I have a batch of vectors $X$ that have row vectors of size $n$, and batch size of $k$, so $$\begin{bmatrix} v_{11} & ... & v_{1n} \\ v_{21} & ... & v_{2n} \\ &\;\;\vdots \notag \\ v_{k1} & v_{k2} & v_{kn} \end{bmatrix}$$

The $k$ is always bigger than $n$ ($k>n$), so the matrix is non-square. What I would like is to calculate in a vectorized manner, if all of the vectors are orthogonal. I tried fallowing $$XX^T$$ and check if it equals $I$ (identity matrix). But I have doubts, if it works for non-square matrix.


1 Answer 1


It works. Let $i$-th row be $x_i^T$: $$\begin{align}XX^T&=\begin{bmatrix}x_1^T\\x_2^T\\\vdots \\x_k^T\end{bmatrix}\begin{bmatrix}x_1 & x_2 & \dots & x_k\end{bmatrix}\\&=\begin{bmatrix}x_1^Tx_1&x_1^Tx_2&\dots&x_1^Tx_k\\x_2^Tx_1&x_2^Tx_2&\dots&x_2^Tx_k\\\vdots&\ddots&&\vdots\\x_k^Tx_1&x_k^Tx_2&\dots&x_k^Tx_k\end{bmatrix}_{k\times k}\end{align}$$ And, the off-diagonal entries will be $0$ if all vectors are orthogonal. Note that, for this being equal to $I$, they should be orthonormal.

However, since $k>n$, you won't be able to get $k$ orthonormal or orthogonal vectors of size $n$, because the dimensionality of the space spanned by these vectors is $n$ at maximum.

  • $\begingroup$ Thanks for your answer @gunes. So in general I'd search for such vectors, that are: $XX^T = \lambda I$ ? Or the normalization constant can be different for every vector and would be a matrix, e.g. $XX^T = N^{-1/2}IN^{-1/2}$ (assuming $N$ is a normalization matrix) ? $\endgroup$ Jun 17, 2021 at 5:42
  • $\begingroup$ Or to check if they're orthogonal $||XX^T - diag(XX^T)||^2_2 = 0$ would be enough ? $\endgroup$ Jun 17, 2021 at 5:54
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    $\begingroup$ checking off diagonals as in your second message is enough. But you wont have orthogonals for $k>n$ because of the dimensionality constraint. $\endgroup$
    – gunes
    Jun 17, 2021 at 6:59
  • $\begingroup$ Thanks for answer. In general those vectors come from the same transformation, i.e. $T(y_k) = x_k$, but I want to make it in a batch manner to avoid for loop. So I think $n$ is enough. $\endgroup$ Jun 17, 2021 at 7:07

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