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I am wondering if there is a way to formulate or generate a matrix $X \in R^{n\times n}$ whose column vectors $\{x_1,\dotsc,x_i,\dotsc,x_n\}$ are such that $x_i$ and $x_j$ are orthogonal iff $\lvert i-j\rvert>m$. The matrix $X^T X$ would be a band matrix with zeros off-band. I do not have a particular requirement for $x_i \cdot x_j$ when $\lvert i-j\rvert\le m$ other than being non-zero. I do require, however, that all vectors are distinct.

One example of $X$ would be a band matrix with bandwidth $2m+1$ and zero off-band. However, I am looking for a more general formula. I would like to somehow link it to SVD, but I am not sure how to. Any ideas would be appreciated!

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  • $\begingroup$ Spelling note: the usual abbreviation for 'if and only if' is 'iff', not 'iif'. Also, when you say that you require the columns to be unique, you probably mean that you require them to be distinct. I have edited accordingly. Note that distinct-ness is automatic: if $i \ne j$, then we can choose $k$ such that $\lvert i -k\rvert$ and $\lvert j - k\rvert$ are on opposite sides of $m$, and then one of $x_i\cdot x_k$ and $x_j\cdot x_k$ is $0$ and the other is not. $\endgroup$
    – LSpice
    Commented Feb 26, 2022 at 0:18
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    $\begingroup$ @LSpice Thanks for the corrections, and also for pointing out that the distinctiveness is automatic. $\endgroup$
    – CWC
    Commented Feb 26, 2022 at 1:12

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It seems that the band matrix example is an essentially sufficient characterization, up to a choice of basis.

The orthogonality constraint forces zeroes in the QR decomposition $X=QR$: $R$ is an upper triangular $m$-band matrix (and $Q$ is our usual orthogonal matrix.)

Conversely, any such pair $Q$, $R$ generates an $X$ where any two column vectors further than $m$ apart are constructed from different column vectors in $Q$ and are thus orthogonal.

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