Question

Let \(A\) be a square matrix of real numbers whose columns are (non-zero) orthogonal vectors.
a) Show that \(A^{T} A\) is a diagonal matrix - whose inverse is thus obvious to compute.
b) Use this observation (or any other method) to discover a simple general formula for the inverse, \(A^{-1}\) involving only its transpose, \(A^{T}\), and \(\left(A^{T} A\right)^{-1}\). In the special case where the columns of \(A\) are orthonormal, your formula should reduce to \(A^{-1}=A^{T}\).

Answer 1

a) If the columns of \(A\) are orthogonal, then \(A^T A\) is symmetric and has diagonal elements equal to the squares of the norms of the columns of \(A\). Thus, \(A^T A\) is a diagonal matrix.

b) We have:
\[A^{-1}=\left(A^T A\right)^{-1} A^T\]
In the special case where the columns of \(A\) are orthonormal, the diagonal elements of \(A^T A\) are equal to 1 , so its inverse is simply the identity matrix, and thus:
\[A^{-1}=I A^T = A^T\]