrix is symmetric, we verify that \( a_{ij} = a_{ji} \) for all \( i \) and \( j \):

\( a_{12} = -1 \) and \( a_{21} = -1 \)
\( a_{13} = 3 \) and \( a_{31} = 3 \)
\( a_{23} = 4 \) and \( a_{32} = 4 \)

Since all the elements satisfy \( a_{ij} = a_{ji} \), matrix \( \mathbf{A} \) is symmetric.

Skew-Symmetric Matrix

A skew-symmetric matrix is a square matrix that is equal to the negative of its transpose. For a matrix \( \mathbf{A} \) to be skew-symmetric, it must satisfy the condition:

\[ \mathbf{A}^\top = -\mathbf{A} \]

This means that for all elements \( a_{ij} \) of the matrix, the following condition holds:

\[ a_{ij} = -a_{ji} \]

Additionally, for a skew-symmetric matrix, the diagonal elements must all be zero. This is because for any diagonal element \( a_{ii} \):

\[ a_{ii} = -a_{ii} \implies 2a_{ii} = 0 \implies a_{ii} = 0 \]

Thus, all diagonal elements \( a_{ii} \) of a skew-symmetric matrix are zero.

Example

Consider the following 3x3 matrix:

\[ \mathbf{A} = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 5 \\ 3 & -5 & 0 \end{bmatrix} \]

To check if this matrix is skew-symmetric, we verify that \( a_{ij} = -a_{ji} \) for all \( i \) and \( j \), and that the diagonal elements are zero:

\( a_{12} = 2 \) and \( a_{21} = -2 \)
\( a_{13} = -3 \) and \( a_{31} = 3 \)
\( a_{23} = 5 \) and \( a_{32} = -5 \)
Diagonal elements: \( a_{11} = 0 \), \( a_{22} = 0 \), \( a_{33} = 0 \)

Since all the conditions are satisfied, matrix \( \mathbf{A} \) is skew-symmetric.

Properties of Symmetric and Skew-Symmetric Matrices

Transpose of a Symmetric Matrix:
- If \( \mathbf{A} \) is symmetric, then \( \mathbf{A}^\top \) is also symmetric.
- Proof: Since \( \mathbf{A} = \mathbf{A}^\top \), transposing \( \mathbf{A} \) again gives \( (\mathbf{A}^\top)^\top = \mathbf{A} = \mathbf{A}^\top \).
Transpose of a Skew-Symmetric Matrix:
- If \( \mathbf{A} \) is skew-symmetric, then \( \mathbf{A}^\top \) is also skew-symmetric.
- Proof: For a skew-symmetric matrix, \( \mathbf{A}^\top = -\mathbf{A} \). Transposing again gives \( (\mathbf{A}^\top)^\top = (-\mathbf{A})^\top = -\mathbf{A} \), confirming \( \mathbf{A} \) is still skew-symmetric.
Scalar Multiplication of a Symmetric Matrix:
- If \( \mathbf{A} \) is symmetric, then \( k\mathbf{A} \) is also symmetric for any scalar \( k \).
- Proof: \( (k\mathbf{A})^\top = k\mathbf{A}^\top = k\mathbf{A} \), which means \( k\mathbf{A} \) is symmetric.
Scalar Multiplication of a Skew-Symmetric Matrix:
- If \( \mathbf{A} \) is skew-symmetric, then \( k\mathbf{A} \) is also skew-symmetric for any scalar \( k \).
- Proof: \( (k\mathbf{A})^\top = k\mathbf{A}^\top = k(-\mathbf{A}) = -k\mathbf{A} \), showing \( k\mathbf{A} \) is skew-symmetric.
Square of a Symmetric Matrix:
- If \( \mathbf{A} \) is symmetric, then \( \mathbf{A}^2 \) is also symmetric.
- Proof: \( (\mathbf{A}^2)^\top = (\mathbf{A} \mathbf{A})^\top = \mathbf{A}^\top \mathbf{A}^\top = \mathbf{A} \mathbf{A} = \mathbf{A}^2 \), proving \( \mathbf{A}^2 \) is symmetric.
Square of a Skew-Symmetric Matrix:
- If \( \mathbf{A} \) is skew-symmetric, then \( \mathbf{A}^2 \) is symmetric.
- Proof: \( (\mathbf{A}^2)^\top = (\mathbf{A} \mathbf{A})^\top = \mathbf{A}^\top \mathbf{A}^\top = (-\mathbf{A})(-\mathbf{A}) = \mathbf{A}^2 \), showing \( \mathbf{A}^2 \) is symmetric.
Decomposition of a matrix into a sum of symmetric and skew-symmetrix matrices:

