Proof that the largest eigenvalue of a stochastic matrix is $1$

The biggest eigenvalue of a stochastic matrix (i.e. a matrix whose access declare and also whose rows amount to $1$) is $1$.

Wikipedia notes this as a grandfather clause of the Perron-Frobenius theorem, yet I ask yourself if there is a less complex (even more straight) means to show this outcome.

2022-06-07 14:36:55
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Say $A$ is a $n \times n$ row stochastic matrix. Currently: $$A \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} = \begin{pmatrix} \sum_{i=1}^n a_{1i} \\ \sum_{i=1}^n a_{2i} \\ \vdots \\ \sum_{i=1}^n a_{ni} \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{pmatrix} $$ Thus the eigenvalue $1$ is acquired.

To show that the this is the biggest eigenvalue you can make use of the Gershgorin circle theorem. Take row $k$ in $A$ . The angled component will certainly be $a_{kk}$ and also the distance will certainly be $\sum_{i\neq k} |a_{ki}| = \sum_{i \neq k} a_{ki}$ given that all $a_{ki} \geq 0$ . This will certainly be a circle with its facility in $a_{kk} \in [0,1]$ , and also a distance of $\sum_{i \neq k} a_{ki} = 1-a_{kk}$ . So this circle will certainly have $1$ on its border. This holds true for all Gershgorin circles for this matrix (given that $k$ was taken randomly). Hence, given that all eigenvalues hinge on the union of the Gershgorin circles, all eigenvalues $\lambda_i$ please $|\lambda_i| \leq 1$ .

2022-06-07 15:01:12

Here is an actually primary evidence (which is a mild alteration of Fanfan's answer to a question of mine). As Calle programs, it is very easy to see that the eigenvalue $1$ is gotten. Currently, intend $Ax = \lambda x$ for some $\lambda > 1$. Given that the rows of $A$ are nonnegative and also amount to $1$, each component of vector $Ax$ is a convex mix of the parts of $x$, which can be no more than $x_{max}$, the biggest part of $x$. On the various other hand, at the very least one component of $\lambda x$ is more than $x_{max}$, which confirms that $\lambda > 1$ is difficult.

2022-06-07 14:55:52