there! I tried myself on the following exercise today and came up with the (hopefully correct) proof below. I would love to get some feedback if everything seems correct and whether there are any notational errors.
Let $V$ be a finite-dimensional, real vector space with inner product $\langle\cdot,\cdot\rangle$ and the norm $\left\lVert\cdot\right\rVert$ induced by it. Show that any projection $\varphi \colon V \to V$ is orthogonal if and only if it is nonexpanding, i.e. if $\left\lVert\varphi(v)\right\rVert \leq \left\lVert v \right\rVert$ holds true for $\forall v\in V$.
"$\Longrightarrow$": Let $\varphi$ be an orthogonal projection and $U = E_1(\varphi)$ the vector subspace such that $V = U \oplus U^\bot$ and $\forall (u + u^\bot) \in U \oplus U^\bot: \varphi(u + u^\bot) = u$. We know that $im(\varphi) = U$ and $ker(\varphi) = U^\bot$. Let $v = (u + u^\bot) \in V$, then:
$$\begin{align} \langle\varphi(v), v - \varphi(v)\rangle &= \langle\varphi(u + u^\bot), (u + u^\bot) - \varphi(u + u^\bot)\rangle \\ &= \langle u, u + u^\bot - u\rangle \\ &= \langle u, u^\bot\rangle \\ &= 0 \end{align}$$
Therefore: $$\begin{align} \varphi(v) \perp v - \varphi(v) &\implies \left\lVert\varphi(v)\right\rVert^2 + \left\lVert v - \varphi(v)\right\rVert^2 = \left\lVert\varphi(v) + v - \varphi(v) \right\rVert^2 \\ &\implies \left\lVert\varphi(v)\right\rVert^2 + \left\lVert v - \varphi(v)\right\rVert^2 = \left\lVert v \right\rVert^2 \\ &\implies \left\lVert\varphi(v)\right\rVert^2 \leq \left\lVert v \right\rVert^2 \\ &\implies \left\lVert\varphi(v)\right\rVert \leq \left\lVert v \right\rVert \end{align}$$
"$\Longleftarrow$": Let $\varphi$ be a projection and nonexpanding. We shall assume that $\varphi$ is not an orthogonal projection. We know that $V = im(\varphi) \oplus ker(\varphi)$ since $\varphi$ is a projection and since we assume $\varphi$ is not an orthogonal projection, $im(\varphi) \neq ker(\varphi)^\bot$. Therefore $\exists b \in im(\varphi), k \in ker(\varphi): b \not\perp k$, i.e. $\langle b, k \rangle \neq 0$. Let $\lambda \in \mathbb{R}$ be a real number and $v = b + \lambda k \in V$. Then:
$$\left\lVert \varphi(v) \right\rVert = \left\lVert \varphi(b + \lambda k )\right\rVert = \left\lVert \varphi(b) + \lambda \varphi(k) \right\rVert = \left\lVert \varphi(b) \right\rVert$$
Therefore:
$$\begin{align} & \left\lVert \varphi(b) \right\rVert > \left\lVert v \right\rVert \\ \Leftrightarrow & \left\lVert b \right\rVert > \left\lVert v \right\rVert \\ \Leftrightarrow & \left\lVert b \right\rVert^2 > \left\lVert v \right\rVert^2 \\ \Leftrightarrow & \langle b, b \rangle > \langle v, v \rangle \\ \Leftrightarrow & \langle b, b \rangle > \langle b + \lambda k, b + \lambda k \rangle \\ \Leftrightarrow & \langle b, b \rangle > \langle b, b \rangle + 2\langle b, \lambda k \rangle + \langle \lambda k, \lambda k \rangle \\ \Leftrightarrow & 0 > 2\langle b, \lambda k \rangle + \langle k, \lambda k \rangle \\ \Leftrightarrow & 0 > 2\lambda\langle b, k \rangle + \lambda^2\langle k, k \rangle \\ \Leftrightarrow & 0 > \lambda \cdot (2\langle b, k \rangle + \lambda \langle k, k \rangle) \end{align}$$
Let $\lambda < 0$, then:
$$\begin{align} \Leftrightarrow & 2\langle b, k \rangle + \lambda \langle k, k \rangle > 0 \\ \Leftrightarrow & 2\langle b, k \rangle > -\lambda\langle k, k \rangle \\ \Leftrightarrow & -\frac{2\langle b, k \rangle}{\langle k, k \rangle} < \lambda \end{align}$$
Since $\langle b, k \rangle \neq 0$ and $\langle k, k \rangle > 0$ a $\lambda$ which satisfies the above inequality exists: $\lambda \in \left( -\frac{2\langle b, k \rangle}{\langle k, k \rangle}, 0 \right)$ (if $\langle b, k \rangle > 0$, otherwise negate $\lambda$).
Therefore: $\left\lVert \varphi(v) \right\rVert > \left\lVert v \right\rVert$ which is a contradiction to $\varphi$ being nonexpanding ($\forall v \in V: \left\lVert \varphi(v) \right\rVert \leq \left\lVert v \right\rVert$). Therefore our assumption must be wrong, and $\varphi$ is an orthogonal projection.
Please let me know if you can spot any mistakes or if this proof seems correct to you and also mention any notational errors I might have made. Thank you for your time working through all this!
I think you have the right idea, but the second part of your proof comes across as clunky. I'm not a fan of how $\lambda$ changes its domain throughout the proof. At first, you insist on $\lambda \in \mathbb{R}$. You then insist on $\lambda < 0$ half way through the equivalent statements. What does this mean? Is the next statement genuinely equivalent to the previous? Or is this being applied retroactively? If so, then why did we not inisist on $\lambda \in (-\infty, 0)$ instead of $\mathbb{R}$ from the start?
You then go on to say parenthetically that "if $\langle b, k \rangle < 0$, then negate $\lambda$". At no point do you assume $\langle b, k \rangle > 0$, so this statement comes out of the blue. Of course, with a little unpacking, one can deduce that $\langle b, k \rangle > 0$ if and only if $\lambda < 0$, but this isn't made clear in the proof.
I prefer the way I mentioned in the comments: force $\langle b, k \rangle > 0$ by negating $b$ or $k$ right from the outset. From that, you can deduce that $\lambda < 0$.
Or, even better, from the step $0 > \lambda \cdot (2\langle b, k \rangle + \lambda \langle k, k \rangle)$, we can see that this quadratic is minimised at $-\frac{\langle b, k \rangle}{\langle k, k \rangle}$. Instead of solving for all the bad $\lambda$, instead pick the worst $\lambda$, and run through the proof in the other direction. That is,
Indeed, the above calculation could even be adapted to prove the other direction too. Just set $b$ and $k$ to be arbitrary vectors in the image and kernel respectively. Then, assuming $\|\phi(v)\| \le \|v\|$ implies $\frac{\langle b, k \rangle^2}{\langle k, k \rangle} \le 0$, which can only be true if $\langle b, k \rangle^2 = 0$.
Other than that, the proof is good.