1.) First, note that :$\underset{\text{dim}(\mathscr{V})\leq d}{\operatorname{argmin}} \sum_{i=1}^n {\left \| \phi(X^{(i)}) - \mathscr{P}_{\mathscr{V}}\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2=\underset{\text{dim}(\mathscr{V})\leq d}{\operatorname{argmax}} \sum_{i=1}^n {\left \| \mathscr{P}_{\mathscr{V}}\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2$
Because of the pythagorean theorem : ${\left \|\phi(X^{(i)}) - \mathscr{P}_{\mathscr{V}}\phi(X^{(i)}) \right \|}^2_{\mathscr{F}} + {\left \| \mathscr{P}_{\mathscr{V}}\phi(X^{(i)}) \right \|}^2_{\mathscr{F}} = {\left \|\phi(X^{(i)})\right \|}^2_{\mathscr{F}}$
Secondly, we have the decomposition :$\mathscr{P}_{\mathscr{V}}\phi(X^{(i)}) = \mathscr{P}_{\mathscr{V \cap \mathscr{L} \mathbb{R}^n}}\phi(X^{(i)}) + \mathscr{P}_{\mathscr{V \cap (\mathscr{L} \mathbb{R}^n)^{\perp}}}\phi(X^{(i)})$
And thanks to the pythagorean theorem : ${\left \| \mathscr{P}_{\mathscr{V}}\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2 = {\left \| \mathscr{P}_{\mathscr{V \cap \mathscr{L} \mathbb{R}^n}}\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2 + {\left \| \mathscr{P}_{\mathscr{V \cap (\mathscr{L} \mathbb{R}^n)^{\perp}}}\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2$
But ${\left \| \mathscr{P}_{\mathscr{V \cap (\mathscr{L} \mathbb{R}^n)^{\perp}}}\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2=0$ because $\phi(X^{(i)}) \in \mathscr{L} \mathbb{R}^n$
So the problem we have to solve is to find : $\underset{\text{dim}(\mathscr{V})\leq d}{\operatorname{argmax}} \sum_{i=1}^n {\left \| \mathscr{P}_{\mathscr{V} \cap \mathscr{L} \mathbb{R}^n }\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2$
That's why we can restrict ourselves to subspace $\mathscr{V}$ included in $\mathscr{L} \mathbb{R}^n$, and any space of this form can be written $\mathscr{L}V$, with $V \subset \mathbb{R}^n$ :
$\underset{\text{dim}(\mathscr{V})\leq d}{\operatorname{argmax}} \sum_{i=1}^n {\left \| \mathscr{P}_{\mathscr{V} \cap \mathscr{L} \mathbb{R}^n }\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2= \mathscr{L} \underset{\text{dim}(\mathscr{L}V)\leq d}{\operatorname{argmax}} \sum_{i=1}^n {\left \| \mathscr{P}_{\mathscr{L} V}\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2$
Moreover because $\left \{ \mathscr{L} V : \text{dim}(V) \leq d \right \} = \left \{ \mathscr{L} V : \text{dim}(\mathscr{L}V) \leq d \right \}$ (we can see that by restricting the application $\mathscr{L}$ on ${\mathbb{R}}^n / \text{Ker}(\mathscr{L}$) ) :
$\mathscr{L} \underset{\text{dim}(V)\leq d}{\operatorname{argmax}} \sum_{i=1}^n {\left \| \mathscr{P}_{\mathscr{L} V}\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2 = \mathscr{L} \underset{\text{dim}(\mathscr{L}V)\leq d}{\operatorname{argmax}} \sum_{i=1}^n {\left \| \mathscr{P}_{\mathscr{L} V}\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2$
So that finally we have (using the pythagorean theorem as in the beginning of the solution):
$\underset{\text{dim}(\mathscr{V})\leq d}{\operatorname{argmin}} \sum_{i=1}^n {\left \| \phi(X^{(i)}) - \mathscr{P}_{\mathscr{V}}\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2 = \mathscr{L} V_d$
Where $V_d=\underset{\text{dim}(V)\leq d}{\operatorname{argmin}} \sum_{i=1}^n {\left \| \phi(X^{(i)}) - \mathscr{P}_{\mathscr{L} V}\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2$
2.) We will prove that ${\left \langle \mathscr{L}K^{-1/2}\alpha,\mathscr{L}K^{-1/2}\beta \right \rangle}_{\mathscr{F}}=\left \langle \alpha,\beta \right \rangle$, for $\alpha, \beta$ two real vector.
