A difficulty in obtaining a formula in the proof of a theorem.

Question

The theorem and part of its proof is given below:

But the last line in the proof did not come up with me after calculation, I do not know where I will use that $\lambda = <Ax_{0},x_{0}>$ , could anyone explain it for me please?

Also in the rest of the proof below:

I can not see how this $\alpha$ give us the final line, could anyone explain this for me please?

Answer 1

$$\langle A(x_0+\alpha v), x_0+\alpha v \rangle \geq \lambda \langle x_0+\alpha v, x_0+\alpha v \rangle\\ \langle Ax_0, x_0 \rangle+\langle Ax_0, \alpha v \rangle+\langle A\alpha v, x_0 \rangle+\langle A\alpha v, \alpha v \rangle \geq \lambda \left( \langle x_0, x_0 \rangle +\langle x_0, \alpha v \rangle+\langle \alpha v, x_0 \rangle+\langle \alpha v, \alpha v \rangle\right)\\ \lambda +\bar{\alpha}\langle Ax_0, v \rangle+\alpha \langle A v, x_0 \rangle+|\alpha|^2\langle Av , v \rangle \geq \lambda \left( 1 +\bar{\alpha}\langle x_0, v \rangle+\alpha\langle v, x_0 \rangle+|\alpha|^2\langle v, v \rangle\right)\\ \bar{\alpha}\langle Ax_0, v \rangle+\alpha \langle A v, x_0 \rangle+|\alpha|^2\langle Av , v \rangle \geq \lambda \left( \bar{\alpha}\langle x_0, v \rangle+\alpha\langle v, x_0 \rangle+|\alpha|^2\langle v, v \rangle\right)\\ $$

Now, using the fact that $A$ is self adjoint and that $\lambda=\bar{\lambda}$, we get $$\bar{\alpha}\langle Ax_0, v \rangle+\alpha \langle v, Ax_0 \rangle+|\alpha|^2\langle Av , v \rangle \geq \bar{\alpha}\langle \lambda x_0, v \rangle+\alpha\langle v, \lambda x_0 \rangle+|\alpha|^2\langle \lambda v, v \rangle\\ \overline{\alpha\langle v, Ax_0 \rangle}+\alpha \langle v, Ax_0 \rangle+|\alpha|^2\langle Av , v \rangle \geq \overline{\alpha\langle v, \lambda x_0 \rangle}+\alpha\langle v, \lambda x_0 \rangle+|\alpha|^2\langle \lambda v, v \rangle\\ $$

Now move everything on the left hand side and you obtain exactly the given inequality.

For the edit Taking $\alpha$ of the given form, the part inside the real part is real and your inequality becomes of the form $$\gamma r+ \delta r^2 \geq 0$$

Now, unless $\gamma=0$, the LHS has two real roots and hence takes both positive and negative values.

Answer 2

Here is another approach (essentially the proof showing the existence of a Lagrange multiplier):

Suppose $x_0$ is as above and $w_0$ is a unit vector that satisfies $x_0 \bot w_0$.

Note that $\|t x_0 + s w_0\|^2 = t^2+s^2$ and if we let $x(s) = \sqrt{1-s^2} x_0 + s w_0$ then $\|x(s) \| = 1$ for all $|s| <1$. In addition, $x(0) = x_0$ and $x'(0) = w_0$.

Now let $f(x) = \langle Ax, x \rangle$ and $\phi(s) = f(x(s))$. Note that $\phi'(0) = 2 \operatorname{re} \langle Ax_0, w_0 \rangle$, from which we conclude (since $x_0$ is a minimiser) that $\langle Ax_0, w_0 \rangle = 0$ for all $w_0 \bot x_0$. In particular, $A x_0 \in \operatorname{sp} \{ x_0 \}$, hence $x_0$ is an eigenvector.