Let $A$ be an $n\times n$ matrix with entries in the set $\{0,1\}$ which has exactly two ones in each column and two ones in each row. Give necessary and sufficient conditions for the rank of $A$ to be $n$.
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What have you tried? Did you try the simple case, n=2? What conditions can you guess for n=3? – Isomorphism Jun 27 '13 at 20:54
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really i do not undertand this problem when n =2 only i have one l.i vector so the rank is 1, so i´m very confused. – Rosa Maria Gtz. Jun 27 '13 at 21:14
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i know too that an n x n matrix A has rank n if and only ifA has an inverse. – Rosa Maria Gtz. Jun 27 '13 at 21:18
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Let $G$ be an undirected bipartite graph with $2n$ nodes $p_1,\ldots,p_n,q_1,\ldots,q_n$, where $p_i$ is connected to $q_j$ if and only if $a_{ij}=1$. Since $A$ has exactly two $1$s on each column and each row, each node $p_i$ is connected to exactly two $q_j$s and vice versa. Therefore $A$ is a bipartite graph and it can be decomposed into disjoint union of cycles of the form $$ p_{i_1}\to q_{j_1}\to p_{i_2}\to q_{j_2}\to \ldots p_{i_k}\to q_{j_k}\to p_{i_1}, $$ where $i_1,\ldots,i_k$ are distinct and $j_1,\ldots,j_k$ are distinct. In other words, by relabeling the rows and columns, $A$ is permutationally equivalent to a block-diagonal matrix whose diagonal blocks are of the form $$ B_k = \pmatrix{1&&&1\\ 1&\ddots\\ &\ddots&\ddots\\ &&1&1}\tag{1} $$ where $B_k$ is $k\times k$ with $k>1$. It is easy to see that $$ \det(B_k)=\begin{cases} 0 &\text{ when } k \text{ is even},\\ 2 &\text{ when } k\neq1 \text{ is odd } \end{cases}. $$ Therefore $A$ has full rank if and only if $A=P(B_{k_1}\oplus\ldots\oplus B_{k_m})Q$, where each block $B_{k_r}$ is of the form $(1)$ with an odd $k_r$ and $P,Q$ are some permutation matrices. Put it another way, $A$ has full rank iff its corresponding graph $G$ can be decomposed into a disjoint union of cycles of length $2$ modulo $4$.
user1551
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The problem that i have is that i know when a matrix has rank n , and the basic definition is equivalent to the linear independence of the columns, this problem depend in how i can choose the order of 1`s,i.e ,I have 2n−1 choices for the first column (it has to be nonzero). Then 2n−2 choices for the second columns (it has to be linearly independent from the first one). Etc. I konow how i can write a prove of this... – Rosa Maria Gtz. Jun 27 '13 at 21:42
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First prove that by a sequence of permutations of rows and columns, you can make sure that all the diagonal entries of the matrix are $1$.
Thus $A=P_1 (I+P_2)P_3$ where $P_1,P_2,P_3$ are permutation matrices and and $P_2$ is a permutation matrix which corresponds to a permutation which fixes no element. $A$ has rank $n \Leftrightarrow I+P_2$ has rank $n$.
I don't think this completely answers the question. But trying to find permutation matrices which do not have $-1$ as an eigenvalue and do not fix any coordinate may be an interesting exercise.
nsoum
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