For any square matrix \( \mathbf{A} \), which may not necessarily be symmetric or skew-symmetric, the matrix \( \mathbf{A} + \mathbf{A}^\top \) is always symmetric, and the matrix \( \mathbf{A} - \mathbf{A}^\top \) is always skew-symmetric. To see why this is true, consider the transpose of \( \mathbf{A} + \mathbf{A}^\top \):

\[ (\mathbf{A} + \mathbf{A}^\top)^\top = \mathbf{A}^\top + (\mathbf{A}^\top)^\top = \mathbf{A}^\top + \mathbf{A}. \]

Since \( (\mathbf{A} + \mathbf{A}^\top)^\top = \mathbf{A} + \mathbf{A}^\top \), the matrix \( \mathbf{A} + \mathbf{A}^\top \) is symmetric. Now, consider the transpose of \( \mathbf{A} - \mathbf{A}^\top \):

\[ (\mathbf{A} - \mathbf{A}^\top)^\top = \mathbf{A}^\top - (\mathbf{A}^\top)^\top = \mathbf{A}^\top - \mathbf{A}. \]

Since \( (\mathbf{A} - \mathbf{A}^\top)^\top = -(\mathbf{A} - \mathbf{A}^\top) \), the matrix \( \mathbf{A} - \mathbf{A}^\top \) is skew-symmetric.

This leads to the decomposition theorem, which states that any square matrix \( \mathbf{A} \) can be uniquely decomposed into the sum of a symmetric matrix and a skew-symmetric matrix:

\[ \mathbf{A} = \frac{1}{2}(\mathbf{A} + \mathbf{A}^\top) + \frac{1}{2}(\mathbf{A} - \mathbf{A}^\top). \]

Here, \( \frac{1}{2}(\mathbf{A} + \mathbf{A}^\top) \) is symmetric, and \( \frac{1}{2}(\mathbf{A} - \mathbf{A}^\top) \) is skew-symmetric. This decomposition shows that any square matrix can be represented as the sum of a symmetric matrix and a skew-symmetric matrix in a unique way.

Uniqueness:

To understand why this decomposition is unique, suppose we claim that \( \mathbf{A} \) has two different decompositions: \( \mathbf{A} = \mathbf{P} + \mathbf{Q} \), where \( \mathbf{P} \) is symmetric and \( \mathbf{Q} \) is skew-symmetric, and \( \mathbf{A} = \mathbf{R} + \mathbf{S} \), where \( \mathbf{R} \) is symmetric and \( \mathbf{S} \) is skew-symmetric.

This would mean that:

\[ \mathbf{P} + \mathbf{Q} = \mathbf{R} + \mathbf{S}. \]

Rearranging the terms, we have:

\[ \mathbf{P} - \mathbf{R} = \mathbf{S} - \mathbf{Q}. \]

Now, notice that \( \mathbf{P} - \mathbf{R} \) is the difference between two symmetric matrices, so it is symmetric. Similarly, \( \mathbf{S} - \mathbf{Q} \) is the difference between two skew-symmetric matrices, so it is skew-symmetric.

