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Chapter 9: Problem 13

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
The kernel of \( \mathbf{X}^{\prime} \mathbf{X} \) is equivalent to the kernel of \( \mathbf{X} \), and if the matrix \( \mathbf{X} \) has full column rank, then the matrix \( \mathbf{X}^{\prime} \mathbf{X} \) is nonsingular.

Step by step solution

01

Equal Kernels

To show that \( \operatorname{ker}(\mathbf{X}^{\prime} \mathbf{X}) = \operatorname{ker}(\mathbf{X}) \), let's start by choosing an arbitrary vector \( \mathbf{v} \) from the kernel of \( \mathbf{X} \), so \( \mathbf{Xv} = \mathbf{0} \). We need to show that this vector also belongs to the kernel of \( \mathbf{X}^{\prime} \mathbf{X} \). We have \( \mathbf{X}^{\prime} (\mathbf{Xv}) = \mathbf{X}^{\prime} \mathbf{0} = \mathbf{0} \), meaning that \( \mathbf{v} \) is in the kernel of \( \mathbf{X}^{\prime} \mathbf{X} \), proving one direction of the equality.
02

Proof for the Other Direction

For the other direction, consider an arbitrary vector \( \mathbf{u} \) from \( \operatorname{ker}(\mathbf{X}^{\prime} \mathbf{X}) \). That means \( \mathbf{X}^{\prime} \mathbf{Xu} = \mathbf{0} \), which implies \( (\mathbf{Xu})^{\prime} (\mathbf{Xu}) = \mathbf{0} \). This means \( \mathbf{Xu} = \mathbf{0} \), implying \( \mathbf{u} \) is in the kernel of \( \mathbf{X} \). These proofs meet the requirements of (a) part.
03

Full Column Rank & Nonsingularity

Assuming that the matrix \( \mathbf{X} \) has full column rank \( p \), any vector in the kernel of \( \mathbf{X} \) is a zero vector. And from the previous steps, this means that any vector in the kernel of \( \mathbf{X}^{\prime} \mathbf{X} \) is also a zero vector. In other words, the matrix \( \mathbf{X}^{\prime} \mathbf{X} \) is nonsingular. This completes the proof for the (b) part.

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Most popular questions from this chapter

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