Question

For any square matrix \( A \) with real entries, if \( A + A' \) is a symmetric matrix then :

• A. \( (A - A') \) cannot be a skew symmetric matrix
• B. \( (A - A') \) is a skew symmetric matrix
• C. \( A \) is always a symmetric matrix
• D. \( A \) is always a skew symmetric matrix

Hint:

Any square matrix \( A \) can be represented as the sum of a symmetric and a skew-symmetric matrix: \( A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A') \).
Standard results: \( (A + A') \) is always symmetric and \( (A - A') \) is always skew-symmetric.

Answer 1

The Correct Option is B

Concept:
• A square matrix \( M \) is defined as symmetric if \( M' = M \).
• A square matrix \( M \) is defined as skew-symmetric if \( M' = -M \).
• For any square matrix \( A \), the sum \( A + A' \) is always symmetric.
• For any square matrix \( A \), the difference \( A - A' \) is always skew-symmetric.

Step 1: Define the matrix and calculate its transpose
Let the matrix be \( X = A - A' \).
To find if it is symmetric or skew-symmetric, we compute its transpose \( X' \):
\[ X' = (A - A')' \]
Applying the transpose property \( (P - Q)' = P' - Q' \):
\[ X' = A' - (A')' \]

Step 2: Simplify the expression using transpose properties
We know that the transpose of a transpose of a matrix is the matrix itself, i.e., \( (A')' = A \).
Substituting this into our equation:
\[ X' = A' - A \]
Factor out the negative sign from the right-hand side:
\[ X' = -(A - A') \]

Step 3: Conclude the type of matrix
Since \( X = A - A' \), we can write:
\[ X' = -X \]
By definition, any matrix \( X \) that satisfies \( X' = -X \) is a skew-symmetric matrix.
Therefore, \( A - A' \) is a skew-symmetric matrix.