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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 $A^{-1}=A^{T}$.
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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:
$A^{-1}=I A^T = A^T$
by Platinum (101k points)

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