Thus, \( \mathbf{P} - \mathbf{R} \) is both symmetric and skew-symmetric. Similarly, \( \mathbf{S} - \mathbf{Q} \) is both symmetric and skew-symmetric. The only matrix that is both symmetric and skew-symmetric is the zero matrix. Therefore, we must have:

\[ \mathbf{P} - \mathbf{R} = 0 \quad \text{and} \quad \mathbf{S} - \mathbf{Q} = 0. \]

This implies \( \mathbf{P} = \mathbf{R} \) and \( \mathbf{Q} = \mathbf{S} \). Hence, it is not possible to have two different decompositions of \( \mathbf{A} \) into a symmetric and a skew-symmetric matrix. Therefore, the decomposition is unique.
Given any matrix \( \mathbf{A} \), the matrices \( \mathbf{A} \mathbf{A}^\top \) and \( \mathbf{A}^\top \mathbf{A} \) are always symmetric.

To see why \( \mathbf{A} \mathbf{A}^\top \) is symmetric, consider its transpose:

\[ (\mathbf{A} \mathbf{A}^\top)^\top = (\mathbf{A}^\top)^\top \mathbf{A}^\top. \]

Since the transpose of a transpose returns the original matrix, \( (\mathbf{A}^\top)^\top = \mathbf{A} \). Thus,

\[ (\mathbf{A} \mathbf{A}^\top)^\top = \mathbf{A} \mathbf{A}^\top. \]

This shows that \( \mathbf{A} \mathbf{A}^\top \) is equal to its transpose and therefore is symmetric.

Similarly, to show that \( \mathbf{A}^\top \mathbf{A} \) is symmetric, consider its transpose:

\[ (\mathbf{A}^\top \mathbf{A})^\top = \mathbf{A}^\top (\mathbf{A}^\top)^\top. \]

Again, since \( (\mathbf{A}^\top)^\top = \mathbf{A} \), we have:

\[ (\mathbf{A}^\top \mathbf{A})^\top = \mathbf{A}^\top \mathbf{A}. \]

Thus, \( \mathbf{A}^\top \mathbf{A} \) is equal to its transpose and is therefore symmetric.

This proves that for any matrix \( \mathbf{A} \), both \( \mathbf{A} \mathbf{A}^\top \) and \( \mathbf{A}^\top \mathbf{A} \) are symmetric matrices.
Given non-zero matrices \( \mathbf{A} \) and \( \mathbf{B} \), several properties arise depending on whether \( \mathbf{A} \) and \( \mathbf{B} \) are symmetric or skew-symmetric.

If \( \mathbf{A} \) and \( \mathbf{B} \) are both symmetric matrices, then their sum \( \mathbf{A} + \mathbf{B} \) and their difference \( \mathbf{A} - \mathbf{B} \) are also symmetric. To see why, consider that if \( \mathbf{A} \) and \( \mathbf{B} \) are symmetric, then \( \mathbf{A}^\top = \mathbf{A} \) and \( \mathbf{B}^\top = \mathbf{B} \). Therefore,

\[ (\mathbf{A} + \mathbf{B})^\top = \mathbf{A}^\top + \mathbf{B}^\top = \mathbf{A} + \mathbf{B}, \]

\[ (\mathbf{A} - \mathbf{B})^\top = \mathbf{A}^\top - \mathbf{B}^\top = \mathbf{A} - \mathbf{B}. \]

Since the transpose of the sum and difference equals the sum and difference themselves, both \( \mathbf{A} + \mathbf{B} \) and \( \mathbf{A} - \mathbf{B} \) are symmetric.

If \( \mathbf{A} \) and \( \mathbf{B} \) are both skew-symmetric matrices, then their sum \( \mathbf{A} + \mathbf{B} \) and their difference \( \mathbf{A} - \mathbf{B} \) are also skew-symmetric. This is because if \( \mathbf{A} \) and \( \mathbf{B} \) are skew-symmetric, then \( \mathbf{A}^\top = -\mathbf{A} \) and \( \mathbf{B}^\top = -\mathbf{B} \). Thus,

\[ (\mathbf{A} + \mathbf{B})^\top = \mathbf{A}^\top + \mathbf{B}^\top = -\mathbf{A} - \mathbf{B} = -(\mathbf{A} + \mathbf{B}), \]

\[ (\mathbf{A} - \mathbf{B})^\top = \mathbf{A}^\top - \mathbf{B}^\top = -\mathbf{A} + \mathbf{B} = -(\mathbf{A} - \mathbf{B}). \]

Since the transpose of the sum and difference is the negative of the sum and difference, both \( \mathbf{A} + \mathbf{B} \) and \( \mathbf{A} - \mathbf{B} \) are skew-symmetric.

