How to show for positive semi-definite, there exist $x \in \mathbb{R}^n$ such that $Bx+c = 0$ and and $\|x\| \leq \Delta$?

Let $$B \in \mathbb{R}^{n \times n}$$ be symmetric and positive semi-definite such that $$B = U\Lambda U^T$$, where $$U = [u_1,\cdots,u_n]^T$$ is an orthogonal matrix with $$u_i \in \mathbb{R}^n$$, and $$\Lambda=\text{diag}(0,\cdots,0,\lambda_k,\cdots,\lambda_n)$$ is the diagonal matrix with $$0<\lambda_k \leq\cdots \leq \lambda_n$$. Show that for $$\Delta >0$$ the following are equivalet:

1- $$u_i^Tc=0,\,\,\,\,\,\, \forall i=1,\cdots,k-1$$, and $$\sum_{i=k}^n \frac{(u_i^T)^2}{\lambda_i^2}\leq \delta^2$$

2- There exist $$x \in \mathbb{R}^n$$ such that $$Bx+c = 0$$ and and $$\|x\| \leq \Delta$$.

I can proof $$(2)$$ implies $$(1)$$ but do not know how to show $$(1)$$ implies $$(2)$$.

• What exactly is $c$ in this context? – Omnomnomnom Apr 24 at 7:34
• A given matrix. I think we need to use the fact that $\lambda_n I -B$ is negative semi-definite. – Saeed Apr 24 at 13:28
• Sorry for the confusion, I was missing a part of the statement which is now $(1)$. As I explained above, from $(2)$ we can get $(1)$ by using eigenvalue decomposition of $B$ and writing $C=UU^TC$ and sum the equalities and apply Cauchy-Schwarz inequality. – Saeed Apr 25 at 3:29