Understanding Skew-Symmetric Matrices

A skew-symmetric matrix is a square matrix \(A\) where its transpose (\(A^T\)) is equal to its negative (\(-A\)). This relationship can be written as:

\(A^T = -A\)

This definition implies that for any element \(a_{ij}\) (the element in the \(i\)-th row and \(j\)-th column), the element \(a_{ji}\) must satisfy \(a_{ji} = -a_{ij}\). We need to check three statements regarding a \(3 \times 3\) skew-symmetric matrix.

Statement 1: Diagonal Elements are Zero

Statement 1 says that all diagonal elements of the matrix are zero.

For any diagonal element, the row index is the same as the column index, meaning \(i=j\). So, we are looking at elements \(a_{ii}\).

Using the property \(a_{ji} = -a_{ij}\) for \(i=j\), we get:

\(a_{ii} = -a_{ii}\)

If we add \(a_{ii}\) to both sides of the equation, we get:

\(2a_{ii} = 0\)

Dividing by 2 yields:

\(a_{ii} = 0\)

This confirms that all diagonal elements (\(a_{11}, a_{22}, a_{33}\)) must be zero. So, Statement 1 is correct.

Statement 2: Sum of Diagonal Elements

Statement 2 claims that the sum of all the diagonal elements is zero.

The sum of the diagonal elements of a matrix is called its trace. For a \(3 \times 3\) matrix \(A\), the trace is \(tr(A) = a_{11} + a_{22} + a_{33}\).

From Statement 1, we established that \(a_{11} = 0\), \(a_{22} = 0\), and \(a_{33} = 0\). Therefore, the sum is:

\(tr(A) = 0 + 0 + 0 = 0\)

This means the sum of the diagonal elements is zero. So, Statement 2 is correct.

Statement 3: Orthogonal Matrix Property

Statement 3 proposes that the skew-symmetric matrix \(A\) is an orthogonal matrix.

A matrix is orthogonal if its transpose equals its inverse, satisfying \(A^T A = I\), where \(I\) is the identity matrix.

For a skew-symmetric matrix, we know \(A^T = -A\). Substituting this into the orthogonality condition:

\((-A)A = I\)

\(-A^2 = I\)

\(A^2 = -I\)

Let's consider a general \(3 \times 3\) skew-symmetric matrix:

Calculating \(A^2\):

\(A^2 = A \times A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix} \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}\)

\(A^2 = \begin{pmatrix} -a^2-b^2 & -bc & ac \\ -bc & -a^2-c^2 & -ab \\ ac & -ab & -b^2-c^2 \end{pmatrix}\)

For \(A\) to be orthogonal, \(A^2\) must equal \(-I\):

\(\begin{pmatrix} -a^2-b^2 & -bc & ac \\ -bc & -a^2-c^2 & -ab \\ ac & -ab & -b^2-c^2 \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}\)

This requires the off-diagonal terms to be zero (\(ac=0, -bc=0, ab=0\)) and the diagonal terms to be -1 (\(-a^2-b^2 = -1\), \(-a^2-c^2 = -1\), \(-b^2-c^2 = -1\)).

If \(a=0\), we need \(b^2=1\), \(c^2=1\), and \(b^2+c^2 = 1\). But \(1+1 = 2 \neq 1\). This shows a contradiction. A general skew-symmetric matrix does not satisfy \(A^2 = -I\). For example, the zero matrix is skew-symmetric but \(0^2 \neq -I\). Therefore, Statement 3 is incorrect.

Final Conclusion

Reviewing the analysis:

• Statement 1 is correct.
• Statement 2 is correct.
• Statement 3 is incorrect.

Thus, the correct statements are 1 and 2 only.