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? 
 A: 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 $v^Tu=0$.
A: $$\pmatrix{Q & 0 \\ 0 & 1}\pmatrix{\mathbf{0} & u \\ v^\top & a}\pmatrix{Q^{-1} & 0 \\ 0 & 1}=\pmatrix{\mathbf{0} & Qu \\ v^\top Q^{-1} & a}$$
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 $$v^\top u = 0$$
A: 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.
A: 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.
