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.
the rank of an interesting matrix
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linear-algebra
matrices
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0Sorry, 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 ? – 2012-07-31
2 Answers
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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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0@JohnSmith: Hmm, well, how to say this... John: did you read my previous comment? – 2012-08-03
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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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0You 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](http://math.stackexchange.com/privileges/comment). – 2012-07-31