Deflation


Consider the solution $\x_0 = 0\in\bC^2$ of

(1)
\begin{align} F = \begin{bmatrix}x_1^3+x_1 x_2^2\\ x_1 x_2^2 + x_2^3\\ x_1^2x_2+x_1x_2^2 \end{bmatrix} = 0 \end{align}

(a) Verify that $\x_0$ is a singular solution.

(2)
\begin{eqnarray} \nonumber DF(x) =  \begin{bmatrix} 3x_{1}^{2} + x_{2}^{2} & 2x_{1}x_{2} \\ x_{2}^{2} & 2x_{1}x_{2} + 3x_{2}^{2} \\ 2x_{1}x_{2} + x_{2}^{2} & x_{1}^{2} + 2x_{1}x_{2} \\ \end{bmatrix} \end{eqnarray}

By inspection, $DF(0,0)$ is identically the zero matrix so $x_{0} = (0,0)$ is a singular solution since the matrix has zero rank.

(b) Construct $\Dfl(F)$ with your favorite choice of random ingredients. Is $(\x_0,\bflambda_0) = \lift(\x_0)$ a regular solution?

We must deflate twice. We will append equations to $F$ to construct $Dfl(F,...)$.

(3)
\begin{eqnarray} \nonumber \nonumber Dfl(F,...) = \begin{bmatrix} F \\ DF \begin{bmatrix} \lambda_{1} \\ \lambda_{2} \end{bmatrix} \\ \lambda_{1} + \lambda_{2} - 3 \\ 2\lambda_{1} + \lambda_{2} - 3 \end{bmatrix} = 0 \end{eqnarray}

Notice that $\lambda_{1} = 2$ and $\lambda_{2} = 1$ supply a solution to $Dfl(F,...)$ and $\begin{bmatrix} 2 \\ 1 \end{bmatrix}$ is in the nullspace of $Df$.
Substituting $\lambda_{1} = 2$ and $\lambda_{2} = 1$ into $Dfl(F,...)$ we obtain:

(4)
\begin{eqnarray} \nonumber Dfl(F,DF,2,1) = \begin{bmatrix} x_{1}^{3} + x_{1}x_{2}^{2} \\ x_{1}x_{2}^{2} + x_{2}^{3} \\ x_{1}^{2}x_{2} + x_{1}x_{2}^{2} \\ 6x_{1}^{2} + 2x_{2}^{2} + 2x_{1}x_{2} \\ 5x_{2}^{2} + 2x_{1}x_{2}\\ 6x_{1}x_{2} + 2x_{2}^{2} + x_{1}^{2} \end{bmatrix} = 0 \end{eqnarray}

We want to check if $(0,0)$ is nonsingular so we compute $D(Dfl)$ at $(0,0)$:

(5)
\begin{eqnarray} \nonumber D(Dfl(F,DF)) = \begin{bmatrix} 3x_{1}^{2} + x_{2}^{2} && 2x_{1}x_{2} \\ x_{2}^{2} && 2x_{1}x_{2} + 3x_{2}^{2} \\ 2x_{1}x_{2} + x_{2}^{2} && x_{1}^{2} + 2x_{1}x_{2} \\ 12x_{1} + 2x_{2} && 4x_{2} + 2x_{1} \\ 2x_{2} && 10x_{2} + 2x_{1} \\ 6x_{2} + 2x_{1} && 6x_{1} + 4 x_{2} \end{bmatrix}\ = 0 \end{eqnarray}

The matrix has corank $2$ at $(0,0)$. We need to deflate again so the form is:

(6)
\begin{eqnarray} \nonumber Dfl^{2}(F,DF,Dlf,...) = \begin{bmatrix} F \\ DF\begin{bmatrix} 2 \\ 1 \end{bmatrix} \\ D(Dfl) \begin{bmatrix} \mu_{1} \\ \mu_{2} \end{bmatrix} \\ \mu_{1} + \mu_{2} - 1 \\ 2 \mu_{1} + \mu_{2} - 2 \end{bmatrix} = 0 \end{eqnarray}

$\mu_{1} = 1$ and $\mu_{2} = 0$ supply a solution to $Dfl^{2}(F,...)$ so we obtain:

(7)
\begin{eqnarray} \nonumber Dfl^2(F,DF,1,0) = \begin{bmatrix} x_{1}^{3} + x_{1}x_{2}^{2} \\ x_{1}x_{2}^{2} + x_{2}^{3} \\ x_{1}^{2}x_{2} + x_{1}x_{2}^{2} \\ 6x_{1}^{2} + 2x_{2}^{2} + 2x_{1}x_{2} \\ 5x_{2}^{2} + 2x_{1}x_{2}\\ 6x_{1}x_{2} + 2x_{2}^{2} + x_{1}^{2} \\ 3x_{1}^{2} + x_{2}^{2} \\ x_{2}^{2} \\ 2x_{1}x_{2} + x_{2}^{2} \\ 12x_{1} + 2x_{2} \\ 2x_{2} \\ 6x_{2} + 2x_{1} \end{bmatrix} = 0 \end{eqnarray}

We can observe the positive rank from this system. The hope is for full rank:

(8)
\begin{eqnarray} \nonumber D(Dfl^{2}) = \begin{bmatrix} 3x_{1}^{2} + x_{2}^{2} && 2x_{1}x_{2} \\ x_{2}^{2} + 3x_{2}^{2} && 2x_{1}x_{2} + 3x_{2}^{2} \\ 2x_{1}x_{2} + x_{2}^{2} && x_{1}^{2} + 2x_{1}x_{2} \\ 12x_{1} + 2x_{2} && 4x_{2} + 2x_{1} \\ 2x_{2} && 10x_{2} + 2x_{1} \\ 6x_{2} + 2x_{1} && 6x_{1} + 4x_{2} \\ 6x_{1} && 2x_{2} \\ 0 && 2x_{2} \\ 2x_{2} && 2x_{1} + 2x_{2} \\ 12 && 2 \\ 0 && 2 \\ 2 && 6 \\ \end{bmatrix} = 0 \end{eqnarray}

After substituting $(0,0)$ we observe there are $2$ linearly independent column vectors so we have deflated sucessfully.

Using the system $Dfl^{2}(F,DF,1,0)$ above we now construct a numerically stable homotopy (at least at $(0,0)$). Observe that $(0,0)$ will still be a solution to $Dfl^{2}(F,...)$. The benefit however is that Newton's method in the predictor-corrector step of the numerical homotopy will converge quadratically in a neighborhood around $(0,0)$.

(c) If so, construct $\Dfl^d(F)$ for $d\geq 2$. Is $\lift^d(\x_0)$ regular?

(See above argument)