If $\mathbf{ABD=0}$, $\mathbf{A}$ and $\mathbf{B}$ are known and $\mathbf{A}$ is $M \times N$, $\mathbf{B}$ is $N \times R$, $\mathbf{B^HB=I}$ and $\mathbf{D}$ is $R \times R$. Does $\mathbf{D}$ exist such that $\mathbf{D^HD=I}$ where $\mathbf{H}$ is the conjugate transpose. $\mathbf{BD}$ is generally known, it can be computed as the null matrix for $\mathbf{A}$, but I'm more interested in the solution the gives $\mathbf{D^HD=I, D\neq I}$.
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If such a $D$ exists in $M_R(\mathbb{C})$, then $D$ is invertible by $D^*D=I_R$. So $ABD=0$ implies $AB=0$.
Now if $AB=0$, then any $D$ such that $D^*D=I_R$ works. That is the set of unitary matrices in $M_R(\mathbb{C})$.
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$\begingroup$ Thank you for the reply, I should've mentined $D \neq I$. I updated the question $\endgroup$ – SoCal93 Mar 25 '13 at 21:02
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1$\begingroup$ @SoCal93 Id does not change much. Your solution set is the set of unitary matrices $D$. $\endgroup$ – Julien Mar 25 '13 at 21:05
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$\begingroup$ Thank you, so it exists only if $AB=0$. In my case $AB \neq 0$, so I can't find a matrix with $D^*D=1$ $\endgroup$ – SoCal93 Mar 25 '13 at 21:21
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$\begingroup$ @SoCal93 Yes, that's what I've observed: such a $D$ exists if and only if $AB=0$, in which case the solution set is the set of all unitary matrices. $\endgroup$ – Julien Mar 25 '13 at 21:23
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$\begingroup$ I dont think it's entirely true that if $D^HD=1$, then $D$ is invertible. there are rectangular (non-invertible) matrices that satisfy $D^HD=I$ $\endgroup$ – SoCal93 Mar 25 '13 at 22:27
