Eigenvalues of a matrix with only one non-zero row and non-zero column. Here is the full question. 


*

*Only the last row and the last column can contain non-zero entries. 

*The matrix entries can take values only from $\{0,1\}$. It is a kind of binary matrix. 
I am interested in the eigenvalues of this matrix. What can we say about them? In particular, when are all of them positive? 

$$\pmatrix{Q & 0 \\ 0 & 1}\pmatrix{\mathbf{0} & u \\ v^\top & a}\pmatrix{Q^{-1} & 0 \\ 0 & 1}=\pmatrix{\mathbf{0} & Qu \\ v^\top Q^{-1} & a}$$
This similarity operation preserves the value of the product of the vectors:
$$v^\top u \leftarrow v^\top Q^{-1} Q u = v^\top u $$
Say that the $Q$ is Gaussian elimination on the column $u$ so that you are left with a zero column except the one element, giving the $2 \times 2$ sub-matrix
$$\pmatrix{0 & 1 \\ v^\top u & a}$$
This represents the only two non-trivial eigenvalues (the rest are zero). It may be transformed with the similarity parameterized with some $k$:
\begin{align}
 \pmatrix{1 & 0 \\ k & 1}\pmatrix{0 & 1 \\ v^\top u & a}\pmatrix{1 & 0 \\ -k & 1} \\
 = \pmatrix{0 & 1 \\ v^\top u & k+a}\pmatrix{1 & 0 \\ -k & 1} \\
 = \pmatrix{-k &  1 \\ v^\top u - k^2 - ka & k+a} \\
\end{align}
and of course you want to solve
$$v^\top u - k^2 - ka = 0$$
giving the two eigenvalues of $$\lambda_0=-k$$ and $$\lambda_1 = k+a$$
$$ k^2 + ka - v^\top u = 0 \Rightarrow k=\frac{-a \pm \sqrt{a^2 + 4 v^\top u}}{2}$$
$$\lambda = \frac{a \pm \sqrt{a^2 + 4 v^\top u}}{2}$$
Since your elements are $0$ or $1$ you have that $v^\top u\ge 0$, and you have eigenvalues both positive and negative when $v^\top u \gt 0$, and another zero eigenvalue ($\lambda_0 = 0$ and $\lambda_1 = a$) when $v^\top u = 0$.
Not all of them are positive since there are zero eigenvalues, and they are only non-negative when $$v^\top u = 0$$
