Minimization of positive quadratic function using gradient descent in at most $ n $ steps For minimization positive quadratic form $$f = \frac{1}{2}\left\langle Ax,x \right\rangle - \left\langle b,x\right\rangle \rightarrow \min_{x\in\mathbb{R}^n},$$ we use gradient descent  $$x^{k+1} = x^{k} - \alpha_k \nabla f(x^k)$$ with step $\alpha_k = \frac{1}{\lambda_{k+1}}$ , where $\lambda_{k+1}$ is an eigenvector of $A$ $(0 < \mu = \lambda_1 \leqslant \cdots \leqslant \lambda_n = L)$. I need to prove that $x^n = x^*$, where $Ax^* = b$.

I have an idea, that I need to go to basis, where $A$ becomes diagonal, $A = P^T \Lambda P$, where $\Lambda = \text{diag} \{\lambda_1, \ldots, \lambda_n \}$ and $P$ consists of eigenvectors.
I tried to  express $x^* - x^0$ and $x^n - x^0$ as a linear combination of basis vectors to compare them, but didn't succeed. Could you please help me with that proof?

You can assume that you have an orthonormal basis of eigenvectors so that $A$ is
diagonal, with the eigenvalues $\lambda_k$ on the diagonal. Let me use $\Lambda$
for $A$ in this basis, just for emphasis.
Note that the solution is given by $x^*= \Lambda^{-1} b$.
Then $x_{k+1} = x_k -{1 \over \lambda_{k+1}} (\Lambda x_k -b) = (I-{1 \over \lambda_{k+1}} \Lambda) x_k + {1 \over \lambda_{k+1}} b$, for $k=0,...,n-1$.
Note that the $k+1$th entry of $x_{k+1}$ satisfies $[x_{k+1}]_{k+1} = 0 + [x^*]_{k+1}$, and if $[x_k]_i = [x^*]_i$ for $i=0,...,k$, then 
$[x_{k+1}]_i = [x^*]_i$ for $i=0,...,k$.
Hence $x_n = x^*$.