However, if \( \mathbf{A} \) is symmetric and \( \mathbf{B} \) is skew-symmetric, then neither \( \mathbf{A} + \mathbf{B} \) nor \( \mathbf{A} - \mathbf{B} \) are symmetric or skew-symmetric. To see this, note that if \( \mathbf{A} \) is symmetric (\( \mathbf{A}^\top = \mathbf{A} \)) and \( \mathbf{B} \) is skew-symmetric (\( \mathbf{B}^\top = -\mathbf{B} \)), then

\[ (\mathbf{A} + \mathbf{B})^\top = \mathbf{A}^\top + \mathbf{B}^\top = \mathbf{A} - \mathbf{B} \neq \mathbf{A} + \mathbf{B}, \]

\[ (\mathbf{A} - \mathbf{B})^\top = \mathbf{A}^\top - \mathbf{B}^\top = \mathbf{A} + \mathbf{B} \neq -(\mathbf{A} - \mathbf{B}). \]

Therefore, \( \mathbf{A} + \mathbf{B} \) and \( \mathbf{A} - \mathbf{B} \) are neither symmetric nor skew-symmetric.
Given two square matrices \( \mathbf{A} \) and \( \mathbf{B} \), if it is claimed that \( \mathbf{A} \) is symmetric, \( \mathbf{B} \) is skew-symmetric, and \( \mathbf{A} + \mathbf{B} \) is symmetric, then \( \mathbf{B} \) must be the null matrix (a matrix with all entries equal to zero).

To see why this is true, recall that a matrix \( \mathbf{A} \) is symmetric if \( \mathbf{A}^\top = \mathbf{A} \), and a matrix \( \mathbf{B} \) is skew-symmetric if \( \mathbf{B}^\top = -\mathbf{B} \). Now consider the transpose of \( \mathbf{A} + \mathbf{B} \):

\[ (\mathbf{A} + \mathbf{B})^\top = \mathbf{A}^\top + \mathbf{B}^\top. \]

Since \( \mathbf{A} \) is symmetric and \( \mathbf{B} \) is skew-symmetric, this becomes:

\[ (\mathbf{A} + \mathbf{B})^\top = \mathbf{A} + (-\mathbf{B}) = \mathbf{A} - \mathbf{B}. \]

If \( \mathbf{A} + \mathbf{B} \) is also symmetric, then it must satisfy \( (\mathbf{A} + \mathbf{B})^\top = \mathbf{A} + \mathbf{B} \). Therefore,

\[ \mathbf{A} - \mathbf{B} = \mathbf{A} + \mathbf{B}. \]

Simplifying this equation:

\[ -\mathbf{B} = \mathbf{B} \implies 2\mathbf{B} = 0 \implies \mathbf{B} = 0. \]

Thus, \( \mathbf{B} \) must be the null matrix.

Similarly, if it is claimed that \( \mathbf{A} \) is symmetric, \( \mathbf{B} \) is skew-symmetric, and \( \mathbf{A} + \mathbf{B} \) is skew-symmetric, then \( \mathbf{A} \) must be the null matrix.

To prove this, consider the transpose of \( \mathbf{A} + \mathbf{B} \):

\[ (\mathbf{A} + \mathbf{B})^\top = \mathbf{A}^\top + \mathbf{B}^\top. \]

Using the properties of \( \mathbf{A} \) and \( \mathbf{B} \), this becomes:

\[ (\mathbf{A} + \mathbf{B})^\top = \mathbf{A} + (-\mathbf{B}) = \mathbf{A} - \mathbf{B}. \]

If \( \mathbf{A} + \mathbf{B} \) is skew-symmetric, then \( (\mathbf{A} + \mathbf{B})^\top = -(\mathbf{A} + \mathbf{B}) \). Therefore,

\[ \mathbf{A} - \mathbf{B} = -(\mathbf{A} + \mathbf{B}). \]

Simplifying this equation:

\[ \mathbf{A} - \mathbf{B} = -\mathbf{A} - \mathbf{B} \implies 2\mathbf{A} = 0 \implies \mathbf{A} = 0. \]

Thus, \( \mathbf{A} \) must be the null matrix.

