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Suppose \(\mathbf{X}\) is an \(n \times p\) matrix with rank \(p\). (a) Show that \(\operatorname{ker}\left(\mathbf{X}^{\prime} \mathbf{X}\right)=\operatorname{ker}(\mathbf{X})\). (b) Use part (a) and the last exercise to show that if \(\mathbf{X}\) has full column rank, then \(\mathbf{X}^{\prime} \mathbf{X}\) is nonsingular.

Short Answer

Expert verified
Part (a) is demonstrated by proving that for any vector in the kernel of \(\mathbf{X}^{\prime} \mathbf{X}\), it must also be in the kernel of \(\mathbf{X}\), and vice versa, thus showing that \(\operatorname{ker}(\mathbf{X}^{\prime} \mathbf{X}) = \operatorname{ker}(\mathbf{X})\). In part (b), using the result from part (a), we know that the size of \(\operatorname{ker}(\mathbf{X}^{\prime} \mathbf{X})\) is \(0\) if \(\mathbf{X}\) has full column rank. Thus, \(\mathbf{X}^{\prime} \mathbf{X}\) is nonsingular.

Step by step solution

01

Defining the Kernels

Let's consider two vectors \(v\) such that \(v \in \operatorname{ker}(\mathbf{X}^{\prime} \mathbf{X})\) and \(v \in \operatorname{ker}(\mathbf{X})\). Then, by definition of a kernel, we have \(\mathbf{X}^{\prime} \mathbf{X} v = 0\) and \(\mathbf{X} v = 0\).
02

Proving \(\operatorname{ker}(\mathbf{X}^{\prime} \mathbf{X}) \subseteq \operatorname{ker}(\mathbf{X})\)

For any \(v \in \operatorname{ker}(\mathbf{X}^{\prime} \mathbf{X})\), we have \(v^{\prime} \mathbf{X}^{\prime} \mathbf{X} v = 0\). This implies \( (\mathbf{X} v)^{\prime} (\mathbf{X} v) = 0\). Since this holds only if \(\mathbf{X} v = 0\), we can conclude that \(v \in \operatorname{ker}(\mathbf{X})\).
03

Proving \(\operatorname{ker}(\mathbf{X}) \subseteq \operatorname{ker}(\mathbf{X}^{\prime} \mathbf{X})\)

For any \(v \in \operatorname{ker}(\mathbf{X})\), we have \(\mathbf{X} v = 0\). This implies \(\mathbf{X}^{\prime} \mathbf{X} v = \mathbf{X}^{\prime} 0 = 0\). Thus, \(v \in \operatorname{ker}(\mathbf{X}^{\prime} \mathbf{X})\).
04

Using part (a) to prove part (b)

From part (a), we know \(\operatorname{ker}(\mathbf{X}^{\prime} \mathbf{X}) = \operatorname{ker}(\mathbf{X})\). Since \(\mathbf{X}\) has full column rank \(p\), the size of its kernel (\(\operatorname{ker}(\mathbf{X})\)) is \(0\). Therefore, the size of \(\operatorname{ker}(\mathbf{X}^{\prime} \mathbf{X})\) is also \(0\), which means that \(\mathbf{X}^{\prime} \mathbf{X}\) is nonsingular.

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Key Concepts

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

Matrix Rank
The concept of matrix rank revolves around determining the dimension of the vector space generated by its rows or columns. The rank of a matrix is essentially a measure of its 'fullness' or how many linearly independent rows or columns it has. For instance, if an \(n \times p\) matrix \(\mathbf{X}\) is of rank \(p\), it means that all \(p\) columns are linearly independent. This concept is crucial because it tells us if the system of linear equations associated with the matrix has a unique solution, infinitely many solutions, or no solution.

Understanding the rank informs us about the behavior of the matrix in various operations such as solving linear systems, finding inverses, and more. The rank condition is also important in determining whether transformations associated with the matrix result in zero information loss. In our example, having a full column rank is especially essential for further proving properties of the kernel and the nonsingularity of a matrix.
Kernel of a Matrix
The kernel of a matrix, also known as the null space, contains all the vectors that when multiplied by the matrix result in the zero vector. In simple terms, these are the vectors that are 'annihilated' by the matrix. If a vector \(v\) satisfies \(\mathbf{A}v = 0\) for a matrix \(\mathbf{A}\), then \(v\) is in the kernel of \(\mathbf{A}\).

From the exercise, showing \(\operatorname{ker}(\mathbf{X}'\mathbf{X}) = \operatorname{ker}(\mathbf{X})\) involves proving that the null space of the matrix product \(\mathbf{X}'\mathbf{X}\) is identical to the null space of \(\mathbf{X}\). When \(\mathbf{X}\) has full column rank, its kernel contains only the zero vector, meaning no non-zero vector can be 'annihilated', simplifying further properties and proofs related to the matrix. Understanding the kernel is essential for grasping ideas related to solution spaces of matrix equations, as well as transformations and projections in vector spaces.
Nonsingular Matrix
A matrix is considered nonsingular or invertible if it has an inverse, that is, there exists another matrix which, when multiplied with it, results in the identity matrix. Matrices that are nonsingular have maximum rank, meaning no column or row is a linear combination of the others.

In the context of our exercise, once it's established that \(\operatorname{ker}(\mathbf{X}'\mathbf{X}) = \operatorname{ker}(\mathbf{X})\), knowing that \(\mathbf{X}\) has full column rank implies \(\mathbf{X}'\mathbf{X}\) is nonsingular. This is because a zero-kernel (i.e., the null space only contains the zero vector) means there are no non-trivial solutions to the homogeneous equation, allowing us to assert that \(\mathbf{X}'\mathbf{X}\) has an inverse. Nonsingularity is fundamental in solving linear systems uniquely and understanding the behaviors of transformations in linear algebra.
Full Column Rank
Full column rank in a matrix means that all columns of the matrix are linearly independent. This is a critical condition when dealing with matrices as it guarantees certain desirable algebraic properties. Specifically, an \(n \times p\) matrix having full column rank means that \(p\) is the maximum possible rank.

For our matrix \(\mathbf{X}\), having full column rank implies it can transform vectors without collapsing information into lower dimensions (i.e., without redundancy). Thus, solving \(\mathbf{X}v = 0\) leads only to the trivial solution because no combination of columns will eliminate any vector except the zero vector. This allows us to conclude that the kernel of \(\mathbf{X}\) is {0}, leading to our simplified proofs, such as proving \(\mathbf{X}'\mathbf{X}\) is nonsingular. Full column rank is a central idea in many applications of linear algebra including data analysis, statistics, and theoretical calculations.

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