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?
At first I thought I understood your question, but after reading those comments and answers here, it seems that people here have very different interpretations. So I'm not sure if I understand it correctly now.
To my understanding, you want to find the eigenvalues of $$ A=\begin{pmatrix}0_{(n-1)\times(n-1)}&u\\ v^T&a\end{pmatrix}, $$ where $a$ is a scalar, $u,v$ are vectors and all entries in $a,u,v$ are either $0$ or $1$. The rank of this matrix is at most $2$. So, when $n\ge3$, $A$ must have some zero eigenvalues. In general, the eignevalues of $A$ include (at least) $(n-2)$ zeros and $$\frac{a \pm \sqrt{a^2 + 4v^Tu}}{2}.$$ Since $u,v$ are $0-1$ vectors, $A$ has exactly one positive eigenvalue and one negative eigenvalue if $v^Tu>0$, and the eigenvalues of $A$ are $\{a,0,0,\ldots,0\}$ if $v^Tu=0$.
$$\pmatrix{Q & 0 \\ 0 & 1}\pmatrix{\mathbf{0} & u \\ v^\top & a}\pmatrix{Q^{-1} & 0 \\ 0 & 1}=\pmatrix{\mathbf{0} & Qu \\ v^\top Q^{-1} & a}$$
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 $$v^\top u = 0$$
Such a matrix, of size $3 \times 3$ or larger, will never have all non-zero eigenvalues. This is because the first two columns of the matrix are linearly independent, hence the determinant is zero, hence zero is an eigenvalue.
I'm not sure whether this is the kind of thing you are looking for but...
Let $n_r$ be the number of ones on the last row and $n_c$ that on the last column. Then, by Gershgorin circle theorem then the eigenvalues all lie in
$B(0,1)\cup B(a,min\{n_c,n_r\}-a)$
where $a$ is bottom right corner entry of the matrix and $B(x,y)$ is the closed ball of radius $y$, centred around $x$.
Also - adding to Jim's answer - the rank of such a matrix will be at most two, thus the matrix will have at most two non-zero eigenvalues.
