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For unit left eigenvector $u$ and right eigenvector $v$ of a real matrix at a simple eigenvalue $\lambda$, its normwise sensitivity isIt measures the first-order amplification of a matrix perturbation into an eigenvalue perturbation.

A square nonnegative matrix $A$ is irreducible if for each pair $(i,j)$ there is an integer $k\ge 0$ with $(A^k{)}_{ij}>0$. This says that every index can reach every other through positive entries. The Perron–Frobenius theorem then gives a positive leading eigenvector and an algebraically simple eigenvalue. Irreducibility alone permits other eigenvalues of the same modulus: the cyclic permutation matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ has eigenvalues $1,-1$.

For the Petersen graph, each vertex has three neighbours, adjacent vertices have no common neighbour, and distinct nonadjacent vertices have exactly one common neighbour. The entries of $A^2$ count length-two walks, so diagonal entries are three, adjacent entries zero, and other off-diagonal entries one. HenceThe connected graph has the simple eigenvalue three with eigenvector $1$. On its orthogonal complement $J=0$, so the remaining eigenvalues satisfy ${\lambda }^2+\lambda -2=0$ and are one or minus two. If their multiplicities are $r,s$, then $r+s=9$ and $\mathrm{tr}A=3+r-2s=0$. Thus

Differentiate the eigenvalue equationat $t=0$, where $A^{\prime }(0)=E$. Suppressing the argument $0$ givesLeft-multiplication by the transpose of the unit left eigenvector $u$ and use of $u^{T}A=\lambda u^T$ cancel the terms involving $v^{\prime }$:Because the eigenvalue is simple, $u^{T}v\ne 0$, and hence the first-order perturbation of a simple eigenvalue isThe definition of the operator norm, followed by the Cauchy-Schwarz inequality, givesTherefore

If $Lf=\lambda f$ and $f_t+Af=0$, differentiation and the Lax equation give $(L-\lambda )f_t=-A(L-\lambda )f$, so ${\lambda }_t=0$ after projection onto the normalized eigenfunction for a nondegenerate eigenvalue.

For $\Psi =(f,f_x{)}^T$,The compatibility ${\Psi }_{xt}={\Psi }_{tx}$ is precisely ${\partial }_t{M}_L-{\partial }_x{M}_A-[M_A,M_L]=0$. If all coefficients are $x$-independent, this becomes ${\dot{M}}_L=[M_A,M_L]$, and cyclicity of trace gives $d\mathrm{Tr}(M_L^k)/dt=0$.

The strictly positive case states that a real square matrix $W$ with $W_{ij}>0$ has a positive simple eigenvalue $\lambda$ and positive left and right eigenvectors, with all other eigenvalues strictly smaller in modulus. The Brouwer fixed-point theorem applied to $v\mapsto Wv/(1^{T}Wv)$ on the nonnegative unit simplex gives $v>0$ and $Wv=\lambda v$. For another eigenvector $z$, maximize $|z_i|/v_i$; the triangle inequality gives $|\mu |\le \lambda$. Equality forces every modulus ratio and complex argument to agree because every entry is positive, so $z$ is proportional to $v$ and $\mu =\lambda$. Apply the same argument to $W^T$ for a positive left eigenvector; its positive pairing with $v$ rules out a generalized eigenvector at $\lambda$, establishing algebraic simplicity. For merely nonnegative matrices a nonnegative leading eigenvector exists. An irreducible nonnegative matrix has a positive leading eigenvector and simple Perron eigenvalue; a primitive nonnegative matrix has the strict modulus gap.