Question

If \(S\) is a symmetric matrix and \(A\) is an antisymmetric matrix (Problem 2), show that \(\operatorname{Tr}(S A)=0\), Hint \(:\) Prove \(\operatorname{Tr}(S A)=-\operatorname{Tr}(S A)\)

Short answer

The trace of the product of a symmetric and an antisymmetric matrix is zero: \( \operatorname{Tr}(SA) = 0 \)

Step-by-step solution

- Identify matrix properties

Recall that a symmetric matrix is one that is equal to its transpose: \[ S^T = S \]An antisymmetric matrix is one where the transpose equals the negative of the matrix:\[ A^T = -A \]

- Use the cyclic property of the trace

The trace of a product of matrices has the cyclic property, meaning that the trace of the product of matrices is invariant under cyclic permutations of the matrices:\[ \operatorname{Tr}(AB) = \operatorname{Tr}(BA) \]We will use this property to show that \( \operatorname{Tr}(SA) \) remains the same when the order is changed.

- Express the trace of SA in terms of its transpose

Using the trace property and the definitions of symmetric and antisymmetric matrices, we can express the trace of \( SA \) as follows:\[ \operatorname{Tr}(SA) = \operatorname{Tr}((SA)^T) \]Because \[ (SA)^T = A^T S^T \] and substituting the properties \( S^T = S \) and \( A^T = -A \), we get:\[ (SA)^T = (-A)S \]

- Simplify and show equality

Simplify the expression \((SA)^T = (-A) S\):\[ (SA)^T = -AS \]Since \( \operatorname{Tr}(AS) = \operatorname{Tr}(SA) \) due to the cyclic property, we have:\[ \operatorname{Tr}(-AS) = -\operatorname{Tr}(AS) = -\operatorname{Tr}(SA) \]Therefore:\[ \operatorname{Tr}(SA) = -\operatorname{Tr}(SA) \]

- Conclude

Since the only solution to \( \operatorname{Tr}(SA) = -\operatorname{Tr}(SA) \) is \( \operatorname{Tr}(SA) = 0 \), it follows that: \[ \operatorname{Tr}(SA) = 0 \]

Key concepts

symmetric matrix

A symmetric matrix is a square matrix that is equal to its transpose. This means if you flip the matrix over its diagonal, it remains the same. Mathematically, a matrix \(S\) is symmetric if \(S^T = S\).
Here are some properties of symmetric matrices:

• The elements on the diagonal can be any number.
• Symmetric matrices have real eigenvalues.
• They are always diagonalizable (can be written in the form \(PDP^{-1}\), where \(D\) is a diagonal matrix).

Understanding these properties helps in visualizing symmetric matrices and how they behave in various operations.

antisymmetric matrix

An antisymmetric matrix, also known as a skew-symmetric matrix, satisfies the condition that its transpose is equal to the negative of the matrix itself. Mathematically, a matrix \(A\) is antisymmetric if \(A^T = -A\).
This implies that all the elements on its diagonal are zero because if \(a_{ii}\) (diagonal element) is equal to \(-a_{ii}\), then \(a_{ii} = 0\). Here are some additional properties:

• An antisymmetric matrix must be square (same number of rows and columns).
• They have purely imaginary eigenvalues or zero.
• The off-diagonal elements must be the opposite in sign (if \(a_{ij}\) is an element, then \(a_{ji} = -a_{ij}\)).

These characteristics help you quickly identify antisymmetric matrices and understand their role in mathematical operations.

trace of a matrix

The trace of a matrix is the sum of the elements on the main diagonal. For a square matrix \(M\), the trace, denoted \(\operatorname{Tr}(M)\), is given by:
\[ \operatorname{Tr}(M) = \sum_{i} m_{ii} \]
Properties of the trace include:

• It is linear: \( \operatorname{Tr}(A+B) = \operatorname{Tr}(A) + \operatorname{Tr}(B)\).
• It is invariant under similarity transformations: if \(M = P^{-1}AP\), then \( \operatorname{Tr}(A) = \operatorname{Tr}(M) \).
• The trace of a product of matrices has a cyclic property, which states \( \operatorname{Tr}(AB) = \operatorname{Tr}(BA) \).

The trace is particularly useful in many areas of linear algebra and quantum mechanics.

cyclic property of trace

The cyclic property of the trace is an essential tool in matrix algebra. It states that the trace of a product of matrices remains the same no matter how the matrices are cyclically permuted. For instance:
\[ \operatorname{Tr}(ABC) = \operatorname{Tr}(BCA) = \operatorname{Tr}(CAB) \]
This property is extremely useful in proofs and simplifications. Here are a few key points:

• The property holds only for the trace of a product, not necessarily for individual entries or other operations.
• This property was crucial in the given exercise to show that \( \operatorname{Tr}(SA) = -\operatorname{Tr}(SA) \), leading to the conclusion that \( \operatorname{Tr}(SA) = 0 \).

Familiarity with this property simplifies the handling of multiple matrix products and their traces. It can be used to switch matrices around within the trace to make calculations easier.