Sufficient optimality conditions hold for almost all nonlinear semidefinite programs

Abstract

We derive a new genericity result for nonlinear semidefinite programming (NLSDP). Namely, almost all linear perturbations of a given NLSDP are shown to be nondegenerate. Here, nondegeneracy for NLSDP refers to the transversality constraint qualification, strict complementarity and second-order sufficient condition. Due to the presence of the second-order sufficient condition, our result is a nontrivial extension of the corresponding results for linear semidefinite programs (SDP) from Alizadeh et al. (Math Program 77(2, Ser. B):111–128, 1997). The proof of the genericity result makes use of Forsgren’s derivation of optimality conditions for NLSDP in Forsgren (Math Program 88(1, Ser. A):105–128, 2000). Due to the latter approach, the positive semidefiniteness of a symmetric matrix G(x), depending continuously on x, is locally equivalent to the fact that a certain Schur complement S(x) of G(x) is positive semidefinite. This yields a reduced NLSDP by considering the new semidefinite constraint $S(x) \succeq 0$, instead of $G(x) \succeq 0$. While deriving optimality conditions for the reduced NLSDP, the well-known and often mentioned “H-term” in the second-order sufficient condition vanishes. This allows us to access the proof of the genericity result for NLSDP.

Appendix

Appendix is devoted to the proofs of Lemmata 4 and 5.

Proof of Lemma 4

First, we derive a representation of DS(x) in terms of DG(x). Note that for x close to $\bar{x}$ and an arbitrary $h \in \mathbb {R}^n$ it holds:

$$\begin{aligned} D A^{-1} (x) [h]=- A(x)^{-1} (DA(x)[h]) A (x)^{-1}. \end{aligned}$$

Thus, using the formula (13) for S(x), we calculate

$$\begin{aligned} DS( x)[h] = W( x)^\top (DG( x)[h]) W( x), \quad W(x):=\left( \begin{array}{c} -A ( x)^{-1} B ( x) \\ {{\mathrm{Id}}}\end{array} \right) . \end{aligned}$$

(28)

Now, let $\omega \in \Lambda ^R$ be given, and let $\Omega := \varOmega (\omega )$. One easily verifies that $\Omega = W \omega W^\top$, where $W := W(\bar{x})$. Since the columns of W span the kernel of $G(\bar{x})$, we have

$$\begin{aligned} \Omega \bullet G(\bar{x}) = \text {tr} [ G(\bar{x}) W \omega W^\top ] = 0. \end{aligned}$$

The application of basic symmetry properties of the trace yields

$$\begin{aligned} \omega \bullet DS = \text {tr}\left[ \omega W^\top (DG) W \right] = \text {tr}\left[ W \omega W^\top DG \right] = (W \omega W^\top ) \bullet DG, \end{aligned}$$

(29)

where we omit the argument $\bar{x}$ in the involved matrices. This implies

$$\begin{aligned} D_x L (\bar{x} , \Omega ) = D_x L^R (\bar{x} , \omega ) = 0, \end{aligned}$$

and, hence, $\Omega \in \Lambda$.

Now, we show that for each $\Omega \in \Lambda$ there exists a unique $\omega \in \Lambda ^R$ with $\Omega = W \omega W^\top$. The uniqueness follows directly from the fact that the columns of W are linearly independent. Further, we set

$$\begin{aligned} \omega := Q (WQ)^\top \Omega (W Q) Q^\top , \end{aligned}$$

where Q is a matrix with the property that the columns of W Q are orthonormal and span the same linear subspace as those of W. First, we show that $W \omega W^\top$ is equal to $\Omega$. Indeed, since the columns of WQ form an orthogonal basis of the kernel of G(x) and $\Omega \bullet G(x) = 0$, there exists a diagonal matrix $\Omega _\Delta$ such that $\Omega = (WQ) \Omega _\Delta (WQ)^\top$. We calculate:

$$\begin{aligned} \begin{array}{rcl} W \omega W^\top &{}=&{} W ( Q (WQ)^\top \Omega (W Q) Q^\top ) W^\top \\ &{}=&{} W ( Q (WQ)^\top (WQ) \Omega _\Delta (WQ)^\top (W Q) Q^\top ) W^\top \\ &{}=&{}W ( Q \Omega _\Delta Q^\top ) W^\top = (W Q) \Omega _\Delta (WQ)^\top =\Omega . \end{array} \end{aligned}$$

