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Given a set of (linearly independent) $d\times d$ complex unitary matrices $\{U_i\}_{i=1}^n \subseteq M_d$ with $n\geq d$, does there exist a vector $v\in \mathbb{C}^d$ such that $$\text{span} \{U_1v, U_2v, \ldots , U_nv\} = \mathbb{C}^d ?$$

The motivation for this question comes from the theory of mixed unitary quantum channels. A quantum channel $\Phi: M_d \rightarrow M_d$ is a completely positive and trace preserving linear map. Any such map admits a Kraus representation of the form $\Phi (X) = \sum_{i=1}^k A_i X A_i^*$, where $\{A_i \}_{i=1}^k \subseteq M_d$ and $\sum_{i=1}^k A_i^* A_i = \mathbb{I}_d$. We say that a quantum channel is mixed unitary if it can be expressed as a convex combination of unitary conjugations: $\Phi (X) = \sum_{i=1}^n p_i U_i X U_i^*$. Our aim then is to look for a rank one input projector $X = vv^*$ for some $v\in \mathbb{C}^d$ such that the output $\Phi (vv^*) = \sum_{i=1}^n p_i (U_i v) (U_i v)^*$ has full rank. This is possible only if $n\geq d$. To avoid trivial counterexamples, we can also assume that $\{ U_i\}_{i=1}^n \subseteq M_d$ is linearly independent.

Follow-up question: Since it has been shown below that the question can be answered in the negative for $d\geq 4$, the natural way of progression would be to ask if one can provide a classification of all the sets of (linearly independent) unitary matrices $\{U_i\}_{i=1}^n \subseteq M_d$ which allow for the existence of $v\in \mathbb{C}^d$ such that $$\text{span}\{U_1v, U_2v, \ldots ,U_nv\}=\mathbb{C}^d.$$ One can also try to answer this question for (linearly independent) sets of arbitrary complex matrices: $\{ A_i \}_{i=1}^k \subseteq M_d$.

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    $\begingroup$ The answer is obviously no, if $U_k = e^{it_k}\,I$ with distinct $t_k$. So, would you like these matrices to be linearly independent, for example? $\endgroup$
    – amsmath
    Commented Mar 1, 2021 at 3:03
  • $\begingroup$ @amsmath Yes I've now added this additional assumption in the question. Thanks for pointing this out! $\endgroup$
    – mathwizard
    Commented Mar 1, 2021 at 3:09
  • $\begingroup$ Proof for $d=2$: Assume the two standard basis vectors do not work as $v$. Then $U_1 = [u,v]$ and $U_2 = [su,tv]$ with $s,t\in\mathbb C$, $|s|=|t|=1$. Now, choose a vector $x$ with non-zero entries. Then$$\alpha U_1x + \beta U_2x = \alpha(x_1u + x_2v) + \beta(x_1 su + x_2 tv) = (\alpha + s\beta)x_1u + (\alpha + t\beta)x_2v.$$ Since $u\perp v$, this implies $\alpha + s\beta = \alpha + t\beta = 0$, which is only possible for $(\alpha,\beta)\neq (0,0)$ if $s=t$. But then $U_2 = tU_1$, contradiction! $\endgroup$
    – amsmath
    Commented Mar 1, 2021 at 3:59
  • $\begingroup$ Does it matter that $M_d$ is a $d^2$-dimensional space, yet you're using as few as $d$ matrices? It seems like there is a high chance of a set of $d$ LI matrices not yielding $d$ LI images for any $v$. $\endgroup$ Commented Mar 10, 2021 at 1:49
  • $\begingroup$ @KevinP.Barry Yes, I think so too. That is why I've included the number $n\geq d$ itself as a parameter when I ask for a classification of sets of matrices $\{A_i\}_{i=1}^n$ which allow the desired vector to exist. If $n$ is close to $d$, then we should expect that our requirement imposes severe constraints on the sets of allowed matrices. If $n> d^2-d$ for instance, the answer by Daniel below shows that the desired vector exists irrespective of what kind of matrices are present in the set! $\endgroup$
    – mathwizard
    Commented Mar 10, 2021 at 2:12

3 Answers 3

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For $d\geq 4$ the answer is no. To see why, let $k=d-2$, so that $k\geq 2$.