(1)3.) First $V=\text{Span}(K^{-1/2}b_1,...,K^{-1/2}b_d)$, because $K^{-1/2}$ is a bijection and so $\mathscr{L} V=\text{Span}(\mathscr{L} K^{-1/2}b_1,...,\mathscr{L} K^{-1/2}b_d)$, because $\mathscr{L}$ is a bijection, because $K$ is nonsingular.
And this is an orthonormal basis because : ${\left \langle \mathscr{L}K^{-1/2}b_i,\mathscr{L}K^{-1/2}b_j \right \rangle}_{\mathscr{F}}=\left \langle b_i,b_j \right \rangle = \delta_{i,j}$
Because of the equality proved in the precedent question.
4.) We have ${\mathscr{P}}_{\mathscr{L}V} \mathscr{L}\alpha = \sum_{i=1}^d {\left \langle \mathscr{L} K^{-1/2} b_k, \mathscr{L}\alpha \right \rangle}_{\mathscr{F}} \mathscr{L} K^{-1/2} b_k = \mathscr{L} K^{-1/2} \sum_{i=1}^d {\left \langle \mathscr{L} K^{-1/2} b_k, \mathscr{L}\alpha \right \rangle}_{\mathscr{F}} b_k$ because $(K^{-1/2}b_1,...,K^{-1/2}b_d)$ is an orthonormal basis of $\mathscr{L} V$ and then using linearity of $\mathscr{L}$ and $K^{-1/2}$.
Secondly, ${\left \langle \mathscr{L} K^{-1/2} b_k, \mathscr{L}\alpha \right \rangle}_{\mathscr{F}}={\left \langle \mathscr{L} K^{-1/2} b_k, \mathscr{L} K^{-1/2} K^{1/2} \alpha \right \rangle}_{\mathscr{F}}={\left \langle b_k, K^{1/2} \alpha \right \rangle}$ because of the equality proved question 2.
So that finally we have ${\mathscr{P}}_{\mathscr{L}V} \mathscr{L}\alpha = \mathscr{L} K^{-1/2} \sum_{i=1}^d {\left \langle b_k, K^{1/2} \alpha \right \rangle} b_k= \mathscr{L} K^{-1/2} \text{Proj}_{K^{1/2} V} K^{1/2} \alpha$ using the same formula for a projection on a subspace using an orthonormal basis (and the fact that $(b_1,...,b_d)$ is an othornormal basis of $K^{1/2} V$).
5.) Recall that $\mathscr{L} e_i=\phi(X^{(i)})$ and by using the equality proved at the precedent questions we have :
$\sum_{i=1}^n {\left \| \phi(X^{(i)}) - \mathscr{P}_{\mathscr{L} V}\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2=\sum_{i=1}^n {\left \| \mathscr{L} e_i - \mathscr{L} K^{-1/2} \text{Proj}_{K^{1/2} V} K^{1/2} e_i \right \|}_{\mathscr{F}}^2$
Using linearity of operand : ${\left \| \mathscr{L} e_i - \mathscr{L} K^{-1/2} \text{Proj}_{K^{-1/2} V} K^{-1/2} e_i \right \|}_{\mathscr{F}}^2={\left \| \mathscr{L} K^{-1/2} (K^{1/2} - \text{Proj}_{K^{1/2} V} K^{1/2}) e_i \right \|}_{\mathscr{F}}^2$
We then have thanks to te equality proved question 2 : ${\left \| \mathscr{L} K^{-1/2} (K^{1/2} - \text{Proj}_{K^{1/2} V} K^{1/2}) e_i \right \|}_{\mathscr{F}}^2={\left \| ( K^{1/2} - \text{Proj}_{K^{1/2} V} K^{1/2}) e_i \right \|}^2$
We have for any $n \times n$ matrix $A$:$\sum_{i=1}^n {\left \|A e_i \right \|}^2=\sum_{i=1}^n \sum_{j=1}^n {(A e_i)_j}^2=\sum_{i=1}^n \sum_{j=1}^n {A_{j,i}}^2={\left \| A \right \|}_{F}^2$
