Question

Exhibit a basis and calculate the dimension of each of the following subspaces of \(\mathbf{M}_{22}\) a. \(\left\\{A \mid A^{T}=-A\right\\}\) b. \(\left\\{A \mid A\left[\begin{array}{rr}1 & 1 \\ -1 & 0\end{array}\right]=\left[\begin{array}{rr}1 & 1 \\ -1 & 0\end{array}\right] A\right\\}\) c. \(\left\\{A \mid A\left[\begin{array}{rr}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\right\\}\) d. \(\left\\{A \mid A\left[\begin{array}{rr}1 & 1 \\ -1 & 0\end{array}\right]=\left[\begin{array}{rr}0 & 1 \\ -1 & 1\end{array}\right] A\right\\}\)

Short answer

a. Dimension = 1, b. Dimension = 2, c. Dimension = 2, d. Dimension = 0.

Step-by-step solution

Identifying condition for skew-symmetric matrices

A matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is skew-symmetric if \( A^T = -A \), which implies that \( a = 0 \), \( d = 0 \), and \( b = -c \). Therefore, every skew-symmetric \( 2 \times 2 \) matrix can be written as \( \begin{bmatrix} 0 & b \ -b & 0 \end{bmatrix} \).

Basis and dimension for skew-symmetric matrices

The set of all skew-symmetric \( 2 \times 2 \) matrices is generated by a single matrix \( \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} \). Thus, the basis for this subspace is \( \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} \), and the dimension of the subspace is 1.

Set up commutation condition for matrices in part b

To find matrices \( A \) that commute with \( \begin{bmatrix} 1 & 1 \ -1 & 0 \end{bmatrix} \), solve the matrix equation \( A \begin{bmatrix} 1 & 1 \ -1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \ -1 & 0 \end{bmatrix} A \). This yields the equations \( b + c = b \), \( a + d = c \), \( -c = d \), and \( d = 0 \).

Solve equations for part b

The conditions simplify to \( b = b + c \) (no new condition), \( d = -c \), and \( a + d = c \). Substituting \( d = 0 \), we get \( a = c \) and \( d = 0 \), giving the general form \( A = \begin{bmatrix} a & b \ 0 & a \end{bmatrix} \).

Basis and dimension for matrices in part b

The basis vectors for the resulting subspace are \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \) and \( \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} \). Therefore, the dimension is 2.

Setup nullification condition for matrices in part c

We need to find matrices \( A \) such that \( A \begin{bmatrix} 1 & 0 \ -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \). This gives the equations \( a - c = 0 \) and \( b - d = 0 \) for the elements of \( A \).

Solve equations for part c

From the equations, we get \( a = c \) and \( b = d \), so \( A = \begin{bmatrix} a & b \ a & b \end{bmatrix} \).

Determine basis and dimension for nullified matrices

The basis vectors are \( \begin{bmatrix} 1 & 0 \ 1 & 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 & 1 \ 0 & 1 \end{bmatrix} \), which show the dimension of the subspace is 2.

Commutation condition for matrices in part d

For the matrix \( A \) to commute with \( \begin{bmatrix} 1 & 1 \ -1 & 0 \end{bmatrix} \) as given, solve \( A \begin{bmatrix} 1 & 1 \ -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \ -1 & 1 \end{bmatrix} A \).

Simplify equations for part d

Solving the conditions from the commutation equation results in no consistent way to resolve \( A \) as it requires solving a set of ambiguous equations, effectively not forming a valid basis.

Analyze basis and dimension for part d

The resulting set of equations for part d leads to contradictions. Hence, there is no consistent subspace, implying dimension is 0.

Key concepts

Skew-Symmetric Matrices

Skew-symmetric matrices are an interesting type of square matrix whose transpose is equal to its negative. Mathematically, this condition is expressed as:

• For a matrix \( A \), the condition is \( A^T = -A \).
• This implies for any \( i, j \): \( a_{ij} = -a_{ji} \).

For a 2x2 matrix example such as \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), if \( A \) is skew-symmetric, the diagonal elements must be zero \( a = 0 \) and \( d = 0 \), while off-diagonals are negatives of each other \( b = -c \).
Consequently, each skew-symmetric 2x2 matrix takes the form \( \begin{bmatrix} 0 & b \ -b & 0 \end{bmatrix} \).
The basis for this set comprises a single matrix \( \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} \), hence the subspace has a dimension of 1.

Matrix Commutation

Matrix commutation implies that two matrices can be multiplied in any order to give the same result. For matrices \( A \) and \( B \), the equation \( AB = BA \) must hold true.
In the context of the exercise, determine which matrices \( A \) commute with \( B = \begin{bmatrix} 1 & 1 \ -1 & 0 \end{bmatrix} \). This involves solving the set of equations derived from the equation \( A \begin{bmatrix} 1 & 1 \ -1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \ -1 & 0 \end{bmatrix} A \).

• This results in conditions such as \( b + c = b \), \( a + d = c \), \( -c = d \).
• Arriving at a general solution: A matrix of the form \( A = \begin{bmatrix} a & b \ 0 & a \end{bmatrix} \).

The basis for this subspace includes matrices like \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \) and \( \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} \), with a dimension of 2.

Basis and Dimension

The concepts of basis and dimension are fundamental in understanding matrix subspaces. A basis is a set of vectors that spans the entire vector space, while the dimension is the number of vectors in that set.
For instance, consider the subspace of skew-symmetric matrices. Here, the basis is the single matrix \( \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} \), implying a dimension of 1.
Similarly, in the commutation example, the basis consists of two matrices, indicating the dimension is 2.

• Basis vectors must be independent and span the space.
• Dimension tells how many independent directions or vectors define the space.

Each type of matrix subspace may have different bases and dimensions.

Nullification Condition

The nullification condition occurs when a product of matrices results in a zero matrix. In the provided exercise, investigate matrices \( A \) such that when multiplied by another matrix, the outcome is zero.
Specifically, find \( A \) where \( A \begin{bmatrix} 1 & 0 \ -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \).

• Resulting equations are \( a - c = 0 \) and \( b - d = 0 \).
• This leads to matrix form \( A = \begin{bmatrix} a & b \ a & b \end{bmatrix} \).

The basis here is expressed by matrices \( \begin{bmatrix} 1 & 0 \ 1 & 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 & 1 \ 0 & 1 \end{bmatrix} \), indicating the dimension is 2.
Thus, this shows how the nullification condition influences the properties and dimensions of certain matrix subspaces.