Let $s_1(A) \geq \cdots \geq s_n(A)$ denote the singular values of a matrix $A$. If $$ 1-\delta \leq s_n(A) \leq s_1(A) \leq 1+\delta $$ I would like to show that $\|A^T A - I\| \leq 3 \max{\delta, \delta^2}$, where $\| \|$ denotes the operator norm. I can show that $x^T (A^T A - I) x$ obeys a similar bound using the observation that $x^T A^T A x = \|A x\|^2$ but it may not be the case that $(A^T A - I)x$ is parallel to x, where $x$ is the vector that achieves the operator norm.
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