Combining all this equality together finally gives :
$\sum_{i=1}^n {\left \| \phi(X^{(i)}) - \mathscr{P}_{\mathscr{L} V}\phi(X^{(i)}) \right \|}_{\mathscr{F}}^2={\left \| K^{1/2} - \text{Proj}_{K^{1/2} V} K^{1/2}\right \|}_F^2$
6.) Because of the precedent equality, to prove that $V_d=\text{Span}(v_1,...,v_d)$, it suffices to prove :
$\text{Span}(v_1,...,v_d)=\underset{\text{dim}(V)\leq d}{\operatorname{argmin}} {\left \| K^{1/2} - \text{Proj}_{K^{1/2} V} K^{1/2}\right \|}_F^2$
Because of the theorem C.5, we know that for the frobenius norm to be minimal, $\text{Proj}_{K^{1/2} V} K^{1/2}$ should be equal to $\sum_{k=1}^d \sigma_k(K^{1/2}) v_k v_k^T$, where $v_i$ are the orthogonal eigenvectors of $K$ (and of $K^{1/2}$) ranked in the non increasing order in respect to their eigenvalue (Note that we have the decomposition : $K^{1/2}=\sum_{k=1}^n \sigma_k(K^{1/2}) v_k v_k^T$).
So we have to prove that for $V=\text{Span}(v_1,...,v_d)$ : $\text{Proj}_{K^{1/2} V} K^{1/2}=\sum_{k=1}^d \sigma_k(K^{1/2}) v_k v_k^T$
First if $V=\text{Span}(v_1,...,v_d)$, then $K^{1/2} V=V$. This is because none of the eigenvalue of $K^{1/2}$ is nul and because every subspace associated with an eigenvector is stable by $K^{1/2}$.
Recall that $\text{Proj}_{V}=\sum_{k=1}^d v_k v_k^T$ so that we have the following equalities : $\text{Proj}_{V} K^{1/2}=(\sum_{k=1}^d v_k v_k^T) (\sum_{k=1}^n \sigma_k(K^{1/2}) v_k v_k^T)=\sum_{k=1}^d \sigma_k(K^{1/2}) v_k v_k^T$
So we have proved that $V_d=\text{Span}(v_1,...,v_d)$
7.) Using the equality proved in 2.) : ${\left \langle f_i , f_j \right \rangle}_{\mathscr{F}}={\left \langle \mathscr{L} K^{-1/2} v_i , \mathscr{L} K^{-1/2} v_j \right \rangle}_{\mathscr{F}}={\left \langle v_i , v_j \right \rangle}=\delta_{i,j}$
Moreover $K^{-1/2} V_d=V_d$, because the subspace associated to the eigenvalues (these are the same for $K^{-1/2}$,$K^{1/2}$,$K$) are stable : $\mathscr{L} K^{-1/2} V_d=\mathscr{L} V_d=\mathscr{V}_d$
So we have proved that $(f_1,...,f_d)$ is an orthonormal basis of the $\mathscr{V}_d$.
This implied : $\mathscr{P}_{\mathscr{V}_d}\phi(X^{(i)})=\sum_{k=1}^d {\left \langle f_k,\phi(X^{(i)}) \right \rangle}_{\mathscr{F}} f_k=\sum_{k=1}^d {\left \langle \mathscr{L} K^{-1/2} v_k, \mathscr{L} e_i \right \rangle}_{\mathscr{F}} f_k$
From question 2 : ${\left \langle \mathscr{L} K^{-1/2} v_k, \mathscr{L} e_i \right \rangle}_{\mathscr{F}}={\left \langle \mathscr{L} K^{-1/2} v_k, \mathscr{L} K^{-1/2} K^{1/2} e_i \right \rangle}_{\mathscr{F}}={\left \langle v_k, K^{1/2} e_i \right \rangle}$
So that we finally have : $\mathscr{P}_{\mathscr{V}_d}\phi(X^{(i)})=\sum_{k=1}^d {\left \langle v_k, K^{1/2} e_i \right \rangle} f_k$