In summary, if \( \mathbf{A} \) is symmetric, \( \mathbf{B} \) is skew-symmetric, and \( \mathbf{A} + \mathbf{B} \) is symmetric, then \( \mathbf{B} \) must be the null matrix. Similarly, if \( \mathbf{A} + \mathbf{B} \) is skew-symmetric, then \( \mathbf{A} \) must be the null matrix.
If \( \mathbf{A} \) and \( \mathbf{B} \) are symmetric matrices, then the product \( \mathbf{AB} \) is symmetric if and only if \( \mathbf{A} \) and \( \mathbf{B} \) commute, that is, \( \mathbf{AB} = \mathbf{BA} \).

Proof:

To prove this, we need to show two things: 1. If \( \mathbf{A} \) and \( \mathbf{B} \) are symmetric and \( \mathbf{AB} \) is symmetric, then \( \mathbf{AB} = \mathbf{BA} \). 2. If \( \mathbf{A} \) and \( \mathbf{B} \) are symmetric and \( \mathbf{AB} = \mathbf{BA} \), then \( \mathbf{AB} \) is symmetric.

1. If \( \mathbf{A} \) and \( \mathbf{B} \) are symmetric and \( \mathbf{AB} \) is symmetric, then \( \mathbf{AB} = \mathbf{BA} \):

Suppose \( \mathbf{A} \) and \( \mathbf{B} \) are symmetric matrices, so \( \mathbf{A}^\top = \mathbf{A} \) and \( \mathbf{B}^\top = \mathbf{B} \). If \( \mathbf{AB} \) is symmetric, then \( (\mathbf{AB})^\top = \mathbf{AB} \). Let's calculate the transpose of \( \mathbf{AB} \):

\[ (\mathbf{AB})^\top = \mathbf{B}^\top \mathbf{A}^\top. \]

Since \( \mathbf{A} \) and \( \mathbf{B} \) are symmetric, this becomes:

\[ (\mathbf{AB})^\top = \mathbf{B} \mathbf{A}. \]

If \( \mathbf{AB} \) is symmetric, then:

\[ \mathbf{AB} = (\mathbf{AB})^\top = \mathbf{B} \mathbf{A}. \]

Therefore, \( \mathbf{AB} = \mathbf{BA} \).

2. If \( \mathbf{A} \) and \( \mathbf{B} \) are symmetric and \( \mathbf{AB} = \mathbf{BA} \), then \( \mathbf{AB} \) is symmetric:

Suppose \( \mathbf{A} \) and \( \mathbf{B} \) are symmetric matrices, so \( \mathbf{A}^\top = \mathbf{A} \) and \( \mathbf{B}^\top = \mathbf{B} \), and suppose \( \mathbf{AB} = \mathbf{BA} \). We want to show that \( \mathbf{AB} \) is symmetric.

Calculate the transpose of \( \mathbf{AB} \):

\[ (\mathbf{AB})^\top = \mathbf{B}^\top \mathbf{A}^\top. \]

Since \( \mathbf{A} \) and \( \mathbf{B} \) are symmetric:

\[ (\mathbf{AB})^\top = \mathbf{B} \mathbf{A}. \]

Given \( \mathbf{AB} = \mathbf{BA} \), we substitute:

\[ (\mathbf{AB})^\top = \mathbf{AB}. \]

Therefore, \( \mathbf{AB} \) is symmetric.

If \( \mathbf{A} \) and \( \mathbf{B} \) are both skew-symmetric matrices and they commute (i.e., \( \mathbf{AB} = \mathbf{BA} \)), then the product \( \mathbf{AB} \) is symmetric.

