Problem 7 Exhibit a basis and calculate th... [FREE SOLUTION]

Chapter 6: Problem 7

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

Expert verified

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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