Question

Let \(A\) be an \((n-1) \times n\) matrix with rows \(\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{n-1}\) and let \(A_{i}\) denote the \((n-1) \times(n-1)\) matrix obtained from \(A\) by deleting column \(i\). Define the vector \(\mathbf{y}\) in \(\mathbb{R}^{n}\) by $$ \mathbf{y}=\left[\operatorname{det} A_{1}-\operatorname{det} A_{2} \operatorname{det} A_{3} \cdots(-1)^{n+1} \operatorname{det} A_{n}\right] $$ Show that: a. \(\mathbf{x}_{i} \cdot \mathbf{y}=0\) for all \(i=1,2, \ldots, n-1 .\) [Hint: Write \(B_{i}=\left[\begin{array}{c}x_{i} \\ A\end{array}\right]\) and show that det \(\left.B_{i}=0 .\right]\) b. \(\mathbf{y} \neq \mathbf{0}\) if and only if \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{n-1}\right\\}\) is linearly independent. [Hint: If some det \(A_{i} \neq 0,\) the rows of \(A_{i}\) are linearly independent. Conversely, if the \(\mathbf{x}_{i}\) are independent, consider \(A=U R\) where \(R\) is in reduced row-echelon form.] c. If \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{n-1}\right\\}\) is linearly independent, use Theorem 8.1 .3(3) to show that all solutions to the system of \(n-1\) homogeneous equations $$ A \mathbf{x}^{T}=\mathbf{0} $$ are given by \(t \mathbf{y}, t\) a parameter.

Short answer

\(\mathbf{x}_{i} \cdot \mathbf{y} = 0\) for all \(i\); \(\mathbf{y} \neq \mathbf{0}\) iff the set is independent; solutions are \(t\mathbf{y}\).

Step-by-step solution

Define the Matrix \(B_{i}\)

For part (a), define \(B_{i}\) as the matrix obtained by adding \(\mathbf{x}_{i}\) as a top row to the matrix \(A\), i.e., \(B_{i} = \begin{bmatrix} \mathbf{x}_{i} \ A \end{bmatrix}\).

Show \(\operatorname{det} B_{i} = 0\)

Since \(B_{i}\) is a square matrix with more rows than columns from \(A\), any square matrix larger than its determinant is 0 when it has linearly dependent rows. For determining \(\mathbf{y}\), expand the determinant along the first row, leading to \(\mathbf{x}_{i} \cdot \mathbf{y} = 0\). This shows part (a): \(\mathbf{x}_{i} \cdot \mathbf{y} = 0\) for each \(i\).

Determine Linearity of \(\mathbf{y}\)

For part (b), analyze when \(\mathbf{y} eq \mathbf{0}\). If the vectors \(\{ \mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_{n-1} \}\) are linearly independent, then at least one of the determinants \(A_{i}\) is non-zero, making \(\mathbf{y} eq \mathbf{0}\). Thus, \(\mathbf{y} eq \mathbf{0}\) if and only if the set is linearly independent.

Show Solution is Scalar Multiple of \(\mathbf{y}\)

In part (c), if \(\{ \mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_{n-1} \}\) is linearly independent, \(A = UR\) where \(R\) has full row-rank, confirming zero row-space for \(\mathbf{y} eq \mathbf{0}\). Theorem 8.1.3(3) confirms any \(\mathbf{x}\) satisfying \(A \mathbf{x}^{T} = \mathbf{0}\) is a scalar multiple of \(\mathbf{y}\).

Key concepts

Linear Independence

Linear independence is a fundamental concept in matrix theory and linear algebra, which describes a situation where a set of vectors does not rely on each other to span the space. In simpler terms, if vectors are linearly independent, none of the vectors can be expressed as a combination of the others.

In the context of our problem, we have a set of vectors \( \{ \mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_{n-1} \} \). These vectors are rows of the matrix \(A\). For these vectors to be linearly independent, at least one of the determinants \(\operatorname{det} A_i\) must be non-zero. This is because each \(A_i\) is formed by removing one column from \(A\), and its determinant reflects the volume spanned by \(n-1\) vectors in \(\mathbb{R}^{n}\).

When vectors are linearly independent, they can fully span the space, indicating there is a unique, non-trivial solution set to the problem. This uniqueness is crucial when solving linear equations, as it ensures the solution cannot be written as a linear combination of other solutions.

Determinant

A determinant is a scalar value that provides insight into the properties of a matrix. It is primarily used to determine whether a set of vectors is linearly independent, as well as to find out if a matrix is invertible.

In our exercise, each submatrix \(A_i\) is crucial for evaluating whether the set of vectors \( \{ \mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_{n-1} \} \) is linearly independent. By computing the determinant of these matrices, we can learn about the span and dimensional characteristics of the matrix \(A\).

When the determinant is zero, it's a sign that the matrix is degenerate, meaning that its rows or columns are linearly dependent. In contrast, a non-zero determinant indicates independence. This aligns with the exercise where the determinant of \(B_i\) being zero reinforces the linear dependency among the vectors \(\mathbf{x}_i\) when appended to \(A\).

In practical terms, calculating the determinant of \(A_i\) allows us to deduce critical insights about the solution space of a matrix equation, especially crucial when dealing with the system's uniqueness and variability.

Homogeneous Equations

A system of homogeneous equations is characterized by having all terms equal to zero. In a matrix notation, this is represented as \(A\mathbf{x}^{T} = \mathbf{0}\). Here, \(A\) is a matrix, and \(\mathbf{x}^{T}\) is the transpose of the solution vector.

In our problem, the matrix \(A\) is paired with a vector \(\mathbf{x}^{T}\), and the condition for solutions is that the dot product must yield a zero vector. Such systems always have at least the trivial solution, where \(\mathbf{x} = \mathbf{0}\).

However, if the set of row vectors in \(A\) is linearly independent, any other solution must be a scalar multiple of \( \mathbf{y} \), as discussed in part (c) of the solution using Theorem 8.1.3(3). Each solution therefore can be expressed as \(t \mathbf{y}\), where \(t\) is a parameter. This parameterization indicates that the solution space of the system is one-dimensional along \(\mathbf{y}\), showing the deep relationship between the determinants, linear independence, and solution behavior of homogeneous systems.