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I have to find an example of a matrix $Q$ that has orthonormal columns, but $QQ^T \neq I$.

If a matrix has orthonormal columns, it does not imply that the matrix is orthogonal, so that it is a square matrix. Therefore, I could simply give a matrix with unit vectors that is not a square matrix, since the $QQ^T$ would not be a identity, right?

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  • $\begingroup$ A matrix of the form $AA^T$ is always square. $\endgroup$ Apr 13 '15 at 22:16
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Your idea is right, although you would still need to have some verification that $QQ^T$ is not, in fact, the identity. It's certainly conceivable (without some theorem to back you up…) that you could get unlucky and choose a non-square $Q$ for which $QQ^T$ was still the identity matrix.

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Yes, any $Q$ which is a thin matrix will work, i.e., $Q \in \mathbb{R}^{m \times n}$, with $m > n$. This is because $\text{rank}(QQ^T) = n$, where $QQ^T \in \mathbb{R}^{m \times m}$, whereas $I \in \mathbb{R}^{m \times m}$ has rank $m$.

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