I cannot normalize a real quadratic form $F(x,y,z)$ \begin{eqnarray} F(x,y,z) &=& ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy + 2lx + 2my + 2nz + d. \end{eqnarray} Here, the principal matrix $A$, and matrix of the quadratic form $\tilde{A}$ are as follows: \begin{eqnarray} A &=& \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{pmatrix} \\ \tilde{A} &=& \begin{pmatrix} a & h & g & l \\ h & b & f & m\\ g & f & c & n\\ l & m & n & d \\ \end{pmatrix} \end{eqnarray}
If rank($A$) = 1 and rank($\tilde{A}$) = 3, it should be parabolic pillar, and the quadratic form becomes \begin{eqnarray} F(x', y', z') = \lambda x'^2 + r z' \end{eqnarray} but I can't show why the coefficient of $y'$ becomes zero.
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My understanding is below. Because rank($A$) is 1, there is a non-zero eigen value $\lambda$ and there is a regular matrix $P$ such that \begin{eqnarray} A' = {}^t P A P = \begin{pmatrix} \lambda & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\\ \end{pmatrix}. \end{eqnarray} Then by using variable conversion ($x',y',z'$) and constants ($p, q, r$) like \begin{eqnarray} \begin{pmatrix} x' \\ y' \\ z' \\ \end{pmatrix} = P^{-1}\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix}, \quad \begin{pmatrix} p \\ q \\ r \\ \end{pmatrix} = {}^t P \begin{pmatrix} l \\ m \\ n\\ \end{pmatrix}, \end{eqnarray} the quadratic form becomes \begin{eqnarray} F(x',y',z') = \lambda x'^2 + 2 p x' + 2 q y' + 2 r z' + d \end{eqnarray} and the matrix of this quadratic form is \begin{eqnarray} \tilde{A'} = \begin{pmatrix} \lambda & 0 & 0 & p \\ 0 & 0 & 0 & q \\ 0 & 0 & 0 & r \\ p & q & r & d \\ \end{pmatrix}. \end{eqnarray} If ${\rm rank}(\tilde{A}) = 3$, ${\rm rank}(\tilde{A'})$ is also 3, therefore either of the determinant of 3x3 minor matrix \begin{eqnarray} \begin{vmatrix} \lambda & 0 & p \\ 0 & 0 & r \\ p & r & d \end{vmatrix}, \quad \begin{vmatrix} \lambda & 0 & p \\ 0 & 0 & q \\ p & q & d \end{vmatrix} \end{eqnarray} is not zero. Hence, $r$ or $q$ is not zero at least. However, I cant show either of $r, q$ is zero.