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I have a matrix $$A(x) = \frac{-1}{(1+\|x\|_2^2)^{\frac{3}{2}}}xx^T + \frac{1}{(1+\|x\|_2^2)^{\frac{1}{2}}}I$$ I see that $$\|A(x)\|_2 \le 1 \ \forall x$$

But is $\|A(x)\| \le 1$ in general $\forall x$ and if so, how to show this?.

I have a matrix $$A(x) = \frac{-1}{(1+\|x\|_2^2)^{\frac{3}{2}}}xx^T + \frac{1}{(1+\|x\|_2^2)^{\frac{1}{2}}}I$$ I see that $$\|A(x)\|_2 \le 1 \ \forall x$$

But is $\|A(x)\| \le 1$ in general $\forall x$ and if so, how to show this?.

Yes indeed [Iosif Pinelis] and [Denis Serre], i switched the order. Yes, I am referring to operator norm. Is it correct that $\|A(x)\| \le 1 \forall x$?

I have a matrix $$A(x) = \frac{-1}{{1+\|x\|_2^2}^{\frac{3}{2}}}xx^T + \frac{1}{(1+\|x\|_2^2)^{\frac{1}{2}}}I$$ I see that $$\|A(x)\|_2 \le 1 \ \forall x$$

But is $\|A(x)\| \le 1$ in general $\forall x$ and if so, how to show this?.

I have a matrix $$A(x) = \frac{-1}{(1+\|x\|_2^2)^{\frac{3}{2}}}xx^T + \frac{1}{(1+\|x\|_2^2)^{\frac{1}{2}}}I$$ I see that $$\|A(x)\|_2 \le 1 \ \forall x$$

But is $\|A(x)\| \le 1$ in general $\forall x$ and if so, how to show this?.

I have a matrix $$A(x) = \frac{-1}{\sqrt{1+\|x\|_2^2}}xx^T + \frac{1}{(1+\|x\|_2^2)^{\frac{3}{2}}}I$$ I see that $$\|A(x)\|_2 \le 1 \ \forall x$$

But is $\|A(x)\| \le 1$ in general $\forall x$ and if so, how to show this?.

I have a matrix $$A(x) = \frac{-1}{{1+\|x\|_2^2}^{\frac{3}{2}}}xx^T + \frac{1}{(1+\|x\|_2^2)^{\frac{1}{2}}}I$$ I see that $$\|A(x)\|_2 \le 1 \ \forall x$$

But is $\|A(x)\| \le 1$ in general $\forall x$ and if so, how to show this?.