# Proving a matrix inequality given another inequality

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

• No, the last step of the your inequality is not true in general – DiegoMath May 16 at 16:45

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.
• Is there a way to do this without SVD? A hint I have is to check that $||(I - A^{T}A) u||_{2} < ||u||_{2}$ holds as long as $Au \neq 0$. Then by invertibility, the bound follows. But, I don't really see why this works. – user666729 May 16 at 17:28
• This is probably easiest if you use that for symmetric matrices also $\|B\|_2=\max_{\|u\|_2=1}|u^TBu|$. Then $|u^T(I-A^TA)u|=|\|u\|_2^2-\|Au\|_2^2|$ should be easy to analyze. – LutzL May 16 at 18:33