Show the operator norm of $A^T A - I_n$ can be bounded by $3\max(\delta, \delta^2)$ 
Let $A$ be an $m\times n$ matrix and $\delta>0.$ If all singular values of $A$ are between $1-\delta$ and $1+\delta$, $$1-\delta\leq s_n(A)\leq s_1(A) \leq 1+\delta,$$ prove $$\Vert A^TA-I_n\Vert \leq 3\max(\delta,\delta^2),$$where $s_1(A),\dots,s_n(A)$ are the singular values of $A$ and $s_1(A) \geq \cdots \geq s_n(A)$. Note that here the definition of the operator norm is $$\Vert A \Vert := \max_{x \in S^{n-1}\\ y \in S^{m-1}}\langle Ax,y\rangle=s_1(A).$$


So far, I have tried 
$$\Vert A^TA-I_n\Vert = \Vert V(S^TS-I_n)V^T\Vert=s_1(A^TA-I_n).$$
But I am not sure how to connect with $3\max(\delta,\delta^2)$ . This is an extension exercise 4.1.6 from page 80 High-Dimensional Probability by Roman Vershynin. Thank you!




 A: We only need to prove for any $z \geq 0$,
\begin{equation*}
  \begin{split}
     1 -\delta \leq z \leq 1+\delta \Rightarrow |z^2 - 1| \leq 3\max(\delta,\delta^2).
  \end{split}
\end{equation*}
When $0<\delta \leq 1$, we have
\begin{equation*}
  \begin{split}
     &(1-\delta)^2 \leq z^2 \leq (1+\delta)^2 \\
     \Rightarrow & -3\delta \leq -2\delta \leq  -2\delta + \delta^2 \leq z^2 - 1 \leq 2\delta + \delta^2 \leq 3\delta \quad (\delta^2 \leq \delta)\\
     \Rightarrow & |z^2 - 1| \leq 3\delta = 3\max(\delta, \delta^2). 
  \end{split}
\end{equation*}
When $\delta > 1$, we have
\begin{equation*}
  \begin{split}
     &1 -\delta < 0\leq  z \leq 1 + \delta \\
     \Rightarrow & 0 \leq z^2 \leq (1 + \delta)^2 \\
     \Rightarrow & -3\delta^2< -\delta^2 < -1 \leq z^2 - 1 \leq 2\delta + \delta^2 \leq 3\delta^2 \quad (\delta \leq \delta^2)\\
     \Rightarrow & |z^2 - 1| \leq 3\delta^2 = 3 \max(\delta,\delta^2).
  \end{split}
\end{equation*}
A: Hint: note that $A^TA$ is symmetric, so has real eigenvalues. Can you bound them using the singular values of $A$ (if this isn’t obvious, use the SVD)? What happens when you subtract off the identity matrix? How does this relate to the operator norm?
