Given a set of vectors given the columns of \(A\), we may often want to find an orthonormal basis that spans the same space. Note this assumes that the vectors are linearly independent, ie. \(A\) is either square or tall and \(A\) has full column rank. If this is not true, we can proceed with the the following construction using a linearly independent spanning subset of the vectors.
Given that \(A\) is full column rank we have that \(A^TA\) is invertible and along with it it's positive definite square root \((A^TA)^{\tfrac{1}{2}}\) (For more details on these matrices see the discussion of shape matrices.) Given this we can define a set of vectors, the columns of \(U\) that form an orthonormal basis for the range of \(A\). $$ U = A(A^TA)^{-\tfrac{1}{2}} $$ Perhaps not obviously, the columns of this matrix are orthonormal. We can check this by taking all pairwise inner producst of the columns: $$ U^TU = \big((A^TA)^{-\tfrac{1}{2}}A^T\big)\big(A(A^TA)^{-\tfrac{1}{2}}\big) $$ $$ = (A^TA)^{-\tfrac{1}{2}}A^TA(A^TA)^{-\tfrac{1}{2}} $$ $$ = (A^TA)^{-\tfrac{1}{2}}(A^TA)^{\tfrac{1}{2}} $$ $$ (A^TA)^{-\tfrac{1}{2}}(A^TA)^{-\tfrac{1}{2}} =I $$ Note: we also know that \(U\) spans the same space as \(A\) since if \(y=Ax\) then there exists (uniquely) \(x'\) such that \(y = Ux' = A(A^TA)^{-\tfrac{1}{2}}x'\) Here \(x'\) is simply given by \(x' = (A^TA)^{\tfrac{1}{2}}x \) This construction is illustrated below.
Of course there are many orthonormal bases for the span of \(A\) The orientation of the columns of \(U\) relative to the columns of \(A\) is related to the unique positive definite "shape matrix" \((A^TA)^{\tfrac{1}{2}}\). Specifically, \((A^TA)^{\tfrac{1}{2}}\) is actually the coordinates of the columns of \(A\) relative to the \(U\) basis since clearly $$ A = U(A^TA)^{\tfrac{1}{2}} = A(A^TA)^{-\tfrac{1}{2}}(A^TA)^{\tfrac{1}{2}} $$ If we think of the columns of \(U\) as a new coordinate system than \(A\) will be a positive definite symmetric matrix relative to that coordinate system.
We can extend this construction to give us any possible orthonormal basis by then taking orthonormal linear combinations of the columns of \(U\), ie. right multiplying \(U\) by any rotation matrix \( R \in \mathbb{R}^{m \times m}\) $$ U' = UR $$ We can check again quickly that we still have orthonormality: \(U'^TU' = R^TU^TUR = R^TR = I\).
We note that this construction in some ways can be thought of as the heart of projection matrices and pseudoinverses. Rewrite your system in orthonormal coordinates for the subspace you care about, project onto that subspace, and then perform all your computations on that subspace. In this vein, the projection matrix onto the range of \(A\) is given by $$ UU^T = A(A^TA)^{-\tfrac{1}{2}}(A^TA)^{-\tfrac{1}{2}}A^T = A(A^TA)^{-1}A^T $$