Question

Deal with the problem of solving \(\mathbf{A x}=\mathbf{b}\) when \(\operatorname{det} \mathbf{A}=0\) Suppose that, for a given matrix \(\mathbf{A}\), there is a nonzero vector \(\mathbf{x}\) such that \(\mathbf{A x}=\mathbf{0 . ~ S h o w ~}\) that there is also a nonzero vector \(\mathbf{y}\) such that \(\mathbf{A}^{*} \mathbf{y}=\mathbf{0} .\)

Short answer

Answer: When a matrix is singular (determinant is zero), the solvability of the linear system is affected, making it not uniquely solvable. It can have either no solution or infinitely many solutions. Given the condition that a nonzero vector x exists such that Ax=0, it implies the columns of matrix A are linearly dependent and the system does not have a unique solution.

Step-by-step solution

Determinant of a matrix

The determinant of a square matrix is a scalar value that can be computed from the elements of the matrix. In this exercise, we are given that the determinant of matrix \(\mathbf{A}\) is zero. This means that the matrix is singular, that is, it does not have a unique inverse or it is not invertible. A matrix with a determinant of zero is also called a singular matrix, and such a matrix cannot be invertible. This has an effect on the solvability of the system of linear equations represented by the matrix equation \(\mathbf{A x}=\mathbf{b}\).

Solving the system when the determinant is zero

When the determinant of a matrix is zero, the system represented by the equation \(\mathbf{A x}=\mathbf{b}\) can either have no solution or infinitely many solutions. A system with no unique solution is called an inconsistent system. In such cases, the rows of the matrix \(\mathbf{A}\) are either linearly dependent (there is a row that can be expressed as a linear combination of the other rows) or the rank of \(\mathbf{A}\) is less than the number of variables in the system. In this exercise, we are given that there exists a nonzero vector \(\mathbf{x}\) such that \(\mathbf{A x}=\mathbf{0}\). This implies that the columns of \(\mathbf{A}\) are linearly dependent, and the system does not have a unique solution.

Existence of a nonzero vector \(\mathbf{y}\) such that \(\mathbf{A}^*\mathbf{y}=\mathbf{0}\)

To show that there exists a nonzero vector \(\mathbf{y}\) such that \(\mathbf{A}^*\mathbf{y}=\mathbf{0}\), we can use the fact that the matrix \(\mathbf{A}\) is singular. Since \(\mathbf{A}\) is singular, its columns are linearly dependent. This means there is a linear combination of the columns of \(\mathbf{A}\) that results in a zero vector, i.e., \(\mathbf{A x}=\mathbf{0}\). Taking the conjugate transpose of this equation, we get: \[\begin{aligned} (\mathbf{A x})^* &= \mathbf{0}^* \\ (\mathbf{x}^* \mathbf{A}^*) &= \mathbf{0}^* \\ \mathbf{x}^* \mathbf{A}^* &= \mathbf{0} \end{aligned}\] Let \(\mathbf{y}=\mathbf{x}^*\). The vector \(\mathbf{y}\) is nonzero since \(\mathbf{x}\) is nonzero. So, there exists a nonzero vector \(\mathbf{y}\) such that \(\mathbf{A}^*\mathbf{y}=\mathbf{0}\). In conclusion, when the determinant of the matrix \(\mathbf{A}\) is zero, the system \(\mathbf{A x}=\mathbf{b}\) can have no solution or infinitely many solutions. For the given condition in the exercise, if there is a nonzero vector \(\mathbf{x}\) such that \(\mathbf{A x}=\mathbf{0}\), then there must also exist a nonzero vector \(\mathbf{y}\) such that \(\mathbf{A}^*\mathbf{y}=\mathbf{0}\).

Key concepts

Determinant of a Matrix

When dealing with matrix equations, understanding the role of the determinant of a matrix is crucial. The determinant is a special number calculated from the matrix's elements that tells us valuable properties of the matrix. For instance, it can indicate whether the matrix is invertible or not.

The determinant is often denoted as \(\operatorname{det}(\mathbf{A})\) for a matrix \(\textbf{A}\). When \(\operatorname{det}(\mathbf{A}) = 0\), the matrix does not have an inverse, and this is what we call a singular matrix. This singular nature has profound implications on the system of equations \(\mathbf{A x} = \mathbf{b}\) that the matrix represents. If the system's determinant is zero, it becomes impossible to directly find a unique solution for \(\mathbf{x}\) using standard methods, such as inverse matrix computation.

Singular Matrix

A singular matrix is one that does not have a unique inverse, or simply, it is not invertible. This is directly related to the concept of the determinant, as a matrix is singular if, and only if, its determinant is zero.

In practical terms, a singular matrix does not allow us to solve \(\mathbf{A x} = \mathbf{b}\) in the typical way because there isn't a unique solution. Instead, the system may have no solution or an infinite number of solutions. This happens because the rows (or columns) of a singular matrix are linearly dependent, meaning that at least one row (or column) can be made by a combination of the others. Understanding when a matrix is singular helps in identifying why certain linear systems do not behave as expected and require alternative methods to find solutions, if they exist.

Linearly Dependent Systems

A system of equations is said to have linearly dependent vectors when one vector in the system can be expressed as a linear combination of others. This concept is a cornerstone in linear algebra as it affects the solvability of the system.

For the matrix equation \(\mathbf{A x} = \mathbf{b}\), the presence of linearly dependent rows in matrix \(\mathbf{A}\) often indicates that multiple solutions exist. If a vector \(\mathbf{x}\) is found such that \(\mathbf{A x} = \mathbf{0}\), it confirms that the system of equations is dependent and does not have a single, unique solution. This dependency directly ties to the concept of a singular matrix, as it also signifies the lack of a unique solution.

Conjugate Transpose

The concept of conjugate transpose, also known as the Hermitian transpose, is applied to complex matrices and is an extension of the regular transpose for real matrices. The conjugate transpose of a matrix \(\textbf{A}\) is denoted as \(\mathbf{A}^*\) and is formed by taking the transpose of the matrix and then taking the complex conjugate of each element.

When applied to a vector or matrix equation, the conjugate transpose can reveal properties similar to those discovered by vector or matrix transposition in real matrices, such as symmetry and orthogonality. In our exercise, the conjugate transpose helps us find a vector \(\mathbf{y}\) such that \(\mathbf{A}^*\mathbf{y} = \mathbf{0}\). Since the vector \(\mathbf{x}\) is non-zero and satisfies \(\mathbf{A x} = \mathbf{0}\), it reinforces the fact that the matrix \(\mathbf{A}\) has linearly dependent columns, which, by taking the conjugate transpose, affects the solutions for \(\mathbf{y}\) as well.