Number of eigenvalues of a matrix with all rows multiples of the first row What theorem tells us that for a matrix $A$ (with positive entries - is that assumption necessary?), whose rows are multiples of the first row, there is exactly one non-zero eigenvalue and the other eigenvalues are zero?
I guess there's a relation: rank of matrix = number of non-zero eigenvalues.
 A: In general what you can say is that the rank of an $n\times n$ matrix is at most equal to the number of nonzero eigenvalues. The simplest counterexample is the matrix
$$
\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}
$$
that has only the zero eigenvalue, but has rank $1$.
You surely know that, for an eigenvalue $\lambda$ of the matrix $A$, the following inequalities hold:
$$
1\le d\le m
$$
where $d$ is the dimension of the eigenspace $E_A(\lambda)=\{v\in\mathbb{C}^n:Av=\lambda v\}$, usually called the geometric multiplicity of $\lambda$, and $m$ is the algebraic multiplicity, that is, the maximum exponent $m$ such that $(\lambda-X)^m$ divides the characteristic polynomial $\det(A-XI_n)$ ($I_n$ is the $n\times n$ identity matrix).
In the particular case of $\lambda=0$, the eigenspace $E_A(0)$ is the null space of $A$, so this dimension is $n-k$, where $k$ is the rank of $A$. The number of nonzero eigenvalues (counted with their algebraic multiplicity) is $n-m$, where $m$ is the algebraic multiplicity of the zero eigenvalue. Since $n-k\le n-m$ by the above inequality, we get $m\le k$.
In the special case where $A$ has rank $1$ and every row is a nonzero multiple of the first row, there is at least a nonzero eigenvalue: indeed, $A=uu^H$ for some vector $u\ne0$ ($H$ denotes the hermitian transpose), so
$$
Au=uu^Hu=(u^Hu)u
$$
and the scalar $\mu=u^Hu$ is a nonzero eigenvalue. Thus, by the above considerations, the geometric multiplicity of the zero vector must be the same as the algebraic multiplicity, both equal to $n-1$.
The characteristic polynomial of a rank $1$ matrix is thus $(0-X)^{n-1}(\mu-X)$.

If you are dealing only with real matrices, just change references to $\mathbb{C}$ with $\mathbb{R}$ and the hermitian transpose with the transpose.
A: Your Guess is correct for such matrices. As 
Characteristic polynomial of a matrix $A$ is given by:
$x^n-A_1x^{(n-1)}+A_2x^{(n-2)}-....+(-1)^nA_n=0$
where $A_i$ denotes the sum of all principal minors of $A$ of order $i$.
Notice that for $A$ as defined in problem $A_2, A_3,...A_n$ are all zero. Thus only non zero eigenvalue is Trace$A$. Further $A$ is assumed to be positive just to insure trace $A$ to be non zero.
If negative entries are allowed, Trace$A$ may turn out zero.
A: As you seem to have deduced correctly, the fact that all rows are multiples of the first row means the rank of you matrix is$~1$ (or $0$, but the zero matrix is excluded by the other condition). By rank-nullity this means the geometric multiplicity of the eigenvalue $0$ (which is just the dimension of the kernel) equals $n-1$ where $n$ -s the size of your matrix. Since the trace $t$ of you matrix gives the sum of all eigenvalues (taken with their algebraic multiplicities), the remaining eigenvalue is $t$, and the characteristic polynomial is $X^{n-1}(X-t)$. The fact that all entries are positive ensures $t>0$ in this case, so there is indeed exactly one nonzero eigenvalue, namely $t$.
Note that the question in your first paragraph has been carefully stated: it does not say exactly two eigenvalues, which would fail in the $1\times1$ case, where $t$ is the only eigenvalue. In your second paragraph you are less careful: the rank gives an upper bound for the number of nonzero eigenvalues, but in general there could be less because (1) zero could have greater algebraic multiplicity than geometric multiplicity (which is what the rank gives), and any access will "eat into" the remaining eigenvalues, and (2) after compensating for that, one gets the number of nonzero complex eigenvalues counted with their algebraic multiplicities (which you did not mention). In htis question the point (1) manes one really needs the hypothesis about positive entries; rank $1$ matrices with zero trace do exist, and they are nilpotent (have only $0$ as eigenvalue).