By this answer every complex $k\times k$ matrix can be written as a linear combination of four unitary matrices. As a consequence, we see that the unitary group $\mathscr U(k)$ spans $M_k(\mathbb C)$, so it is certainly possible to find a linear independent set formed by $k^2$ unitary $k\times k$ matrices, say $\{U_i\}_{1\leq i\leq k^2}$. Now consider the linearly independent set $$ \{I_2\oplus U_i:1\leq i\leq k^2\}\subseteq \mathscr U(k+2) = \mathscr U(d), \tag 1 $$ where $I_2$ is the $2\times 2$ identity matrix.

Given any $$ x=(x_1, x_2, \ldots , x_{k+2})\in \mathbb C^{k+2} = \mathbb C^d, $$ notice that every vector of the form $$ (I_2\oplus U_i)x $$ is orthogonal to $$ y: = (\overline{x_2}, -\overline{x_1}, 0, 0, \cdots , 0) $$ so the set $\{(I_2\oplus U_i)(x):1\leq i\leq k^2\}$ cannot span $\mathbb C^d$.

As pointed out by @amsmath, if $x_1=x_2=0$, we need to argue instead that the vectors $(I_2\oplus U_i)x$ have vanishing first two coordinates, hence cannot span $\mathbb C^d$ either.

Since $k\geq 2$, we have that $$ k^2 \geq k+2 = d, $$ so the set in (1) is a linearly independent set with (precisely $k^2$, and hence) at least $d$ unitary $d\times d$ matrices, but there is no vector $x$ such that the images of $x$ under our matrices span $\mathbb C^d$.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Commented Mar 2, 2021 at 13:32
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There is a simple necessary and sufficient condition for the absence of such $v$.

Let $A_1,\ldots,A_n$ be linearly independent $d\times d$ matrices. Define the completely positive map $T(X)=\sum_{i=1}^nA_iXA_i^*$.

Now, let $u=\sum_{i=1}^de_i\otimes e_i$, where $e_1,\ldots,e_d$ stands for the canonical basis of $\mathbb{C}^d$.

It is not hard to prove that for every $A,B\in M_{d\times d}$, $$tr(A\otimes B\ uu^*)=tr(AB^t).$$

Thus, we have the following equivalent conditions:

  1. $tr(T(vv^*)ww^*)=0$
  2. $tr((T(vv^*)\otimes \overline{w}w^t) uu^*)=0$
  3. $tr((vv^*\otimes \overline{w}w^t)B)=0$, where $B$ is the non-normalized state $B=T^*(\cdot)\otimes id (uu^*) $.

Since $A_1,\ldots,A_n$ are linearly independent, $\operatorname{rank}{B}=n$.

So the absence of such $v$ is equivalent to the fact that for every $v\in\mathbb{C}^d$ there is a vector $w\in \mathbb{C}^d\setminus\{\vec{0}\}$ such that $v\otimes w\in \ker(B)$.

In particular, it implies that $\operatorname{rank}(B)\leq d^2-d$, since there are at least $d$ linearly independent product vectors in its kernel.

Therefore, if the number of matrices above, $n$, is greater than $d^2-d$ then there exists such $v$.

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  • $\begingroup$ Very interesting! But, if I understand your answer correctly, the condition $n\leq d^2 -d$ is just a necessary one implied by the absence of such a $v$. I mean this is not sufficient right because even if $n\leq d^2 - d$, such a vector $v$ might exist for some special sets of matrices $\{A_i \}_{i=1}^n$. This is where the follow-up question above becomes important: For which such sets of matrices does the desired vector $v$ exist? $\endgroup$
    – mathwizard
    Commented Mar 10, 2021 at 2:06
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Here is an example of a set of $d$ unitary matrices $\{U_i \}_{i=1}^d \subseteq M_d$ for which the desired vector exists.

For each $i\in \{1,2,\ldots ,d \}$, we define $U_i$ to be the unitary shift matrix: $U_i e_k = e_{k+i}$, where the index addition happens modulo $d$ and $\{e_k \}_{k=1}^d$ is the standard basis of $\mathbb{C}^d$. Then, it should be clear that if we choose $v=e_1$ for instance, the set $\{U_i v \}_{i=1}^d$ is precisely the standard basis of $\mathbb{C}^d$ and hence spans $\mathbb{C}^d$.

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