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Assume that ${\mathbf H}$ is a $N \times M$ matrix. The following parameter is called orthogonality deficiency and describes how much orthogonal the columns of ${\mathbf H}$ are. $$ od({\mathbf H}) = 1 - \frac{\det({{\mathbf H}^H{\mathbf H})}}{\Pi_{n=1}^M\|{{\mathbf h}_n}\|^2}$$ where ${\mathbf h}_n$ is the $n$th column of matrix ${\mathbf H}$. It can be seen that $0\leq od({\mathbf H})\leq 1$. If ${\mathbf H}$ is singular then $od({\mathbf H})=1$ and when $od({\mathbf H})=0$ the colums of mathrix ${\mathbf H}$ are orthogonal. This is, indeed a very useful criterion for matrix orthogonality and has many applications in communications engineering and signal processing.

My question:

When the matrix ${\mathbf H}$ has the form ${\mathbf H}={\mathbf A}+e{\mathbf B}$, where $e$ is a scalar, how can we approximate $od({\mathbf H})$ when $e \ll 1$ and what is the approximation of $od({\mathbf H})$ when $e \ll 1$ (Probably using Taylor expansion)?

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I don't have an answer, but I am curious about this concept of "orthogonality deficiency" and how people use it. Do you have some references on it? It looks odd since it's basis dependent, my first guess would have been using something like the condition number instead. –  Federico Poloni Aug 17 '13 at 12:30
    
Thank you for your interest. The channel between the receiver and the transmitter is modeled by a matrix, say ${\mathbf H}$, in multiple antenna systems. Here, I am trying to model the estimation error at the receiver. –  Mamal Aug 18 '13 at 3:57
    
Generally, orthogonality of the channel matrix is very important and orthogonality means kind of independece between antennas in multiple antenna systems. The parameter $Od$ is a very powerful criterion and is very useful orthogonality assessment. For instance check: "ieeexplore.ieee.org/xpl/…; –  Mamal Aug 18 '13 at 4:00
    
Thanks for the pointer! –  Federico Poloni Aug 18 '13 at 9:33

1 Answer 1

up vote 2 down vote accepted

If $e$ is a small number, then $od(H)\approx od(A)+e(od)'_A(B)$. $od(A)=1-\dfrac{u(A)}{v(A)}$ where $u(A)=\det(A^*A),v(A)=\Pi_i||Ae_i||^2$ and $(e_i)_i$ is the canonical basis. $(od)'_A=-\dfrac{1}{v(A)}u'_A+\dfrac{u(A)}{v^2(A)}v'_A$. $u'_A(K)=trace((A^*K+K^*A)adjoint(A^*A))$. $v'_A(K)=\sum_i((e_i^*K^*Ae_i+e_i^*A^*Ke_i)\Pi_{j\not= i}||Ae_j||^2)$.

Edit: 1. The proof is based on this fact: if $\phi:A\rightarrow \det(A)$, then $\phi'_A:K\rightarrow trace(K.adjoint(A))$. 2. Simplifications: $u(A)=|\det(A)|^2$ and $e_i^*K^*Ae_i+e_i^*A^*Ke_i=2Real((K^*A)_{i,i})$. 3. In the sake of simplicity, assume that $K$ is real. Let $(E_{i,j})$ be the canonical basis of $\mathcal{M}_n(\mathbb{R})$. Then the matrix of $(od)'_A$ is in the form $U=[u_1,\cdots,u_{n^2}]$. If you want a large variation of $od(A)$, then choose $B=U^T$ (in the orthogonal of $\ker(od'_A)$). Then $(od)'_A(B)=||U||^2$. For instance, if $A=\begin{pmatrix}8&-5-6I&-3-I\\-5+4I&2+3I&7+7I\\6-I&-4+6I&8+2I\end{pmatrix}$, then $U\approx [-0.0029,-0.0154,-0.0120,-0.0010,-0.0064,-0.0119,-0.0110,-0.0111,0.0124]$. $od(A)\approx 0.9489$ and $od(A-100.U^T)\approx 0.8147$.

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Thank you so much... It was a good idea. Could you please explain a little bit more about your answer? Or can it be a little be more simplified? –  Mamal Aug 25 '13 at 2:38

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