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 $(2)$.

  • 1
    $\begingroup$ What exactly is $c$ in this context? $\endgroup$ – Ben Grossmann Apr 24 '19 at 7:34
  • $\begingroup$ A given matrix. I think we need to use the fact that $\lambda_n I -B$ is negative semi-definite. $\endgroup$ – Saeed Apr 24 '19 at 13:28
  • $\begingroup$ 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. $\endgroup$ – Saeed Apr 25 '19 at 3:29

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