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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?.

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

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    $\begingroup$ What norms are you using? $\endgroup$
    – Willie Wong
    Commented Feb 21 at 7:49
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    $\begingroup$ Are you sure of the formula ? This matrix sounds familiar to me, but only if the exponents $1/2$ and $3/2$ are exchanged. $\endgroup$
    – Denis Serre
    Commented Feb 21 at 8:09
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    $\begingroup$ It's of course false as written (compare Denis's comment) because it doesn't scale right: $\langle x, A(\lambda x) x\rangle \simeq -\lambda \|x\|$ as $\lambda\to\infty$ $\endgroup$
    – Christian Remling
    Commented Feb 21 at 14:52

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$\newcommand\R{\mathbb R}$As noted by Denis Serre and Christian Remling, the exponents $1/2$ and $3/2$ must be interchanged -- otherwise, $\|A(x)\|_2\to\infty$ if $x=tu$, $u$ is a fixed unit vector, and $t\to\infty$.

So, the meaningful re-interpretation of the question is as follows:

Let $$A(x) := \frac{-xx^T}{(1+\|x\|_2^2)^{3/2}} + \frac{I}{(1+\|x\|_2^2)^{1/2}}$$ for $x\in\R^n$. Is it true that $\|A(x)\|\le1$ for all $x\in\R^n$, where $\|\cdot\|$ is an operator norm?

The answer to this question is no, even if the same norm is used on the domain and the co-domain of the linear operator corresponding to the matrix $A(x)$. (If different norms are allowed to be used for the domain and the co-domain, then the no answer is quite obvious.)

For instance, suppose that $n=5$, $\|\cdot\|$ is the operator norm induced by the $\ell^\infty$ norm on $\R^5$, $x=\frac1{1000}(160, 159, 197, -159, -160)$, and $y=(1, 1, -1, -1, -1)$. Then $\|y\|_\infty=1$ but $$\|A(x)\|\ge\|A(x)y\|_\infty=1.0076\ldots>1.$$

However, since $0\le xx^T\le\|x\|_2^2\,I$, we have $0\le(1+\|x\|_2^2)^{-3/2}\,I\le A(x)\le(1+\|x\|_2^2)^{-1/2}\,I$ and hence $\|A(x)\|_2\le(1+\|x\|_2^2)^{-1/2}\le1$, for all $x\in\R^n$.

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  • $\begingroup$ @ChristianRemling : What do you mean by the $\|\cdot\|_{2,2}$ norm? $\endgroup$
    – Iosif Pinelis
    Commented Feb 21 at 19:21
  • $\begingroup$ $\|A\|_{2,2}=\max_{\|x\|_2=1}\|Ax\|_2$ $\endgroup$
    – Christian Remling
    Commented Feb 21 at 19:22
  • $\begingroup$ Ah, I see. I think this is what the OP denoted simply by $\|\cdot\|_2$, with the $\cdot$ standing for a matrix. This norm was considered in the last paragraph of my answer, and apparently concerning this case the OP said "I see that [...]". So, it does not appear to me that the question was about $\|\cdot\|_{2,2}$. $\endgroup$
    – Iosif Pinelis
    Commented Feb 21 at 19:30
  • $\begingroup$ Yes indeed [Iosif Pinelis] and [Denis Serre], i switched the order. Yes, I am referring to operator norm. Is it correct that $\|A(x)\| \le 1 \forall x$? $\endgroup$
    – Shoeb
    Commented Feb 21 at 20:47
  • $\begingroup$ @Shoeb : Again, your question has now been answered (negatively). Have you seen and read this answer? $\endgroup$
    – Iosif Pinelis
    Commented Feb 21 at 21:01

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