$|R(H)|\leq C||H||^2$ I am asked to show the following:
Consider the function $R:$Mat$(n,\mathbb{R})\to\mathbb{R}$ given by $$R(H)=\det(I+H)-1-Tr(H)$$
Proof that there exists $C>0$ such that $|R(H)|\leq C||H||^2$ for all $H\in$Mat$(n,\mathbb{R})$ with $||H||\leq 1$.
I don't know how to tackle this question, since I don't come any further by taking absolute values and using the triangle inequality. Can someone help me please?
 A: I don't know what you mean with $\|H\|$; I will use the norm $\|H\|=\max_{\lambda_i\in \sigma(H)}|\lambda_i|$ where $\sigma(H)=\left\{\lambda_i\right\}_{i=1}^{k}$ is the set of the eigenvalues of $H$. Anyway, even if we used a different norm, the validity of your claim does not change. This is because $Mat(n,\mathbb{R})$ is finite dimensional and so all norms on it are equivalent (two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent iff $C_1\|\cdot\|_1\leq \|\cdot\|_2\leq C_2\|\cdot\|_1$ for some positive constants $C_1,C_2$).
Suppose $\|H\|\leq 1$.  Now, let $\lambda_1,\dots,\lambda_n\in \mathbb{C}$ be the eigenvalues of $H$, counted with their multiplicity. Hence $|\lambda_i|\leq 1$ for all $i=1,\dots,n$. We have
$$R(H)=\prod_{i=1}^{n}(1+\lambda_i)-1-\sum_{i=1}^{n}\lambda_i$$
Notice that $\prod_{i=1}^{n}(1+\lambda_i)$ is a $n$-variable polynomial in $(\lambda_1,\dots,\lambda_n)$. Substracting $1$ removes the term of degree $0$. Likewise, subtracting $\sum_{i=1}^{n}\lambda_i$ removes all terms of degree $1$. That leaves a polynomial where all the terms are of degree at least $2$. More precisely,
$$R(H)=\sum_{i_1\neq i_2}\lambda_{i_1}\lambda_{i_2}+\dots + \sum_{i_1\neq i_2\neq \dots \neq i_n}\lambda_{i_1}\dots \lambda_{i_n} $$
Here, $\sum_{i_1\neq i_2}\lambda_{i_1}\lambda_{i_2}$ are the terms of degree $2$, and so on, until $\sum_{i_1\neq i_2\neq \dots \neq i_n=1}\lambda_{i_1}\dots \lambda_{i_n} $ which are the terms of degree $n$.
Now we just apply the triangle inequality, and the fact that $|\lambda_i|\leq\|H\|\leq 1$:
$$|R(H)|\leq \sum_{i_1\neq i_2}\|H\|^2+\dots +\sum_{i_1\neq i_2 \neq \dots \neq i_n}\|H\|^n= \sum_{k=2}^{n}\alpha_k\|H\|^k=\\ 
=\|H\|^2\sum_{k=0}^{n-2}\alpha_{k+2}\|H\|^k\leq \|H\|^2\sum_{k=0}^{n-2}\alpha_{k+2}
 $$
where for each $k=2,\dots,n$, $\alpha_k$ is a positive (integer) constant depending exclusively on $n$ ( and not on $H$). Therefore the quantity $\sum_{k=0}^{n-2}\alpha_{k+2}$ is a positive constant that does not depend on $H$. Set $C:=\sum_{k=0}^{n-2}\alpha_{k+2}$ and you are done.
Bonus fact: if $n=2$ then the polynomial $R(H)(\lambda_1,\dots,\lambda_n)$ is of homogeneous degree $2$ and thus the claim is true for all $H$, not just for those with $\|H\|\leq 1$.
EDIT: There is a minor mistake in the proof. The map
$$H\mapsto \max_{\lambda \in \sigma(H)}|\lambda_i| $$
is NOT a norm. What I should have written was
$$\|H\|=\max_{\lambda_i\in \sigma(H^*H)}\sqrt{\lambda_i} $$
However, it is still true that $|\lambda_i|\leq \|H\|$ for all $\lambda_i\in \sigma(H)$, so the proof does not break down.
