Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a full rank matrix $A \in \mathbb{R}^{M \times N^2}$ where the rank of ${A}$ is ${\rm rk}(A)= M \leq N^2$ and the commutation matrix $K_{NN}$. I need to find the rank of a matrix product $rk(A-AK_{NN}) = rk(A(I-K_{NN}))$. My conjecture (I am actually pretty sure) is that the rank should be $min(M, \frac{1}{2}N(N-1))$ where $\frac{1}{2}N(N-1)$ is the rank of $(I-K_{NN})$ known from the literature. Does anybody know how this can be done or at least give me a possible routine to follow?

The elements of A are generated from Gaussian distributions.

share|improve this question
add comment

1 Answer

Your conjecture is wrong. Here is a counterexample: $A=\pmatrix{1&1&1&0\\ 0&1&1&1}$. We have $A(I-K_{2,2})=0$.

When $A$ is a random matrix whose entries are generated from i.i.d. Gaussian distributions, since $A(I-K_{N,N})\operatorname{vec}(X)=0$ for all symmetric matrix $X$, the rank of $L=A(I-K_{N,N})$ is the rank of $L$ restricted on the $\frac12N(N-1)$-dimensional subspace $\{\operatorname{vec}(X): X=-X^T\}$. Hence $\operatorname{rank}(L)=\min\left\{M,\ \frac12N(N-1)\right\}$ almost everywhere (but not everywhere, as the above counterexample illustrates).

share|improve this answer
    
Thank you for your important comments. I think I miss one important assumption here. What if the elements of A are generated from Gaussian distributions. –  user111614 Nov 28 '13 at 9:48
    
I am sorry. But can you please explain in more details regarding " the rank of L=A(I−KN,N) is the rank of L restricted on the 12N(N−1)-dimensional subspace {vec(X):X=−XT}. Hence rank(L)=min{M, 12N(N−1)}"? I did not get the causality inside this sentence. Thank you again for your efforts. –  user111614 Nov 29 '13 at 16:05
    
For all symmetric $X$, $\mathrm{vec}(X)$ lies inside $\ker A(I-K)$. So, with an appropriate change-of-basis matrix $T$, we may assume that $I-K=T[C,0]T^{-1}$ where $C$ has rank $N(N-1)/2$ and the trailing zero block has $N(N+1)/2$ columns. Hence $$\mathrm{rank}(A(I-K))=\mathrm{rank}(AT[C,0]T^{-1})=\mathrm{rank}(AT[C,0])= \mathrm{rank}(ATC).$$ As $TC$ has full rank, $ATC$ has full rank almost everywhere. but $ATC$ is $M\times\frac12N(N-1)$. So, when it has full rank, its rank is $\min\{M,\,\frac12N(N-1)\}$. –  user1551 Nov 30 '13 at 0:20
    
Is the key point to transform $I-K$ into a full rank matrix? If yes, can I simply use the economic sized EVD? E.g., assume that $(I-K)(I-K)^{\rm T} = [Us, Un][Cs 0; 0,0][Us, Un]^{\rm T}$ and I have Us is $N^2 \times \frac{1}{2}N(N-1)$. Then $rank(A(I-K)) = rank(A(I-K)(A(I-K))^{\rm T}) = rank(AUsCs^{0.5}Cs^{0.5}Us^{\rm T}A^{\rm T} =rank(AUsCs^{0.5})$ where $UsCs^{0.5}$ has a full rank. Another question is which reference can I refer to for the conclusion $rank(AC) = rank(A)rank(C)$ given $A$ is random and $C$ is fixed but both of them is full rank. –  user111614 Dec 2 '13 at 16:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.