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Suppose I have a family of $n$ linearly-independent elements $v_i$ of the Hilbert space $\mathbb{C}^m$, which are not necessarily orthogonal. Can I always find a partial isometry $f: \mathbb{C} ^m \to \mathbb{C} ^n$ such that the vectors $f(v_i)$ are orthogonal and nonzero?

If not, under what conditions on the family $v_i \in \mathbb{C}^m$ is such a map available?

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up vote 5 down vote accepted

The answer is no in general: if $n=m$ a necessary and sufficient condition is that the $v_i$'s are already orthogonal.

The answer is yes if $n \leq m/2$ (proof: if $m \geq 2n$ without loss of generatlity we can write $\mathbb C^m \simeq \mathbb C^n \oplus \mathbb C^n \oplus \mathbb C^{m-2n}$ and assume that the first copy of $\mathbb C^n$ is the linear span of the $v_i$'s. Take $g : \mathbb C^n \to \mathbb C^n$ any norm $1$ map that sends $v_i$ to an orthogonal family, and consider the partial isometry $f$ defined by its block-matrix decomposition $f=(g\,\, \sqrt{1-g g^*}\,\, 0):\mathbb C^m \to \mathbb C^n$.) A slight modification of the proof gives the same conclusion if $m=2n-1$, since $\sqrt{1-gg^*}$ does not have full rank if $g$ is norm $1$. For $n>(m+1)/2$ I bet that the answer is no in general but I do not see any simple iff condition (the iff condition that I see is that there exists a norm $1$ $g:\mathbb C^n \to \mathbb C^n$ that sends the family $v_i$ to an orthogonal family, and such that $g$ has at least $2n-m$ singular values equal to $1$, but this is not simple).

As a consequence, the answer to your question becomes yes in general if you allow to enlarge the space $\mathbb C^m$.

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