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I have a real-valued column vector $v$ and the following equation $$ \frac{v v^T}{1+\alpha v^T v} $$ with $\alpha>0$, and I'm trying to find an upper limit not including $v$.

I have found that the following holds for a 2-element vector $v$: \begin{align} \frac{v v^T}{1+\alpha v^T v} &\leq \frac{\lambda_{max}(v v^T)}{1+\alpha v^T v} I\\ &= \frac{v^T v}{1+\alpha v^T v} I\\ &\leq \frac{1}{\alpha} I \end{align} Since the (only non-zero) eigenvalue of the outer product of $v$ is its own inner product, at least for a two-element vector.

Two questions:

  • Does the result generalize to any number of elements in $v$?
  • Does there exist a less conservative upper limit not involving $v$?

1 Answers 1

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First, yes: the spectrum of the matrix $v v^T$ is $\{ v^T v \}$ (euclidean norm). That's because it is a rank-one matrix, so has at most one eigenvalue; and an eigenvector is given by $v$.

Second, consider the function $t \mapsto \frac{t}{1 + \alpha t} = \frac1\alpha \frac{t \alpha}{1 + t \alpha}$. The last fraction is bounded by $1$ for $t \to \infty$. So, I don't think you can bound the function by something not involving $t$.