It remains to show that $\omega$ is an element of $\Lambda ^R$. Since $\Omega$ is positive semidefinite, also $\omega$ is. Analogously as in (29), one shows that $D_x L^R (\bar{x},\omega ) = D_x L (\bar{x}, \Omega ) = 0$. The fact that $\omega \bullet S(\bar{x}) = 0$ follows from $S(\bar{x}) = W^\top G(\bar{x}) W$. Overall, we conclude that $\omega \in \Lambda ^R$. $\square$

Proof of Lemma 5 (a)

By definition, the transversality constraint qualification means that ${{\mathrm{Im}}}DG(\bar{x}) + T_{G(\bar{x})}\mathbb {S}^m_r$ is equal to $\mathbb {S}^m$. This is equivalent to the fact that the orthogonal complement $( {{\mathrm{Im}}}DG(\bar{x}) + T_{G(\bar{x})} \mathbb {S}^m_r )^\perp$ – w.r.t. the trace inner product “$\bullet$” – is equal to $\{ 0 \}$. Since the orthogonal complement is given by the intersection $({{\mathrm{Im}}}DG(\bar{x}))^\perp \cap (T_{G(\bar{x})} \mathbb {S}^m_r )^\perp$, transversality constraint qualification is equivalent to $({{\mathrm{Im}}}DG(\bar{x}))^\perp \cap (T_{G(\bar{x})} \mathbb {S}^m_r )^\perp = \{0\}$. In view of $({{\mathrm{Im}}}DG(\bar{x}))^\perp = {{\mathrm{Ker}}}DG(\bar{x})^*$ and $(T_{G(\bar{x})}\mathbb {S}^{m}_r)^\perp = {{\mathrm{Im}}}D\varphi (G(\bar{x}))^*$, where $\varphi : G \mapsto C - B^\top A^{-1} B$ (cf. (11)), the transversality constraint qualification holds if and only if

$$\begin{aligned} {{\mathrm{Ker}}}DG(\bar{x})^* \cap {{\mathrm{Im}}}D\varphi (G(\bar{x}))^* = \{0\}. \end{aligned}$$

(30)

On the other hand, $DS(\bar{x})$ is regular if and only if

$$\begin{aligned} {{\mathrm{Ker}}}DS(\bar{x})^* = {{\mathrm{Ker}}}\left[ DG(\bar{x})^* \cdot D\varphi (G(\bar{x}))^* \right] = \{0\}. \end{aligned}$$

The latter is equivalent to (30). $\square$

Proof of Lemma 5 (b)

The rank of $G(\bar{x})$ is equal to r. Therefore, ${{\mathrm{rank}}}G(\bar{x}) + {{\mathrm{rank}}}\Omega = m$ if and only if the rank of $\Omega$ equals $m-r$. Now, the assertion follows directly from the fact that $\Omega$ is given by $W \omega W^\top$. $\square$

Proof of Lemma 5 (c)

Since the kernel of $S(\bar{x}) = 0 \in \mathbb {S}^{m-r}$ is the whole $\mathbb {R}^{m-r}$, it holds:

$$\begin{aligned} T^R (\bar{x}) = \{ h \in \mathbb {R}^n \mid DS(\bar{x})[h] = 0\}. \end{aligned}$$

