What is the inverse of a square matrix with orthogonal columns? I know that for an orthogonal matrix, i.e. a square matrix with orthonormal columns, the inverse is the transpose. What about the general case when the columns of a square matrix are non-zero and orthogonal, but not necessarily of length $1$? (The matrix is invertible since non-zero orthogonal vectors are linearly independent.) Is there a simple characterization/formula for the inverse, or a simple way to compute it?
What if we additionally require that the columns are orthogonal, and the rows are orthogonal? (I know there exist invertible matrices with orthogonal columns, but not orthogonal rows.)
 A: A matrix with orthogonal columns (which is easily seen to be invertible) may be written as $QD$ where $Q$ is orthogonal and $D$ is invertible diagonal. Its inverse is then $D^{-1}Q^T$.
A: If the columns of $A$ are orthogonal then $$A^TA=D^2$$ where the diagonal elements of $D$ are the lengths of the columns. Thus $$A^{-1}=D^{-2}A^T=(AD^{-2})^T$$ The recipe is therefore to divide the column vectors of $A$ by their lengths squared and then take the transpose.
A: I am answering my own question, to build on the other answers and comments, in case anyone needs more details.
In general, if $A$ is a square matrix, and $D$ a diagonal matrix with diagonal entries $d_1, d_2, ..., d_n$, then it is readily seen that the $i$th column of $AD$ equals $d_i$ times the $i$th column of $A$.
In our case, if $A$ has orthogonal columns, let $Q$ be the orthonormal matrix that is created from $A$ by dividing each $i$th column of $A$ by its length $a_i := \lVert A_{\text{col $i$}} \rVert$. Put these $a_i$ as the entries of a diagonal matrix $D$. Then $A = QD$, and so $A^{-1} = D^{-1}Q^{-1} = D^{-1}Q^T$.
Here $D$ is indeed invertible because the norms $a_i$ of the columns of $A$ are nonzero. The inverse of a diagonal matrix is, of course, the diagonal matrix formed by taking the reciprocals.
