Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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. – 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
up vote 6 down vote accepted

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.

share|cite|improve this answer
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

share|cite|improve this answer
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

Your Answer


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.