The determinant is Lipschitz on bounded sets

Idea. Change one column at a time. Each change costs a determinant with one column equal to a difference $a_k-b_k$ and the other columns bounded by $R$; Hadamard bounds it by $R^{n-1}|a_k-b_k|$.

Step 1: the hybrid matrices. Let $H_k$ be the matrix with columns $(b_1,\dots,b_k,a_{k+1},\dots,a_n)$, for $k=0,\dots,n$. Then $H_0=A$ and $H_n=B$, so
$$\det A-\det B=\sum_{k=1}^n\big(\det H_{k-1}-\det H_k\big).$$

Step 2: multilinearity. $H_{k-1}$ and $H_k$ differ only in column $k$, namely $a_k$ versus $b_k$. By linearity of $\det$ in that column,
$$\det H_{k-1}-\det H_k=\det(b_1,\dots,b_{k-1},\,a_k-b_k,\,a_{k+1},\dots,a_n).$$

Step 3: Hadamard. The columns of this matrix have lengths $|b_j|\le\|B\|\le R$ (for $j<k$), $|a_j|\le\|A\|\le R$ (for $j>k$), and $|a_k-b_k|=|(A-B)e_k|\le\|A-B\|$. Here we used $|Me_j|\le\|M\|$. Hence
$$|\det H_{k-1}-\det H_k|\le R^{n-1}\|A-B\|.$$

Step 4: sum. Adding the $n$ terms, $|\det A-\det B|\le nR^{n-1}\|A-B\|$.

Sanity check. For $n=1$ the bound reads $|a-b|\le|a-b|$. For $A=\operatorname{diag}(R,\dots,R)$, $B=(R-\delta)I$ the exact difference is $R^n-(R-\delta)^n\approx nR^{n-1}\delta$ and $\|A-B\|=\delta$: the constant is asymptotically sharp.

Remark. This gives a quantitative proof of the continuity of $\det$ and shows that $\det$ is *locally Lipschitz*, with Lipschitz constant on the ball $\{\|A\|\le R\}$ at most $nR^{n-1}$. The derivative $\operatorname{tr}(\operatorname{adj}(A)H)$ computed later in the book has norm at most $nR^{n-1}$ on that ball, consistent with this bound.