1) Pretty much yes. The reason is that the $x_i$'s are going to end up being a linear combination of the $z_i$'s. That works out nicely for Gaussian deviates because any linear combination of Gaussian deviates is, itself, a Gaussian deviate. Unfortunately, this is not necessarily true of other distributions.
2) It's a little puzzling, I know, but they are equivalent. Let $\Sigma$ be your covariance matrix and suppose you have both the Cholesky factorization, $\Sigma=L L^T$ and the eigendecomposition, $\Sigma=U \lambda U^T$. The covariance of $L z$ is given by:
$$
\begin{array}{}
E[L z (L z)^T] & = & E[L z z^T L^T] \\
& = & L \ E[z z^T] \ L^T \\
& = & L \ I \ L^T \\
& = & L L^T \\
& = & \Sigma
\end{array}
$$
Similarly, the covariance of $U \lambda^\frac{1}{2} z$ is given by:
$$
\begin{array}{}
E[U \lambda^\frac{1}{2} z (U \lambda^\frac{1}{2} z)^T] & = & E[U \lambda^\frac{1}{2} z z^T \lambda^\frac{1}{2} U^T] \\
& = & U \lambda^\frac{1}{2} \ E[z z^T] \ \lambda^\frac{1}{2} U^T \\
& = & U \lambda^\frac{1}{2} \ I \ \lambda^\frac{1}{2} U^T \\
& = & U \lambda^\frac{1}{2} \lambda^\frac{1}{2} U^T \\
& = & U \lambda U^T \\
& = & \Sigma
\end{array}
$$
For purposes of computation, I suggest you stick with the Cholesky factorization unless your covariance matrix is ill-conditioned/nearly singular/has a high condition number. Then it's probably best to switch to the eigendecomposition.
I think I got the answer by myself but wish some experts can confirm.
The confusion is that, in CVX book we are converting one optimization problem with constraints to another optimization problem without constraints and solve the dual problem. But in PCA optimization we cannot.
For example, page 227, we convert
$$
\underset{x}{\text{minimize}}~~ x^\top x \\
\text{s.t.}~~~~~~ Ax=b
$$
into maximize the dual function $g(v)=-(1/4)v^\top A A^\top v -b^\top v$, which is
$$
\underset{x}{\text{maximize}}~~\left(-(1/4)v^\top A A^\top v -b^\top v \right)\\
$$
In PCA optimization problem, problem has Lagrangian (for equality constraint we can use $-\lambda$)
$$
\mathcal{L}(\mathbf w,\lambda)=\mathbf w^\top \mathbf{Cw}-\lambda(\mathbf w^\top \mathbf w-1)
$$
For fixed $\lambda$, we get partial derivative and set to $\mathbf 0$.
$$
\frac{\partial \mathcal{L}}{\partial \mathbf w}=\mathbf 0=2\mathbf {Cw}-2\lambda\mathbf w
$$
which is the eigenvector equation
$$
\mathbf {Cw}=\lambda\mathbf w
$$
As pointed out by Matthew Gunn in the comment, PCA problem the objective is not convex see this discussion. Therefore we should not try to minimize dual function to solve the original problem.
Best Answer
1) The definition of eigenvector $Ax = \lambda x$ is ambidextrous. If $x$ is an eigenvector, so is $-x$, for then
$$A(-x) = -Ax = -\lambda x = \lambda (-x)$$
So the definition of an eigenbasis is ambiguous of sign.
2) It's hard to know for sure, but I have a strong suspicion of what is happening here. Your equation
$$ (A - \lambda)x = 0 $$
is technically incorrect. The correct equation is
$$ (A - \lambda I)x$$
The first equation is often used as a shorthand for the second. In general, this is unambiguous, because there is no real mathematical way to subtract a vector from a square matrix, but it is abuse of notation. In
Rthough, you have broadcasting. So if you doits not really what you want. The proper way would be
The reason you are getting zero solutions is that the matrix you are starting with $A$ is invertible. More than likely (almost surely), the matrix you get by subtracting the same number from every entry will also be invertible. For invertible matrices, the only solution to $Ax = 0$ is the zero vector.