Trouble understanding proof on Milnor's proof of Brouwer's fixed-point theorem I am trying to understand the proof from this paper [A Less Strange Version of Milnor's Proof of Brouwer's Fixed-Point Theorem][1]  [![image][2]][2] , Lemma 2.   Here's what I've got so far.

As the partial differential coefficients of $g$ with respect to $x_1,x_2,\ldots x_n$, are uniformly bounded, the  Jacobian matrix
$$
\left(\frac{\partial \mathbf{f_t}}{\partial x_1}, \frac{\partial \mathbf{f_t}}{\partial x_2}, \ldots , \frac{\partial \mathbf{f_t}}{\partial x_n}\right)= I + t \left(\frac{\partial \mathbf{g}}{\partial x_1}, \frac{\partial \mathbf{g}}{\partial x_2}, \ldots , \frac{\partial \mathbf{g}}{\partial x_n}\right)
$$
is dominated by its diagonal and so is nonsingular provided $0< t< t_o$,  with $t_o$ a sufficiently small positive positive number.

I am not sure what does the last line above means? As hard as I tried to to
$$
\left(\frac{\partial \mathbf{f_t}}{\partial x_1}, \frac{\partial \mathbf{f_t}}{\partial x_2}, \ldots , \frac{\partial \mathbf{f_t}}{\partial x_n}\right) = I \left(\frac{\partial \mathbf{f_t}}{\partial x_1}, \frac{\partial \mathbf{f_t}}{\partial x_2}, \ldots , \frac{\partial \mathbf{f_t}}{\partial x_n}\right),
$$
but without a way out.
Any help is welcome.
[1]: https://people.math.aau.dk/~cornean/index.html/newindex_files/brouwer.pdf
[2]: https://i.stack.imgur.com/ANTNW.png

The collection $M_n(\mathbb{R})$ of all $n\times n$ real matrices form a Banach space (a complete normed vector space). There are many choices of matrix norms for this space, all of which depend on the entries of the matrices. If the entries of some matrices are uniformly bounded, then so are their matrix norms.
Given a matrix $A$ with matrix norm $\|A\|<1$,
\begin{align}
I+A+A^2+\cdots+A^n
\end{align}
converges by completeness. Observe that
\begin{align}
(I-A)(I+A+\dots+A^n)=I-A^{n+1}.
\end{align}
Taking the limit $n\to\infty$, we get
\begin{align}
(I-A)(I+A+A^2+\cdots)=I,
\end{align}
so $I-A$ is invertible. In your case, $\|A\|$ is bounded. For small enough $t$ we have $t\|A\|<1$, so $I-tA$ is invertible.
