Show that $A$ is positive definite via the Cholesky decomposition

Question

I have calculated the Cholesky decomposition of the matrix \begin{equation*}A=\begin{pmatrix}1 & -1 & -1 & 0 \\ -1 & 5 & 5 & -4 \\ -1 & 5 & 6 & -3 \\ 0 & -4 & -3 & 6\end{pmatrix}\end{equation*} which is \begin{equation*}A=\tilde{L}\tilde{L}^T=\begin{pmatrix}1 & 0 & 0 & 0 \\ -1 & 2 & 0 & 0 \\ -1 & 2 & 1 & 0 \\ 0 & -2 & 1 & 1\end{pmatrix}\begin{pmatrix}1 & -1 & -1 & 0 \\ 0 & 2 & 2 & -2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix}\end{equation*}

Now I want to show that $A$ is positiv definite.

Can we do the following?

For $x\neq 0$ we have: $$x^TAx=x^T\tilde{L}\tilde{L}^Tx=x^T\tilde{L}I\tilde{L}^Tx=(\tilde{L}^Tx)^TI\tilde{L}^Tx>0$$ since the eigenvalues of $I$ are positive.

Is this proof correct?

Answer 1

No, you need to show that $x^T A x > 0$ for all $x \not = 0$ and you are not using that $L = \tilde{L}$ is nonsingular.

You have $$x^TAx = x^T LL^T x = (L^Tx)^T(L^Tx) = \|L^Tx\|_2^2 \ge 0$$ for any $x$. Moreover, if $x^T A x = 0$, then $L^Tx = 0$ which implies that $x=0$ because $L$ and hence $L^T$ are nonsingular.