Prove that the Cholesky Decomposition for a positive definite matrix is unique i'm working on a homework problem and I have a few ideas where to start but i'm still a bit confused.
The problem is:
Prove that the Cholesky decomposition for a positive denite matrix is always unique.
I know that when we're trying to prove that something is unique we can use a proof by contradiction something like:
Let A not equals to B, then go through the proof to show that A = B, but i'm not sure where to start. 
Can anyone help?
 A: I can outline a general proof strategy: Let $A$ be our positive definite matrix. Suppose $A$ has Cholesky factorizations $A = R^{\ast} R = S^{\ast} S$, for $R,S$ upper-triangular matrices with positive diagonal entries. Then we can write
$$ \langle Ax, x \rangle = \langle Rx, Rx \rangle = \langle Sx, Sx \rangle $$
Pick $x = e_1$, the first coordinate vector. Then $\langle Ax, x \rangle = A_{11} = \|Re_1\|^2 = \|Se_1 \|^2$, and since $R,S$ are upper triangular, this uniquely defines the upper left entry of $R,S$, so they must be the same. Now, the $k$th entry in the first row of $R$ is given by
$$ \langle Re_k, e_1 \rangle = \frac{1}{\sqrt{A_{11}}} \langle Re_k, Re_1 \rangle = \frac{1}{\sqrt{A_{11}}} \langle Ae_k, e_1 \rangle $$
so we in fact know the whole first row of $R$ and $S$, and they must be the same. Now, we reduce our $A$ to a new $(n-1) \times (n-1)$ submatrix, which will also be positive definite, and then repeat the procedure to uniquely determine the subsequent rows. This inductive process demonstrates that $R = S$. I've left out a good number of details that you'll need to sort out, but it's worth learning this proof.
