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The successive differences of the constant entries give a simple LDL decomposition:For example, entry $(i,j)$ of $LD{L}^T$ is the sum of the first $\min (i,j)$ entries of $D$, directly verifying every entry of $A$. Because $L$ is an invertible matrix, $x^{T}Ax=(L^{T}x{)}^{T}D(L^{T}x)$ is strictly positive for every nonzero vector exactly when every diagonal entry of $D$ is positive. ConsequentlyAt $\lambda =14$ it is a positive semidefinite matrix and singular; for $\lambda <14$ its quadratic form takes both positive and negative values.

When $\lambda =30$, taking positive square roots of $D$ yields the Cholesky decomposition

Apply symmetric Gaussian elimination without row exchanges. At step $k$, let $d_k$ be the leading diagonal entry of the remaining symmetric Schur complement. If $d_k\le 0$, stop and report that $A$ is not positive definite. If $d_k>0$, use it to eliminate the rest of its row and column. If all steps succeed, this constructs an LDL decompositionwhere $L$ is unit lower triangular and $D=\mathrm{diag}(d_1,\dots ,d_n)$ has positive diagonal.

The test is correct from first principles. If all $d_k>0$, then for every nonzero $x$,because $L$ is invertible. Conversely, if $A$ is positive definite, its first pivot is $a_{11}>0$, and completing the square givesChoosing $s=-b^{T}y/a$ shows that the Schur complement is positive definite. Induction forces every pivot to be positive. This is also Sylvester's criterion.

At step $k$, updating the remaining matrix costs $O((n-k{)}^2)$ arithmetic operations. Hence the total iswhich proves the existence of the required algorithm.