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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: $$\forall t, \sigma_t < 1$$

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Why not? Let

$$ X = \mathrm{diag}(1/2, 1/3, 1/4, \ldots, 1/n) $$

be an $n\times n$ matrix. Its singular value decomposition has $\Sigma = X$...

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  • $\begingroup$ OK!Thank you for your answer, it made me realize how my question was stupid :) $\endgroup$ – Aslan986 Jan 8 '13 at 22:56
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Try $X:\mathbb{R} \to \mathbb{R}$, $X(h) = \frac{1}{2}h$.

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