Recall from (28) that for all $h \in \mathbb {R}^n$ and x close to $\bar{x}$:

$$\begin{aligned} DS( x) [h] = W( x)^\top (DG( x) [h]) W( x), \quad W(x):=\left( \begin{array}{c} -A ( x)^{-1} B ( x) \\ {{\mathrm{Id}}}\end{array} \right) . \end{aligned}$$

(31)

Further, from (10) we have

$$\begin{aligned} T(\bar{x}) = \{h \in \mathbb {R}^n \mid E^\top DG(\bar{x})[h] E = 0\}, \end{aligned}$$

where the columns of E form a basis of the kernel of $DG(\bar{x})$. Note that the columns of $W(\bar{x})$ form also a basis of the kernel of $DG(\bar{x})$. Then, in view of (31), $T^R (\bar{x})=T(\bar{x})$.

Now, we show that for all $h \in T (\bar{x})$ it holds:

$$\begin{aligned} h^\top \left( D^2_{xx} L^{R} (\bar{x} , \omega ) \right) h = h^\top \left( D^2_{xx} L(\bar{x} , \Omega ) + H(\bar{x}, \Omega ) \right) h. \end{aligned}$$

We calculate:

$$\begin{aligned} h^\top \left( D^2_{xx} L^R (\bar{x} , \omega ) \right) h = h^\top D^2 f(\bar{x}) h - \omega \bullet D^2 S(\bar{x})[h,h]. \end{aligned}$$

We proceed by calculating the term $D^2 S(\bar{x}) [h,h]$ using the formula in (31):

$$\begin{aligned} \begin{array}{rcl} D^2S(\bar{x})[h,h]&{}=&{}(D^2W(\bar{x})[h,h])^\top G(\bar{x})W(\bar{x})+(W(\bar{x})[h])^\top (DG(\bar{x})[h])W(\bar{x})\\ &{}&{} +\,(W(\bar{x})[h])^\top G(\bar{x})(DW(\bar{x})[h]) + (DW(\bar{x})[h])^\top (DG(\bar{x})[h])W(\bar{x}) \\ &{}&{} +\, W(\bar{x})^\top (D^2G(\bar{x})[h,h])W(\bar{x})+W(\bar{x})^\top (DG(\bar{x})[h])(DW(\bar{x})[h])\\ &{}&{}+ \,(DW(\bar{x})[h])^\top G(\bar{x})(DW(\bar{x})[h])\\ &{}&{}+\,W(\bar{x})^\top (DG(\bar{x})[h])(DW(\bar{x})[h])\\ &{}&{}+\,W(\bar{x})^\top G(\bar{x})(D^2W(\bar{x})[h,h]). \end{array} \end{aligned}$$

A straightforward matrix multiplication shows that if $S(x) = 0$ (for x close to $\bar{x}$), then also $G(x)W(x) = 0$. Hence, $D \left( G W \right) (\bar{x})[h]=0$ for any direction $h \in T(\bar{x})$, i.e.,

$$\begin{aligned} (DG(\bar{x})[h])W(\bar{x})=-G(\bar{x})(DW(\bar{x})[h]). \end{aligned}$$

(32)

Using this and $G(\bar{x}) W(\bar{x})=0$, the formula for $D^2 S(\bar{x})[h,h]$ reduces to

$$\begin{aligned} D^2 S(\bar{x})[h,h] = W(\bar{x})^\top (D^2G(\bar{x})[h,h])W(\bar{x}) - 2 (DW(\bar{x})[h])^\top G(\bar{x})(DW(\bar{x})[h]). \end{aligned}$$

