# the rank of an interesting matrix

Let $A$ be a square matrix whose off-diagonal entries $a_{i,j} \in (0,1)$ when $i \neq j$. The diagonal entries of $A$ are all 1s. I am wondering whether $A$ has a full rank.

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Note that all the diagonal entries of A are ones, but the off-diagonal entries are strictly less than 1 (nonnegative) – John Smith Jul 31 '12 at 9:08
Probability matrix or stochastic matrix should have row summed up to 1. In your case, when diagonal entries are 1, the off-diagonal entries should be zero, as it can be negative. Hence A will have full rank. en.wikipedia.org/wiki/Stochastic_matrix – Learner Jul 31 '12 at 9:13
Sorry, A is actually not a probability matrix, but my desciption of A is correct. So for my defintion of A, is it a full rank matrix ? – John Smith Jul 31 '12 at 9:15

If $$A=\begin{pmatrix} 1 & \frac34 & \frac12 &\frac34 \\ \frac34 & 1 & \frac34 & \frac12 \\ \frac12 &\frac34 & 1 & \frac34 & \\ \frac34 & \frac12 & \frac34 & 1 \end{pmatrix}$$ then $(1,-1,1,-1)A=0$, so the matrix is singular.

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A bit of context might help: note that $A=\frac12I+\frac32P+P^2$ where $P$ is the transition matrix of the simple random walk on the graph $\mathbb Z/4\mathbb Z$. Since this graph is bipartite for $P$, with $V_0=\{0,2\}$ and $V_1=\{1,3\}$, one knows that $Pv=-v$ where $v_i=+1$ if $i\in V_0$ and $v_i=-1$ if $i\in V_1$. Hence $P^2v=v$ and $Av=\frac12v+\frac32(-v)+v=0$. This suggests some generalizations to other matrices and to other dimensions. In particular, one can replace every $\frac12$ in $A$ by $a$ and every $\frac34$ by $\frac12(1+a)$, for any $0\lt a\lt 1$. – Did Jul 31 '12 at 11:07
And @Martin, I forgot: +1. :-) – Did Jul 31 '12 at 11:31
Amazing! How can Martin contruct such a counterexample? Is there any intuition behind this? – John Smith Jul 31 '12 at 14:06
@JohnSmith: Hmm, well, how to say this... John: did you read my previous comment? – Did Aug 3 '12 at 23:51

I think it always should. If you think of the colums of being vectors in space, then no vector can be written as a combination of the other vectors without going beyond $(0,1)$ for the off diagonal elements since there is always at least one direction different (because of the diagonal element being one) this is not realy a mathematical proof, rather it is my imagination.

ps. how can i comment, instead of giving an answer? because that is what i intended to do

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You need to have reputation>=50 to be able to comment (with the exception of adding a comment under your own post - you don't need reputation points for that), see here. – Martin Sleziak Jul 31 '12 at 9:29
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