TITLE: Probe permutationally matrix extreme properties
QUESTION [3 upvotes]: Suppose given $M\in\{0,1\}^{n\times n}$ of rank $r$.
Assume that changing even a single $1$ to $0$ in $M$ raises rank. Does it follow that $M$ is permutationally equivalent to a block diagonal matrix with each block of rank $1$?
Assume that changing even a single $1$ to $0$ in $M$ lowers rank. Does it follow that $M$ is permutationally equivalent to a diagonal matrix?
posted: https://math.stackexchange.com/questions/1171495/probe-permutationally-matrix-extreme-properties

REPLY [3 votes]: 1) No. Idea is to take all columns satisfying certain linear relation (i.e. lying in a certain subspace.) Then changing 1 to 0 violates this relation, and there are no immediate reasons that new relation appears. Namely, we may take, say, $n=k^2$, $k\geq 2$, and columns $e_i+e_j$ where $1\leq i\leq k$, $k+1\leq j \leq 2k$. They all satisfy a relation $x_1+\dots+x_k=x_{k+1}+\dots+x_{2k}$, and any $k^2-1$ of them generate the space generated by this relation. Thus rank equals $2k-1$, and when we change 1 to 0 columns form a space of dimension $2k$.
2) Yes for matrices over the field of characterstic 0. Consider non-zero $r\times r$ minor. Clearly there are no 1's outside it. Thus all non-zero columns are linearly independent and we have to prove that each of them is a basic vector. Assume that there are $m$ 1's in some column  (say, column $c$) and $m>1$. Let $c=\sum_{i\in I} e_i, |I|=m$. Replacing each of those 1's to 0 lowers rank, hence column $c-e_i$ is a linear combination of other columns of our matrix for any $i\in I$. Then $c=\frac1{m-1}\sum_{i\in I} (c-e_i)$ is also a linear combination of other columns. A contradiction.
On the other hand, if characteristic of the field equals $p>0$, we may consider invertible $(p+2)\times (p+2)$ matrix with 0-s on diagonal and 1-s outside diagonal. If you replace any 1 to 0 it becomes singular.


TITLE: Normal Distribution Third Step
QUESTION [0 upvotes]: I am getting stuck at the 3rd stage in a process, of calculating based on Normal Distribution.
The following equations are used:

$N(μ,σ)$
$f(x) = P(x \leq 3)$
$z = (x - μ) /σ$
$N(8, 3)$

So, I have calculated as follows:
$\mu= 8, σ = 3, z = (3 - 8 / 3)\simeq -1.67$
But, then I get completely stuck. As next answer, must be: $0.0475$ and $1 - z$ does not give this value. How, can I arrive at that value?
EDIT: $z-1.67 = 0.0475$, BUT how??
EDIT 2:

Where does '0.9525'? I do find the '0.0475' in the Normal Probability Table, looking up z-score, "Areas Under Normal Curve"

REPLY [1 votes]: This notation means that $\sigma^2 = 3$, thus $\sigma = \sqrt{3}$.
$$P(X \leq 3)= P\left(\frac{X-8}{\sqrt{3}}\leq \frac{-5}{\sqrt{3}}\right)= P(Z \leq \frac{-5}{\sqrt{3}})= P(Z \leq -2.88675...)$$
where $Z \sim N(0,1)$.