Help with the proof that $\min \{ (Ax,x):|x| = 1\}$ is an eigenvalue of $A$ if $A$ is self-adjoint

Question

Given a self-adjoint operator $A$ defined on a finite-dimensional inner product space, the problem is to prove $\min \{ (Ax,x):|x| = 1\}$ is actually an eigenvalue of $A$. Here $|x|$ denotes the norm induced from the inner product.

The following is a proof from a text book which I have trouble with.

This is the statement of the formal theorem and the first part, which I understand. It just proves $\inf \{ (Ax,x):|x| = 1\} = \min \{ (Ax,x):|x| = 1\}$. My problem is in the second part below.

This is the second part where I have a problem. My confusion is that the minimum is restricted over $|x|=1$, while I think $|v_1+tw| > 1$ for $t>0$ if $|v_1|=1$, so in plain words, $v_1+tw$ is not inside the restriction of the minimum for any $t>0$. In this case how can we take derivative of that function $f$ and say the derivative is zero at the minimum?

Answer 1

You can rewrite $f$ using the properties of the scalar product as

$$ f(t) = \left< A \left( \frac{v_1 + t w}{|v_1 + tw|} \right), \frac{v_1 + tw}{|v_1 + tw|} \right>. $$

Then for each $t$, the vector $x = \frac{v_1 + tw}{|v_1 + tw|}$ is of unit length.