There are a lot of details to get right in this proof, so let's try to walk through it together.
mathmari said:
$\Rightarrow$ :
Let $x\neq 0$. Then we have that:
\begin{equation*}x^{T}Ax=x^{T}L^{T}DLx=(Lx)^TDLx\end{equation*}
It holds that $(Lx)^TDLx>0$ since the eigenvalues of $D$, so the elements of the diagonal, are positiv.
So $x^{T}Ax>0$, and therefore $A$ is positive definite, right? :unsure:
The basic idea you've got here is correct, nicely done. Let's walk through it a bit more carefully though. Essentially there are three matrix multiplications that must happen:
- $Lx$
- $D(Lx)$
- $(Lx)^{T}D(Lx)$
As you've correctly indicated, the fact that $D$ is a diagonal matrix with positive entries is important for parts (2) and (3). However, how do we know that $Lx\neq 0$? We need to come up with a reason why $Lx\neq 0$, otherwise $(Lx)^{T}D(Lx)$ could equal zero, even though $D$ is a diagonal matrix with positive entries.
mathmari said:
$\Leftarrow$:
If $A$ is invertible then the diagonal elements have to be non-zero, right? For this reason must the diagonal elements of $L_0$ be non-zero?
I got stuck right now. We want to show that there is a matrix $L_0$ such that $A=L_0^TL_0$, or not? Can we either way have the result that the diagonal elements of $L_0$ must be non-zero?
We need to be careful here as well. To answer your question about $A$, consider the matrix $$A=\begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix},$$ which corresponds to a reflection about the line $y=x$. This matrix swaps the basis vectors with one another and has non-zero determinant, so it is invertible. However, its main diagonal is zero. Hence, we arrive at an important conclusion: $A$ invertible does not imply the diagonal elements of $A$ need to be non-zero.
To answer your question about $L_{0}$: The matrix $L_{0}$ comes to us from the Cholesky factorization theorem, we do not need to show its existence. The Cholesky factorization theorem for a symmetric, positive definite matrix -- which we are now assuming $A$ is -- says that there is $L_{0}$ such that $A = L_{0}^{T}L_{0}$. See the "Statement" section of
Cholesky Decomposition Theorem; specifically "When $A$ is symmetric,..."
Going back to step (1) in my previous post, we need to establish a few things:
- $A$ positive definite implies $A$ invertible;
- $A$ invertible implies the diagonal elements of $L_{0}$ are non-zero.
For the first bullet point, suppose $A$ is not invertible. Then there is a non-zero $x$ such that $Ax = 0.$ Can you use the assumption that $A$ is positive definite to derive a contradiction from here?
For the second bullet point, the fact that $A$ is invertible means that its determinant is non-zero. Hence, $0\neq \det(A) = \det(L_{0}^{T}L_{0}) = \det(L_{0})^{2}$ (because $\det(L_{0}^{T}) = \det(L_{0})$). Taking a square root, we see it follows that $\det(L_{0})\neq 0$. So we now know $\det(L_{0})\neq 0$ and that $L_{0}$ is an upper triangular matrix. Do you know what the relationship between these two facts is? In other words, given that $\det(L_{0})\neq 0$ and $L_{0}$ is upper triangular, can we conclude anything about the diagonal elements of $L_{0}$?
As I mentioned above, there are many details to go through on this proof. In an effort to not overload you, I will stop here for now and wait to hear what other questions you have. We'll get there!