If \( \mathbf{A} \) is symmetric and \( \mathbf{B} \) is skew-symmetric, and they commute (i.e., \( \mathbf{AB} = \mathbf{BA} \)), then the product \( \mathbf{AB} \) is skew-symmetric.
If \( \mathbf{A} \) is a symmetric (or skew-symmetric) matrix and \( \mathbf{P} \) is any matrix, then the matrices \( \mathbf{PAP}^\top \) and \( \mathbf{P}^\top \mathbf{A} \mathbf{P} \) are also symmetric (or skew-symmetric).
Proof:

Let \( \mathbf{P} \) be any matrix and \( \mathbf{A} \) be a symmetric (or skew-symmetric) matrix. We want to show that the matrices \( \mathbf{PAP}^\top \) and \( \mathbf{P}^\top \mathbf{A} \mathbf{P} \) are also symmetric (or skew-symmetric).

Case 1: Symmetric Matrices

If \( \mathbf{A} \) is symmetric, then \( \mathbf{A}^\top = \mathbf{A} \).
1. To prove \( \mathbf{PAP}^\top \) is symmetric:
  
  Consider the transpose of \( \mathbf{PAP}^\top \):
  
  \[ (\mathbf{PAP}^\top)^\top = (\mathbf{P}^\top)^\top \mathbf{A}^\top \mathbf{P}^\top. \]
  
  Since \( (\mathbf{P}^\top)^\top = \mathbf{P} \) and \( \mathbf{A}^\top = \mathbf{A} \) because \( \mathbf{A} \) is symmetric, we have:
  
  \[ (\mathbf{PAP}^\top)^\top = \mathbf{P} \mathbf{A} \mathbf{P}^\top = \mathbf{PAP}^\top. \]
  
  Therefore, \( \mathbf{PAP}^\top \) is symmetric.
2. To prove \( \mathbf{P}^\top \mathbf{A} \mathbf{P} \) is symmetric:
  
  Consider the transpose of \( \mathbf{P}^\top \mathbf{A} \mathbf{P} \):
  
  \[ (\mathbf{P}^\top \mathbf{A} \mathbf{P})^\top = \mathbf{P}^\top \mathbf{A}^\top (\mathbf{P}^\top)^\top. \]
  
  Since \( \mathbf{A}^\top = \mathbf{A} \) and \( (\mathbf{P}^\top)^\top = \mathbf{P} \), we get:
  
  \[ (\mathbf{P}^\top \mathbf{A} \mathbf{P})^\top = \mathbf{P}^\top \mathbf{A} \mathbf{P} = \mathbf{P}^\top \mathbf{A} \mathbf{P}. \]
  
  Therefore, \( \mathbf{P}^\top \mathbf{A} \mathbf{P} \) is symmetric.
Case 2: Skew-Symmetric Matrices

If \( \mathbf{A} \) is skew-symmetric, then \( \mathbf{A}^\top = -\mathbf{A} \).
1. To prove \( \mathbf{PAP}^\top \) is skew-symmetric:
  
  Consider the transpose of \( \mathbf{PAP}^\top \):
  
  \[ (\mathbf{PAP}^\top)^\top = (\mathbf{P}^\top)^\top \mathbf{A}^\top \mathbf{P}^\top. \]
  
  Since \( (\mathbf{P}^\top)^\top = \mathbf{P} \) and \( \mathbf{A}^\top = -\mathbf{A} \) because \( \mathbf{A} \) is skew-symmetric, we have:
  
  \[ (\mathbf{PAP}^\top)^\top = \mathbf{P} (-\mathbf{A}) \mathbf{P}^\top = -\mathbf{P} \mathbf{A} \mathbf{P}^\top. \]
  
  Therefore, \( \mathbf{PAP}^\top \) is skew-symmetric.
2. To prove \( \mathbf{P}^\top \mathbf{A} \mathbf{P} \) is skew-symmetric:
  
  Consider the transpose of \( \mathbf{P}^\top \mathbf{A} \mathbf{P} \):
  
  \[ (\mathbf{P}^\top \mathbf{A} \mathbf{P})^\top = \mathbf{P}^\top \mathbf{A}^\top (\mathbf{P}^\top)^\top. \]
  
