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Consider a Matrix $A \in \mathbb C^{m \times n}$, $m<n$ which is build by vectors like $$ A = \begin{pmatrix} | & | & & | \\ \vec a_1 & \vec a_2 & \cdots & \vec a_n \\ | & | & & | \end{pmatrix} $$ where every vector is written like $$ \vec a_i = \begin{pmatrix} a_{1i} \\ a_{2i} \\ \vdots \\ a_{mi} \end{pmatrix} $$ If I calculate the matrix product $A^\dagger A$ I find $$ A^\dagger A = \begin{pmatrix} \langle \vec a_1 , \vec a_1 \rangle & \langle \vec a_1 , \vec a_2 \rangle & \cdots & \langle \vec a_1 , \vec a_n \rangle \\ \langle \vec a_2 , \vec a_1 \rangle & \langle \vec a_2 , \vec a_2 \rangle & \cdots & \langle \vec a_2 , \vec a_n \rangle \\ \vdots & \vdots & \ddots & \vdots\\ \langle \vec a_n , \vec a_1 \rangle & \langle \vec a_n , \vec a_2 \rangle & \cdots & \langle \vec a_n , \vec a_n \rangle \end{pmatrix}\ $$ where I use the standard scalar product $$ \langle \vec a_i , \vec a_j \rangle = \sum_{k=1}^m a_{ki}^* a_{kj} $$

My question is: Is it possible to require $A^\dagger A \overset{!}{=} \mathbb 1$ ? This would imply that all vectors $\vec a_1,...,\vec a_n$ are mutually orthogonal. However, I am not sure if this is possible since $n>m$. Or is it possible because we deal with complex vectors?

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  • $\begingroup$ Just to clarify: when you say you want $A^\dagger A \overset{!}{=} \mathbb 1$ do you want this to equal the identity matrix, or the matrix of all 1s? $\endgroup$ – Aweygan Jul 21 '16 at 13:23
  • $\begingroup$ What does the ! represent? $\endgroup$ – Alex Provost Jul 21 '16 at 13:48
  • $\begingroup$ I want it to be the identity matrix. And the ! just means that I require that. $\endgroup$ – thyme Jul 21 '16 at 14:29
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This is not possible because you can't have a collection of $n$ orthonormal vectors in $\mathbb{C}^m$ if $n > m$. Alternatively,

$$ \operatorname{rank}(A^{\dagger} A) \leq \min ( \operatorname{rank}(A^{\dagger}), \operatorname{rank}(A) ) \leq \operatorname{rank}(A) \leq m $$

since $A$ is an $m \times n$ matrix with $m < n$ and in particular, we can't have $A^{\dagger} A = I_n$ as this would imply that $\operatorname{rank}(A^{\dagger} A) = n$.

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