A simple proof for the relationship between the eigenvalues of a positive definite matrix and its Cholesky decomposition

Question

Can anyone present to me an elegant elementary proof of the relationship between the eigenvalues of a positive definite matrix and its Cholesky decomposition?

More formally, suppose $\mathbf{A}$ is an $n\times n$ positive definite matrix and let $\mathbf{A} = \mathbf{R}^\top \mathbf{R}$ be its Cholesky decomposition. Establish the relationship between the eigenvalues of $\mathbf{A}$ and that of $\mathbf{R}$.

EDIT (Additional remarks): My question specifically wants to find, if possible, an equation or function, say $f$, that relates the eigenvalues, i.e., $f\left(\lambda_i(\mathbf{R})\right) = \lambda_i(\mathbf{A})$, with uniqueness up to order being considered if necessary.

Answer 1

There is no such relation. If the spectrum of $A$ is a function of the spectrum of $R$, it would imply that $$ A=\pmatrix{1&0\\ t&1}\pmatrix{1&t\\ 0&1}=\pmatrix{1&t\\ t&t^2+1} $$ has a constant spectrum, but this is obviously not the case because our $A$ here has a non-constant trace.

In general, if $A=R^TR$ (regardless of whether this is a Cholesky decomposition or not) for a real square matrix $R$, the eigenvalues of $A$ are the squared singular values of $R$.

Answer 2

For a positive definite matrix $A$, with $Q$ as eigenvector matrix and $\Lambda$ as eigenvalue matrix, we have

$$ A = Q \Lambda Q^T $$

This can be rewritten as (since all eigenvalues of $A$ are positive) :

$$ A = (Q \sqrt{\Lambda}) (\sqrt{\Lambda} Q^T) $$

So for $A = R^TR$, $R$ can be a matrix such that,

$$ R = \sqrt{\Lambda} Q^T $$

Also we can multiply any orthogonal matrix $Q$ to this $R$ without changing the original $A = R^TR$ condition, because,

$$ A = (QR)^TQR = R^T(Q^TQ)R = R^TR $$

So rewriting $R$ as $Q\sqrt{\Lambda} Q^T$, we see that eigenvalues of $R$ are square roots of eigenvalues of $A$

Answer 3

Start with Cholesky form: $$A = R^TR$$ let $R = DL$ where $D$ is a diagonal matrix such that $D_{i, i} = R_{i, i}$, then $L_{i, i} = 1$ and $L$'s rows are $R$ divided by diagonals of $D$, which means $\text{det}(L) = 1$

Consider $A = (DL)^T(DL) = L^TD^2L$, then $\text{det}(A) = \text{det}(L)^2\text{det}(D)^2 = \lambda_1\lambda_2\cdots\lambda_n$.

Basically: $\prod_{i = 1}^n(D_{i, i})^2 = \prod_{i = 1}^n\lambda_i = \prod_{i = 1}^n(R_{i, i})^2$