Why are eigenspaces associated with different eigenvalues orthogonal? My teacher stated the following proposition:

Let $f:V\to V$ be an orthogonal/unitary Endomorphism and $V$ a finite-dimensional vector space. Then, If $F\leq V$ and $f(F)\subseteq F$ we have: $$f(F^\perp) \subseteq F^ \perp$$

I understand this proposition and I was able to prove it myself, but then my teacher stated the following as a corollary:

Let $f:V\to V$ be an orthogonal/unitary Endomorphism and $V$ a finite-dimensional vector space. Then eigenspaces associated with different eigenvalues are orthogonal.

There are two things I don't understand here:

What does it mean for two subspaces $A,B$ to be orthogonal?
How can this be proven using the proposition above?

Thank you.
We say $A\perp B$ if $a\perp b$ for all $a\in A$, $b\in B$.
Note that if $A$, $B$ are eigenspaces of eigenvalues $\lambda_A$, $\lambda_B$, then for $a\in A$, $b\in B$, $$\lambda_A\langle a,b\rangle = \langle \lambda_A a,b\rangle=\langle f(a),b\rangle =\langle a,f(b)\rangle=\langle a,\lambda_Bb\rangle=\lambda_B\langle a,b\rangle $$
We say $A\perp B$ if $a\perp b$ for all $a\in A$, $b\in B$.
Note that if $A$, $B$ are eigenspaces of eigenvalues $\lambda_A$, $\lambda_B$, then for $a\in A$, $b\in B$, $$\lambda_A\langle a,b\rangle = \langle \lambda_A a,b\rangle=\langle f(a),b\rangle =\langle a,f(b)\rangle=\langle a,\lambda_Bb\rangle=\lambda_B\langle a,b\rangle $$