  Since \( \mathbf{A}^\top = -\mathbf{A} \) and \( (\mathbf{P}^\top)^\top = \mathbf{P} \), we get:
  
  \[ (\mathbf{P}^\top \mathbf{A} \mathbf{P})^\top = \mathbf{P}^\top (-\mathbf{A}) \mathbf{P} = -(\mathbf{P}^\top \mathbf{A} \mathbf{P}). \]
  
  Therefore, \( \mathbf{P}^\top \mathbf{A} \mathbf{P} \) is skew-symmetric.
Conclusion

If \( \mathbf{A} \) is symmetric (skew-symmetric), then both \( \mathbf{PAP}^\top \) and \( \mathbf{P}^\top \mathbf{A} \mathbf{P} \) are also symmetric (skew-symmetric).
If \( \mathbf{P} \) is a row matrix (1 by \( n \)) and \( \mathbf{A} \) is a skew-symmetric matrix ( \( n \times n \) ), then the matrix \( \mathbf{PAP}^\top \) is a \( 1 \times 1 \) matrix, which means it is a scalar. We want to prove that \( \mathbf{PAP}^\top = [0] \).

Proof:

Given that \( \mathbf{A} \) is a skew-symmetric matrix, it satisfies the property \( \mathbf{A}^\top = -\mathbf{A} \).

Now, consider \( \mathbf{PAP}^\top \). Since \( \mathbf{P} \) is a \( 1 \times n \) row matrix, and \( \mathbf{A} \) is an \( n \times n \) skew-symmetric matrix, the product \( \mathbf{PAP}^\top \) results in a \( 1 \times 1 \) matrix, or a scalar.

From the previous property, we know that if \( \mathbf{A} \) is skew-symmetric, then \( \mathbf{PAP}^\top \) is also skew-symmetric. Since \( \mathbf{PAP}^\top \) is a \( 1 \times 1 \) matrix, and a \( 1 \times 1 \) skew-symmetric matrix must satisfy \( x = -x \), it follows that:

\[ \mathbf{PAP}^\top = -\mathbf{PAP}^\top. \]

Adding \( \mathbf{PAP}^\top \) to both sides, we get:

\[ 2\mathbf{PAP}^\top = 0. \]

Therefore, \( \mathbf{PAP}^\top = 0 \), which means \( \mathbf{PAP}^\top = [0] \).

This proves that if \( \mathbf{P} \) is a row matrix and \( \mathbf{A} \) is a skew-symmetric matrix, then \( \mathbf{PAP}^\top = [0] \).

Conversely, if for a square matrix \( \mathbf{A} \), the matrix \( \mathbf{PAP}^\top = [0] \) for all possible row matrices \( \mathbf{P} \), then \( \mathbf{A} \) must be skew-symmetric.

Example:

Let

\[ \mathbf{A} = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} , \quad \mathbf{P} = \begin{bmatrix} x & y & z \end{bmatrix}. \]

To calculate \( \mathbf{PAP}^\top \):

\[ \mathbf{PAP}^\top = \begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}. \]

Multiplying out the matrices:

\[ \mathbf{PAP}^\top = \begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} xa_1 + yb_1 + zc_1 \\ xa_2 + yb_2 + zc_2 \\ xa_3 + yb_3 + zc_3 \end{bmatrix}. \]

This gives:

\[ \mathbf{PAP}^\top = x(xa_1 + yb_1 + zc_1) + y(xa_2 + yb_2 + zc_2) + z(xa_3 + yb_3 + zc_3). \]

Expanding and rearranging, we get:

\[ \mathbf{PAP}^\top = x^2a_1 + xy(a_2 + b_1) + xz(a_3 + c_1) + y^2b_2 + yz(b_3 + c_2) + z^2c_3. \]

For \( \mathbf{PAP}^\top = 0 \) for all possible row matrices \( \mathbf{P} = [x \ y \ z] \), the coefficients of all the terms must be zero. This implies that \( a_1, b_2, c_3 = 0 \) and \( a_2 = -b_1 \), \( a_3 = -c_1 \), \( b_3 = -c_2 \), which are the conditions for \( \mathbf{A} \) to be skew-symmetric.

We can extend this example as a general proof.