Suppose that for the $2$-norm, we have $||A||_{2} < 1$. Show that $||I - A^{T}A||_{2} < 1.$ Assume $A$ is invertible.
I don't know how to solve this problem. I'm studying for an exam. I know that, by definition,
$$||A||_{2} = \max_{v \in \mathbb{R}^{n} \setminus \{0\}} \frac{||Av||_{2}}{||v||_{2}}.$$
Also, for any $M \in \mathbb{R}^{n\times n}$ and $v \in \mathbb{R}^{n}$, we have the relation $||Mv|| \leq ||M|| \cdot ||v||$. Finally, I know that $||I||_{2} = 1$, always.
So I tried, like,
$$||I - A^{T}A||_{2} \leq ||I|| + ||-A^{T}A|| = 1 - ||A^{T}A||.$$
Is this correct so far? If so, how can I show $0 < ||A^{T}A|| < 1$ to complete the proof? Thanks
The operator norm of the euclidean norm gives the largest singular value $\sigma_1$, that is the largest diagonal entry in $\Sigma$ in the SVD $A=UΣV^T$. Per assumption this is smaller than $1$. Then the target matrix has a decomposition $$ I-A^TA=I-VΣ^2V^T=V(I-Σ^2)V^T $$ so that the singular values are inside the interval $[1-σ_1^2, 1-σ_n^2]\subset[0,1]$, so that the operator norm of it is $$ \|I-A^TA\|_2=1-σ_n^2\le 1. $$
You would get the strict inequality for $σ_n>0$, that is, when $A$ is non-singular.