Writing a convex quadratic constraint in SOC form I am trying to apply the following transformation for a problem:
Consider a quadratic constraint of the form
$$ x^TA^TAx+b^Tx+c\leq0. $$
This is equivalent to the SOC constraint
$$ \left\| \begin{array}{c}(1+b^Tx+c)/2\\Ax\end{array}\right\|_2 \leq (1-b^Tx-c)/2. $$
The challenge for me is, my problem is in the form $x^T Q x + bx +c \leq 0$ where $Q$ is not positive definite (it is semidefinite). So I can't apply Cholesky decomposition $Q = L^T L$ and I can't use the transformation above.
I am now trying to find another method. I tried eigenvalue-eigenvector decomposition: $Q = VDV^T$ where $D$ is a diagonal matrix. But that doesn't give the same solution as Cholesky does. 
How can I write the constraint in SOC form?
 A: As mentioned in the comments, your approach using the eigenvalue decomposition is correct. In this case, we have that
$$ Q=VDV^T$$
which can be written as 
$$ Q=(VD^{1/2})(VD^{1/2})^T. $$
(As noted in the comments, other decompositions could work as well, but I'll use this one, since it's the one you mention). Your concern is that, if $Q$ is a positive definite matrix, then the matrix $L$ obtained by taking the Cholesky decomposition isn't the same as the matrix $VD^{1/2}$ obtained above. However, this doesn't mean that the resulting reformulation doesn't yield the same result in both cases.
Consider your example 
$$ Q=\begin{pmatrix}2&-5/2\\-5/2&4\end{pmatrix}.$$
If we take the Cholesky decomposition, then, to a few digits of accuracy,
$$ L=\begin{pmatrix}1.414&0\\-1.768&0.935\end{pmatrix}. $$
If we take the eigenvalue decomposition, then, to a few digits of accuracy,
$$ VD^{1/2}=\begin{pmatrix}-0.459&1.338\\-0.311&-1.976\end{pmatrix}. $$
Indeed, $L\neq{VD^{1/2}}$. However, let's look at the plots of the level sets for all three formulations:

Voilà
Edit: A note on why this is true: essentially, we are employing this theorem:
Theorem: Let $Q\in\mathbb{R}^{n\times{n}}$ be positive semidefinite, and let $b\in\mathbb{R}^n$, $c\in\mathbb{R}$. Let $L\in\mathbb{R}^{n\times{n}}$ be any matrix such that $Q=L^TL$. Then the sets
$$ S_1=\bigg\{x\in\mathbb{R}^n\ \bigg|\ x^TQx+b^Tx+c\leqslant0\bigg\} $$
and
$$ S_2=\bigg\{x\in\mathbb{R}^n\ \bigg|\ \left\|\begin{array}{c}(1+b^Tx+c)/2\\Lx\end{array}\right\|_2\leqslant(1-b^Tx-c)/2\bigg\} $$
are equal.
This result makes no reference to the form of $L$, only that $Q=L^TL$. Hence, if there exist $L_1$ and $L_2$ such that $Q=L_1^TL_1$ and $Q=L_2^TL_2$, then the sets 
$$ F_i=\bigg\{x\in\mathbb{R}^n\ \bigg|\ \left\|\begin{array}{c}(1+b^Tx+c)/2\\L_ix\end{array}\right\|_2\leqslant(1-b^Tx-c)/2\bigg\} $$
for $i=1,2$ are the same (as illustrated above), even though $L_1\neq{L_2}$.
