SVD - can the non-zero-singular-values be all lower than 1?

Let $$X \in \mathbb{R}^{n \times m}$$ be a (not-zero) matrix.

Consider its Singular Value Decomposition $$X = U\Sigma V^T = U[\text{diag}(\sigma_1, \sigma_2, \dots, \sigma_k )]V^T$$

Is it possible to get a $\Sigma$ matrix such that: $$\forall t, \sigma_t < 1$$

$$X = \mathrm{diag}(1/2, 1/3, 1/4, \ldots, 1/n)$$
be an $n\times n$ matrix. Its singular value decomposition has $\Sigma = X$...
Try $X:\mathbb{R} \to \mathbb{R}$, $X(h) = \frac{1}{2}h$.