Understanding Skew-Symmetric Matrices and Their Rank

A skew-symmetric matrix is a special type of square matrix. A matrix $A$ is called skew-symmetric if its transpose is equal to its negative, i.e., $A^T = -A$. This property has significant implications for the elements of the matrix. For example, the diagonal elements of a skew-symmetric matrix must always be zero, because $a_{ii} = -a_{ii}$, which means $2a_{ii} = 0$, so $a_{ii} = 0$.

The Key Property of Skew-Symmetric Matrix Rank

A crucial property regarding the rank of a real skew-symmetric matrix is that its rank is always an even number. The rank of matrix $A$ is the dimension of the vector space spanned by its columns (or rows). This property, that the rank of a real skew-symmetric matrix must be an even rank, is derived from advanced linear algebra concepts, particularly related to the eigenvalues and structure of such matrices. Any real skew-symmetric matrix with odd dimensions greater than 1 is singular (determinant is zero), meaning its rank must be less than the dimension and also even. For example, a 3x3 skew-symmetric matrix will have determinant 0 and thus rank < 3, and its rank must be even, so it can be 0 or 2.

Analyzing the Options for Skew-Symmetric Matrix Rank

We are given several possible values for the rank of a skew-symmetric matrix and asked which one cannot be its rank. Based on the property discussed, the rank of a real skew-symmetric matrix must always be an even number. Let's look at the options:

• Option 1: Rank is 1.
• Option 2: Rank is 2.
• Option 3: Rank is 3.
• Option 4: Rank is 6.

Comparing these options with the rule that the rank must be an even number:

• Rank 1 is an odd number.
• Rank 2 is an even number.
• Rank 3 is an odd number.
• Rank 6 is an even number.

Therefore, any odd number cannot be the rank of a real skew-symmetric matrix. Both 1 and 3 are odd numbers, so neither can be the rank.

Why Rank 1 is Not Possible for a Skew-Symmetric Matrix

The question asks for a rank that a skew-symmetric matrix cannot have, and option 1 suggests rank 1. Since the rank of a real skew-symmetric matrix must be an even rank, an odd rank like 1 is impossible. For instance, consider a 2x2 skew-symmetric matrix: $$ A = \begin{pmatrix} 0 & a \\ -a & 0 \end{pmatrix} $$ If $a=0$, the matrix is the zero matrix, and its rank is 0 (which is even). If $a \neq 0$, the determinant is $0 \cdot 0 - a \cdot (-a) = a^2$. If $a \neq 0$, $a^2 \neq 0$, so the determinant is non-zero, and the rank is 2 (which is even). A 2x2 skew-symmetric matrix can have rank 0 or 2, both even. It cannot have rank 1. For a general $n \times n$ real skew-symmetric matrix, its rank of matrix will always be even. Thus, 1 is a rank that a skew-symmetric matrix cannot have.