Let $A \in M_{n,n}(\Bbb R)$.
In the lecture notes given to us, it's mentioned that:
If $A$ is invertible, then it has an LU decomposition iff its leading minors are non-zero.
Then, as a corollary, that:
If $A$ is symmetric and PD, then it has an LU factorization.
I do not see why "symmetric" is involved here. We know that a PD is invertible. Isn't it true that if a matrix is PD, then its leading minors are strictly positive? If we look at:
For all $1 \le k \le n$ and for all $x \in \Bbb R^k – \{0\}$, with $x^T = (x_1,…,x_k)$, we have:
$$\langle A_k x, x \rangle = \langle A \tilde x, \tilde x \rangle > 0$$,
Where $\tilde x^T = (x_1,…,x_k,0,…,0)$.
So why is "symmetric" involved? Can't we just say:
If $A$ is PD, then $A$ is LU-factorable?
Best Answer
Yes, it's true that the corollary also holds for non-symmetric PD matrices. However, it probably turns out that the author/teacher is interested in symmetric PD matrices in particular, so that's all that is being mentioned. In a lot of times, we "don't want to bother thinking about" non-symmetric PD matrices.
I expect that this is leading up to the Cholesky decomposition, which only applies to symmetric PD matirces.