Consider the following relation: $$ v_1 = v_2 + \Lambda v_3$$ where $v_1$, $v_2$ and $v_3$ are vectors of a common dimension say, $n$ and $\Lambda$ is a diagonal matrix with proper dimensions and $\lambda_i$ are positive constants as its diagonal elements. In a textbook, the following inequality has been presented as a Lemma without any proof: $$ \Vert v_2 \Vert \le \Vert v_1 \Vert + \bar\sigma\left(\Lambda\right) \Vert v_3 \Vert$$ in which $\bar\sigma\left(\Lambda\right)$ is the maximum singular value of $\Lambda$. Using this relation it has been concluded that if $v_1$ and $v_3$ are bounded then $v_2$ is bounded, too.

My question is: Is there any similar relation for $v_3$ using inequality relations between vector norms and maximum or minimum singular values of matrix $\Lambda$ such that I can conclude that the boundedness of $v_1$ and $v_2$ leads to the boundedness of $v_3$.


1 Answer 1


Sure. Since $\Lambda$ is invertible you have

$$ \|v_3\| = \|\Lambda^{-1}(v_1-v_2)\| \leq \|\Lambda^{-1}\|\cdot\|v_1-v_2\|. $$

Using that $\|\Lambda^{-1}\| \leq {\sigma_\min}^{-1}$, where $\sigma_\min > 0$ is the smallest singular value of $\Lambda$, we have

$$ \|v_3\| \leq \frac{1}{\sigma_\min}\|v_1-v_2\|. $$


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