I was reading the Bulirsch and Stoer's Introduction to Numerical Analysis proof for the Cholesky decomposition, you can find a copy here, and I got stuck.
My problem is that, in my opinion, there's a point in this proof where it uses a false argument. Shortly, in the proof we have a positive-definite matrix $A \in \mathbb C^{n \times n}$ and we are looking for a lower triangular matrix $L$ which satisfies $L·L^* = A$. With this objective, we write $$ L = \begin{pmatrix} \widetilde L & 0\\ c^* & \alpha \\ \end{pmatrix} $$ such that $$ L\cdot L^*= \begin{pmatrix} \widetilde L\cdot\widetilde L^* & \widetilde Lc \\ c^*\widetilde L^* & c^*c+\alpha^2 \end{pmatrix} = \begin{pmatrix} \widetilde A & b \\ b^* & a_{nn} \end{pmatrix} =A $$
We know $\widetilde L$ exists because of the induction hypothesis and $c^*$ exists because $\widetilde L$ is nonsingular, so $\widetilde Lc =b$ has a unique solution. Therefore, we need only find an $\alpha$ such that $c^*c + \alpha^2 = a_{nn}$, $\space \alpha > 0$. However, to show that such an $\alpha$ exists, this is to say, to show that $a_{nn} - c^*c>0$, it argues that $\det(A) = |\det(\widetilde L)|^2\alpha^2 > 0$, and therefore $\alpha^2 = a_{nn} - c^*c>0$. To me, this argument is invalid because it is first supposing that such an $\alpha$ exists, otherwise we couldn't decompose $\det(A)$ in such a way, and under that assumption it follows that $\alpha^2>0$, but it is precisely that very assumption that we want to show is true, so I don't understand how simply supposing it does the job.
Thanks for the help!
You are right that the authors' argument is invalid. One may prove that $a_{nn}>\|c\|^2$ by using the positive definiteness of $A$: $$ \begin{aligned} 0&<\pmatrix{c^\ast \widetilde{L}^{-1}&-1} \pmatrix{\widetilde{A}&b\\ b^\ast&a_{nn}} \pmatrix{(\widetilde{L}^\ast)^{-1}c\\ -1}\\ &=\pmatrix{c^\ast \widetilde{L}^{-1}&-1} \pmatrix{\widetilde{L}\widetilde{L}^\ast&\widetilde{L}c\\ c^\ast\widetilde{L}^\ast&a_{nn}} \pmatrix{(\widetilde{L}^\ast)^{-1}c\\ -1}\\ &=\pmatrix{c^\ast \widetilde{L}^{-1}&-1} \pmatrix{0\\ \|c\|^2-a_{nn}}\\ &=a_{nn}-\|c\|^2. \end{aligned} $$ A different but closely related approach is to consider the Schur complement $S$ of $\widetilde{A}$ in $A$. Since $A$ is congruent to $\widetilde{A}\oplus S$, the Schur complement $S$ must be positive. Now the result follows from the fact that $S=a_{nn}-b^\ast\widetilde{A}^{-1}b=a_{nn}-\|c\|^2$.