Which matrix with positive entries has only one positive eigenvalue

linear algebramatricesmatrix equationsreal-analysis

Let $A$ be a strictly positive matrix, by strictly positive matrix I mean that the entries of $A$ are strictly positive.

Also, we assume that the entries of the matrix are from Natural numbers.

Is there any theorem that gives us information about how many positive eigenvalues the matrix has?

By Perron Frobenius theorem, we can say that there is a positive eigenvalue that is the spectral radius, but what more conditions do we need to conclude that that perron eigenvalue is the only positive eigenvalue?

Let me try to make the question more specific:

Let $A$ be a matrix of the form $D+B$, where $D$ is a diagonal matrix with negative entries and $B$ is a rank one matrix.

Can we conclude from here that $A$ has only one positive eigenvalue?

Best Answer

For your second question, suppose $A$ is at least $2\times2$ and $A=D+uv^T$ where $D$ is a negative diagonal matrix. If $A$ is positive, $u$ and $v$ must be both positive or both negative, otherwise $A$ will possess a non-positive off-diagonal element. Therefore we may assume that $u,v>0$. Define two positive vectors $x$ and $y$ such that $x_i=\sqrt{u_iv_i}$ and $y_i=\sqrt{\frac{u_i}{v_i}}$. Then

  • $A$ is similar to the symmetric matrix $S=\operatorname{diag}(y)^{-1}A\operatorname{diag}(y)=D+xx^T$;
  • $S$ is congruent to $T=|D|^{-1/2}S|D|^{-1/2}=-I+|D|^{-1/2}xx^T|D|^{-1/2}$, where $|D|$ is the entrywise absolute value of $D$.

Since $-1$ is an eigenvalue of $T$ of multiplicity $n-1$, each of $S$ and $A$ must also have $n-1$ negative eigenvalues. However, as $A$ is a positive matrix, $\rho(A)$ is a positive eigenvalue. Therefore $A$ has exactly one positive eigenvalue.

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