I am trying to understand how you determine if LU decomposition is possible on a given matrix. I believe the way to calculate this is to check if the leading-matrices have non-zero determinants. I have performed the calculations for the matrix below to determine if we can utilise LU decomposition. I am wondering if someone can confirm if this is the correct way of determining if LU decomposition can be used on the given matrix.
\begin{equation} A= \left[ { \begin{array}{ccc} 5 & 6 & 2.3 & 6\\ 9 & 2 & 3.5 & 7\\ 3.5 & 6 & 2 & 3\\ 1.5 & 2 & 1.5 & 6\\ \end{array} } \right] \end{equation}
Determine if LU decomposition is possible by checking determinant of leading submatrices are not $0$
\begin{equation} A_1 = 5 \end{equation} \begin{equation} A_2= \left[ { \begin{array}{ccc} 5 & 6 \\ 9 & 2 \\ \end{array} } \right] = |A_2| = (5*2) - (6*9) = -44 \end{equation}
\begin{equation} A_3= \left[ { \begin{array}{ccc} 5 & 6 & 2.3 \\ 9 & 2 & 3.5 \\ 3.3 & 6 & 2 \\ \end{array} } \right] = |A_3| = -14.68 \end{equation}
\begin{equation} |A_4|= 5 \left[ { \begin{array}{ccc} 2 & 3.5 & 7 \\ 6 & 2 & 3 \\ 2 & 1.5 & 6 \\ \end{array} } \right], -6 \left[ { \begin{array}{ccc} 9 & 3.5 & 7 \\ 3.5 & 2 & 3 \\ 1.5 & 1.5 & 6 \\ \end{array} } \right], 2.3 \left[ { \begin{array}{ccc} 9 & 2 & 7 \\ 3.5 & 6 & 3 \\ 1.5 & 2 & 6 \\ \end{array} } \right], -6 \left[ { \begin{array}{ccc} 9 & 2 & 3.5 \\ 3.5 & 6 & 2 \\ 1.5 & 2 & 1.5 \\ \end{array} } \right] = 116.1 \end{equation}
In this instance, since none of the leading matricies have a determinant of $0$ this would be a valid matrix to perform LU decomposition on?
Suppose that $A$ is an $n \times n$ real nonsingular matrix.
If all the $k \times k$ principal minors $A_k$, ($k = 1, 2, \ldots, n$) of $A$ are non-singular, then $A$ has a $LU$-decomposition $$ A = L U $$ where $L$ is lower-triangular and $U$ is upper-triangular.
In this $LU$ decomposition of $A$, $L$ is unique if we demand that the diagonal elements of $L$ consist of all $1$'s, i.e. $\mbox{diag}(L) = (1, 1, \ldots, 1)$.
For the given matrix $A$,
$A_1 = 5 \neq 0$.
$A_2 = -44 \neq 0$.
$A_3 = -11.4 \neq 0$.
$A_4 = \mbox{det}(A) = -116.1 \neq 0$.
Thus, $A$ has a $LU$ factorization which can be easily computed as in the Gaussian elimination procedure.