We have a cone $y^\top D y=0$ ($[y]_1 \geq 0$), where $y \in \mathbb R^{n+1}$, $[y]_1$ is the first element of $y$, and \begin{equation} D=\left[ {\begin{array}{*{20}{c}} 1&{{0_{1 \times n}}}\\ {{0_{n \times 1}}}&{ - {I_n}} \end{array}} \right] . \end{equation} $y^*$ and $\tilde y^*$ are two points on the cone and we can decompose $\tilde y^*-y^*$ into a weighted sum of s set of unit vectors $$ \tilde y^*-y^*=\beta_1 z_1+\beta_2 z_2+\dots+\beta_l z_l+\beta^\bot z^\bot, $$ where $z_i,i=1,\ldots,l$ ($l \in \{1,\ldots,n\}$, i.e., $l$ can vary from $1$ to $n$ ) are orthogonal to each other and we can calculate them beforehand, and $z^\bot$ is orthogonal to $z_i,i=1,\ldots,l$.

Now we can already bound $\beta^\bot z^\bot$ by a constant $\theta$ $$ \|\beta^\bot z^\bot\| \leq \theta. $$ My question is can we bound $\|\tilde y^*-y^*\|$ in this problem?

Thanks for your valuable comments!

  • $\begingroup$ How are your unit vectors related to the cone? (Your decomposition is possible for any choice of an orthonormal basis $\{z_1,\dots z_n\}$)? $\endgroup$
    – GReyes
    Dec 12, 2019 at 1:54
  • $\begingroup$ @GReyes $\{z_1,\dots,z_l\}$ are the eigenvectors of singular matrix $(A^\top A+\lambda D)$ associated with eigenvalue $0$, where $A$ can be any full column rank matrix. $\endgroup$ Dec 12, 2019 at 2:40

1 Answer 1


The vector $t=\bar y^*-y^*$ can be unbounded even when $z^\perp=0$, $n=1$, and $l=1$. Indeed, if $n=1$ then the cone (which we denote by $C$) is a union of two rays ${(x,x):x\ge 0}$ and ${(x,-x):x\ge 0}$. Graphically it is easy to guess and then to check that $C-C=\{(x,y)\in\Bbb R^2: |x|\le |y|\}$. It follows that for each vector $z^\perp\in\Bbb R^2$ a set $\{t\in C-C: (t,z^\perp)=0\}$ is unbounded.


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