Upper bound of the difference of two determinants of $2\times2$ matrices

Question

Proposition. There exists a contant $C>0$ such that $$|\det A-\det B|\le C(\|A\|+\|B\|)\|A-B\|$$ for all matrices $A,B \in \mathbb{R}^{2\times 2}$, where $\|\cdot\|$ denotes the Frobenius norm.

What I have done so far: $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $B=\begin{pmatrix} e & f \\ g & h \end{pmatrix}\in \mathbb{R}^{2\times 2}$, then

$|\det A-\det B|=|ad-bc-eh+fg|$ and $\|A\|=\sqrt{a^2+b^2+c^2+d^2}$, $\|A-B\|=\sqrt{(a-e)^2+(b-f)^2+(c-g)^2+(d-h)^2}$. Thus, I want to define a $C>0$ such that

$|ad-bc-eh+fg|\le C(\sqrt{a^2+b^2+c^2+d^2}+\sqrt{e^2+f^2+g^2+h^2})\sqrt{(a-e)^2+(b-f)^2+(c-g)^2+(d-h)^2}$

for all $a,b,c,d,e,f,g,h\in\mathbb{R}$ is satisfied.

But I don't see how to define $C$, so my question is: how to define $C$?

Answer 1

Note that $2(ad-bc-eh+fg)$ is equal to $$ (a-e)(d+h)+(e+a)(d-h)-(b-f)(c+g)-(f+b)(c-g). $$ Applying the Cauchy–Schwarz inequality we obtain $$ 2\lvert\det A-\det B\rvert\le\|A-B\|\sqrt{(d+h)^2+(e+a)^2+(c+g)^2+(f+b)^2}. $$ Now note that each term $$ |d+h|,\quad |e+a|,\quad |c+g|,\quad |f+b| $$ is bound by the sum $\|A\|+\|B\|$ up to some multiplicative constant (since all norms on a finite-dimensional space are equivalent). Hence, there exists $C>0$ such that each of them is at most $C(\|A\|+\|B\|)$ and so $$ \lvert\det A-\det B\rvert\le C(\|A\|+\|B\|)\|A-B\|. $$

For $n\times n$ matrices you can proceed as follows. Let me first recall Jacobi's formula $$ \frac{d}{dt}\det C(t)={\rm tr}\left({\rm adj\,} C(t)\frac{d}{dt}C(t)\right). $$ Now take $C(t)=\det(A+t(B-A))$. Then $$ \begin{split} \lvert\det A-\det B\rvert &=|C(0)-C(1)|\\ &\le {\rm tr}\left({\rm adj\,} C(t)\frac{d}{dt}C(t)\right)\\ &= {\rm tr}\left(({\rm adj\,} C(t))(B-A)\right) \end{split} $$ for some $t$. This yields easily your inequality for arbitrary $n\times n$ matrices.