# Proving that $\|Av\|\geq \lvert \langle u, v\rangle\rvert\cdot \|A\|$ for $\|Au\| = \|A\|$ in a Hilbert space

I've shown that for a matrix $$A\in \mathbb{R}^{n\times n}$$ and an arbitrary $$v\in \mathbb{R}^n$$, we have the inequality $$\|Av\|\geq \lvert \langle u, v\rangle\rvert\cdot \|A\|$$ where $$\|\cdot\|$$ is the $$2$$-norm and $$u$$ satisfies $$\|u\| = 1$$ and $$\|Au\| = \|A\|$$ (the operator norm of $$A$$ corresponding to the vector $$2$$-norm). To do this, I used the singular value decomposition. However, I was wondering if there's a different proof that relies on more general properties of Hilbert spaces (under the assumption that $$u$$ exists in the more general setting where $$\langle \cdot, \cdot\rangle$$ and $$\|\cdot\|$$ are the Hilbert space inner product and norm). My own short proof is below:

We first handle the case where $$A = \operatorname{diag}(d_1, \ldots, d_n)$$ for some $$d_1\geq \cdots\geq d_n\geq 0$$. Then, $$u = e_1$$, so $$\lvert \langle u, v\rangle\rvert^2\|A\|^2 = d_1^2v_1^2\leq \sum_{i=1}^n d_i^2v_i^2 = \|Av\|^2$$ Now, we consider general $$A$$, which has a singular value decomposition $$A = U\Sigma V^{\mathrm{T}}$$ for $$U$$ and $$V$$ orthogonal and $$\Sigma$$ diagonal in the above form. First, we recall that multiplication by orthogonal matrices preserves the dot product, i.e. $$\langle Qx, Qy\rangle = \langle x, y\rangle$$ for orthogonal $$Q$$. This allows us to write $$\|Av\|^2 = \langle U\Sigma V^{\mathrm{T}}v, U\Sigma V^{\mathrm{T}}v\rangle = \langle \Sigma V^{\mathrm{T}}v, \Sigma V^{\mathrm{T}}v\rangle = \|\Sigma V^{\mathrm{T}}v\|^2$$ for arbitrary $$v\in \mathbb{R}$$. Furthermore, as $$\|Au\|^2 = \|\Sigma V^{\mathrm{T}}u\|^2$$ and $$\|V^{\mathrm{T}}u\|^2 = \|u\|^2 = 1$$, we have that $$\|A\| = \|\Sigma\|$$ and that $$y = V^{\mathrm{T}}u$$ satisfies $$\|\Sigma y\| = \|\Sigma\|$$. Combining our earlier results, $$\|Av\|^2 = \|\Sigma V^{\mathrm{T}}v\|^2\geq \lvert\langle y, V^{\mathrm{T}}v\rangle\rvert^2\|\Sigma\|^2 = \lvert\langle V^{\mathrm{T}}u, V^{\mathrm{T}}v\rangle\rvert^2\|A\|^2 = \lvert\langle u, v\rangle\rvert^2\|A\|^2$$

An easier proof is to use the fact that for any $$y\in\mathbb R^n$$, if $$y\perp u$$, then $$Ay\perp Au$$. Now we can write $$v=\langle u,v\rangle u+y$$ for some $$y\perp u$$. Let $$x_1=\langle u,v\rangle Au$$ and $$x_2=Ay$$. Then we have that $$x_1\perp x_2$$ and $$Av=x_1+x_2$$. So $$\|Av\|\ge\|x_1\|=|\langle u,v\rangle|\|Au\|=|\langle u,v\rangle|\|A\|$$.
Hint for proof of the fact: Suppose that there is some $$y\perp u$$ with $$\langle Ay,Au\rangle\neq0$$. We may assume that $$\|y\|=1$$. Show that there exists some $$\theta$$ such that $$u':=u\cos\theta+y\sin\theta$$ satisfies $$\|Au'\|>\|Au\|$$. This will give us a contradiction.