(CONTEXT) I self-study proof-based linear algebra through the use of Linear Algebra by Serge Lang. And yesterday, I tried to prove a statement that was presented in page 241 of the aforementioned book as the second property of Markov matrices, but despite my effort, I failed to produce a proof. Thoroughly frustrated and now critical of the truth of the statement, I then tried to find a disproof of the statement, and I actually managed to find what seems to be a counterexample to the statement.
(PURPOSE) Personally, I do not think the third edition of a book as widely used as Linear Algebra by Serge Lang would contain such a major error. In addition, I am not proficient in proof-based mathematics, and I thus believe that it is totally possible for my supposed disproof to be erroneous; and therefore, I am desirous to ascertain the validity (or the invalidity) of my disproof of the previously mentioned result. Furthermore, if my disproof is indeed correct, then I hope that this post would benefit those who are also confused by the erratum in page 241 of Linear Algebra by Serge Lang.
(REQUEST) Hence, can you please check the validity of my disproof? In addition, if you so desire, please critique the stylistic aspect of my post. Any constructive feedback would be appreciated.
(DEFINITION) In his Linear Algebra, Serge Lang manually defines a Markov matrix as a square complex matrix such that the sum of the elements of each column is equal to 1.
(STATEMENT) Suppose that $A$ and $B$ are $n\times n$ Markov matrices such that the modulus of each component of $A$ and $B$ is less than or equal to 1. And suppose that $C=AB$ . Then the modulus of each component of $C$ is also less than or equal to 1.
(DISPROOF) We shall let $$A=\begin{pmatrix} e^{-\frac{i\pi}{4}} & e^{\frac{i\pi}{4}} & -1 \\ -e^{-\frac{i\pi}{4}} & -e^{\frac{i\pi}{4}} & 1 \\ 1 & 1 & 1\end{pmatrix}\space,$$ and let $$B=\begin{pmatrix} e^{\frac{i\pi}{4}} & 0 & 0 \\ e^{-\frac{i\pi}{4}} & 0 & 0 \\ 1-\sqrt{2} & 1 & 1\end{pmatrix}\space,$$ and also let $C=AB$ . Through direct computation, one can easily verify that $A$ and $B$ are both Markov matrices, and that the modulus of each component of $A$ and $B$ is indeed less than or equal to $1$; however, we have $$\begin{aligned} \vert c_{11} \vert &= \vert A_1 \cdot B^1 \vert \\ &= \vert a_{11}b_{11}+a_{12}b_{21}+a_{13}b_{13} \vert \\ &= \vert e^{-\frac{i\pi}{4}}e^{\frac{i\pi}{4}}+e^{\frac{i\pi}{4}}e^{-\frac{i\pi}{4}}+(-1)(1-\sqrt{2}) \vert \\ &= \vert 1 + 1 - 1 + \sqrt{2} \vert \\ &= \vert 1 + \sqrt{2} \vert \\ &= 1 + \sqrt{2} \ge 1\space, \end{aligned}$$ which does seem to be a counterexample to the statement above.
Q.E.D.
Your counterexample is valid. Here is a simpler one: let $\omega$ be a complex number on the unit circle that is close to but not equal to $1$. Then $0<|1-\omega|\le1$. Now let $A=\pmatrix{\omega&1-\omega\\ 1-\omega&\omega}$ and $B=\overline{A}$, the complex conjugate of $A$. Then $c_{11}=|\omega|^2+|1-\omega|^2=1+|1-\omega|^2>1$.
(If we take $\omega=e^{i\pi/3}$, we even obtain a counterexample where all elements of $A$ and $B$ lie on the unit circle but $c_{11}$ lies outside the closed unit disc.)