Convergence of the LR algorithm for $2\times 2$ SPD matrices I've been asked to prove that the following iterations converge to the eigenvalues of SPD $A_0 \in \mathbb{R}^{n \times n}$
$A_0 = \begin{bmatrix}a & b\\ b & c \end{bmatrix}$  
with $a \geq c$ and $\lambda_1 \geq \lambda_2 > 0$
for k = 1,2,3 $\ldots$
$\qquad A_{k-1} = G_kG_k^T$
$\qquad A_{k} = G_k^TG_k$
end
I can show that this iteration is valid and that every step is a similarity transform. My initial attempt was to brute force and calculate $G_k$ for every iteration but this gets ugly very quickly. What is a good proof strategy to show that $A_k \rightarrow diag(\lambda_1,\lambda_2)$ ?
 A: Consider $A_k:=\pmatrix{a_k&b_k\\b_k&c_k}$ with $a_0:=a$, $b_0:=b$, $c_0:=c$. Since (with a slight change in the original notation) the Cholesky factor $L_k$ of $A_k$ is
$$
L_k=\pmatrix{a_k^{1/2}&0\\b_k/a_k^{1/2}&(c_k-b_k^2/a_k)^{1/2}},
$$
$$
A_{k+1}=L_k^TL_k=\pmatrix{a_k+b_k^2/a_k&b_k(c_k-b_k^2/a_k)^{1/2}/a_k^{1/2}\\b_k(c_k-b_k^2/a_k)^{1/2}/a_k^{1/2}&c_k-b_k^2/a_k}.
$$
Note that if $a_k\geq c_k$ (this is true for $A_0$), then $a_{k+1}\geq c_{k+1}$ since $b_k^2/a_k\geq 0$. This is hence true for all matrices $A_k$ (using induction). It also shows that if $A_k$ converge to a diagonal matrix, the limit is given by $\mathrm{diag}(\lambda_1,\lambda_2)$. 
We show that the off-diagonal entry is monotonically decreasing, that is, $(A_{k+1})_{2,1}=\alpha_k(A_{k})_{2,1}$ with $\alpha_k<1$. Indeed,
$$
\alpha_k^2=(c_k-b_k^2/a_k)/a_k\leq 1-\frac{b_k^2}{a_kc_k}=\frac{\det(A)}{a_kc_k}<1
$$
since $b_k^2-a_kc_k>0$ ($A_k$ is SPD) and $a_k\geq c_k$. This does not yet show that the method converges. We can, however, bound $\alpha_k$ independently of $k$. Note that $a_kc_k=(a+t)(c-t)=ac+(c-a)t-t^2$ for some $t\geq 0$. Since $a\geq c$, we have $a_kc_k\geq ac$ and therefore
$$
(A_k)_{2,1}\leq\left(1-\frac{b^2}{ac}\right)^k\rightarrow 0\quad\text{as}\quad k\rightarrow\infty.
$$
