A transpose problem

Problem

\(X\) is an \(n×k\) matrix. For this problem, you may want to review Matrix transpose and Matrix Inverse

  1. What is the dimension of \({\left( X'X \right)}^{-1}\) ?
  2. Show that \(X'X\) is symmetric, \({\left( X'X \right)}'=X'X\)
  3. Show that \({\left( X'X \right)}^{-1}\) is symmetric, \({\left( {\left( X'X \right)}^{-1} \right)}'={\left( X'X \right)}^{-1}\)
  4. What is the dimension of \({\left( X'X \right)}^{-1}X'\) ?
  5. Show that the transpose of \({\left( X'X \right)}^{-1}X'\) is \(X{\left( X'X \right)}^{-1}\)

Solution

  1. \(k×k\)
  2. The general rule is \({\left( AB \right)}'=B'A'\) . To find \({\left( X'X \right)}'\) you need to flip \(X'\) and \(X\) around and take the transpose of each. The transpose of \(X'\) is \(X\) . Thus, \({\left( X'X \right)}'=X'X\) .
  3. The general rule is \({\left( A^{-1} \right)}'={\left( A' \right)}^{-1}\) for an invertible matrix \(A\) . Thus, \({\left( {\left( X'X \right)}^{-1} \right)}'={\left( {\left( X'X \right)}' \right)}^{-1}\) . From previous part, \({\left( X'X \right)}'=X'X\) , Thus, \({\left( {\left( X'X \right)}' \right)}^{-1}={\left( X'X \right)}^{-1}\)
  4. \(k×n\)
  5. This expression is a product of \({\left( X'X \right)}^{-1}\) and \(X'\) . The transpose of \({\left( X'X \right)}^{-1}X'\) will be equal to the transpose of \(X'\) multiplied by the transpose of \({\left( X'X \right)}^{-1}\) , that is, \(X{\left( X'X \right)}^{-1}\) with dimension \(n×k\) .