Using this, we have the following:

$$\begin{aligned} \begin{array}{rcl} h^\top \left( D^2_{xx} L^R (\bar{x} , \omega ) \right) h &{} = &{} h^\top D^2 f(\bar{x}) h - \omega \bullet W(\bar{x})^\top (D^2G(\bar{x})[h,h])W(\bar{x}) \\ &{}&{} + \omega \bullet 2 (DW(\bar{x})[h])^\top G(\bar{x})(DW(\bar{x})[h]),\end{array} \end{aligned}$$

which by the properties of the trace goes to

$$\begin{aligned} \begin{array}{rcl} h^\top \left( D^2_{xx} L^R (\bar{x} , \omega ) \right) h &{}=&{} h^\top D^2 f(\bar{x})h-(W(\bar{x}) \omega W(\bar{x})^\top )\bullet (D^2G(\bar{x})[h,h]) \\ &{}&{}+ \omega \bullet 2 (DW(\bar{x})[h])^\top G(\bar{x})(DW(\bar{x})[h]), \end{array} \end{aligned}$$

and finally, using $W(\bar{x}) \omega W(\bar{x})^\top = \Omega$, to the expression

$$\begin{aligned} \begin{array}{rcl} h^\top \left( D^2_{xx} L^R (\bar{x} , \omega ) \right) h &{}=&{} h^\top D^2 f(\bar{x})h-{\Omega }\bullet (D^2G(\bar{x})[h,h]) \\ &{} &{} +\, 2 \omega \bullet (DW(\bar{x})[h])^\top G(\bar{x})(DW(\bar{x})[h]) \\ &{}=&{} h^\top D^2_{xx} L(\bar{x},\Omega ) h +\, 2 \omega \bullet (DW(\bar{x})[h])^\top G(\bar{x})(DW(\bar{x})[h]). \end{array} \end{aligned}$$

By definition of the Moore-Penrose pseudo inverse, we have $G(\bar{x}) G(\bar{x})^\dag G(\bar{x})=G(\bar{x})$. Thus, we obtain

$$\begin{aligned} \begin{array}{rcl} h^\top \left( D^2_{xx} L^R (\bar{x} , \omega ) \right) h &{}=&{} h^\top D^2_{xx} L(\bar{x},\Omega ) h \\ &{}&{} +\, 2 \omega \bullet (DW(\bar{x})[h])^\top G(\bar{x})G(\bar{x})^{\dag }G(\bar{x})(DW(\bar{x})[h])\\ &{}=&{}h^\top D^2_{xx} L(\bar{x},\Omega ) h \\ &{}&{}+\, 2 \omega \bullet (G(\bar{x})DW(\bar{x})[h])^\top G(\bar{x})^{\dag }(G(\bar{x})DW(\bar{x})[h]). \end{array} \end{aligned}$$

Taking again into account (32), it follows that

$$\begin{aligned} \begin{array}{rcl} h^\top \left( D^2_{xx} L^R (\bar{x} , \omega ) \right) h &{}=&{} h^\top D^2_{xx} L(\bar{x},\Omega ) h \\ &{}&{}+\, 2 \omega \bullet W(\bar{x})^\top (DG(\bar{x})[h])G(\bar{x})^{\dag }(DG(\bar{x})[h])W(\bar{x})\\ &{}=&{}h^\top D^2_{xx} L(\bar{x},\Omega ) h \\ &{}&{}+\, 2(W(\bar{x}) \omega W(\bar{x})^\top )\bullet (DG(\bar{x})[h])G(\bar{x})^{\dag }(DG(\bar{x})[h])\\ &{}=&{}h^\top D^2_{xx} L(\bar{x},\Omega ) h \\ &{}&{}+\, 2{\Omega }\bullet (DG(\bar{x})[h])G(\bar{x})^{\dag }(DG(\bar{x})[h]). \end{array} \end{aligned}$$

Recalling the definition of $H(\bar{x}, \Omega )$ in (9), we proved that for all $h \in T (\bar{x})$:

$$\begin{aligned} h^\top \left( D^2_{xx} L^R (\bar{x} , \omega ) \right) h = h^\top \left( D^2_{xx} L(\bar{x},\Omega ) + H(\bar{x},{\Omega }) \right) h. \end{aligned}$$

$\square$