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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?

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Most ot them are zero, so you mean maybe nonnegative? – Alex Mar 27 '13 at 16:24
If $M$ is your matrix, you should try to develop the determinant of $(M - \lambda I)$ using cofactors – Vincent Nivoliers Mar 27 '13 at 16:32
up vote 1 down vote accepted

$$\pmatrix{Q & 0 \\ 0 & 1}\pmatrix{\mathbf{0} & u \\ v^\top & a}\pmatrix{Q^{-1} & 0 \\ 0 & 1}=\pmatrix{\mathbf{0} & Qu \\ v^\top Q^{-1} & a}$$

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 $$v^\top u = 0$$

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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 $v^Tu=0$.

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Hello user1551, Do you remember this question… – dineshdileep Mar 28 '13 at 4:33
I was trying to prove it on my own. The idea was to use induction. For a $n+1 \times n+1$ matrix $B_{n+1}=[[B_{n},0];[0,0]]+A$, $B_{n}$ is a binary matrix with all eigenvalues as 1, $A$ is as you defined in the answer. Now, we have to prove only that $B_{n+1}$ will also have all eigenvalues as 1. – dineshdileep Mar 28 '13 at 4:36
Thanks for the answer btw. – dineshdileep Mar 28 '13 at 4:48
@dineshdileep I remember that question. Did you find a solution? – user1551 Mar 28 '13 at 11:36
No, I try it once in a while. It is tagged in my favorites. Today, I was again at it. That is how this question came out. Please, check the previous comment. – dineshdileep Mar 28 '13 at 15:24

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.

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What if the first columns are zero? – dineshdileep Mar 28 '13 at 4:38
If there's a row of zeros then the determinant is zero. – Jim Mar 28 '13 at 5:39

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.

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that is one way to look at it I guess. Thanks – dineshdileep Mar 28 '13 at 4:38

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