# Vector norm and maximum and minimum singular value decomposition relationship

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 $v_3$.

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