Distribution of a sum let us assume $f<k$ and 
let us assume we have a vector z , which contains $\frac{k^2 -k}{2}$ entries --- z follows a multivariate normal distribution and a vector d which contains $kf - \frac{f(f-1)}{2}$ entries but we dont know them. A is a Matrix such that $Ad$ makes sense.
We are looking at 
$$(z-Ad)^{T}(z-Ad) \to min$$ 
where the solution (if A is regular) must be : $$d=(AA^{T})^{-1}A^{T}z$$
but the minimization problem $$S=(z-Ad)^{T}(z-Ad) \to min$$ should follow a generalized chi square distribution -- we can write 
$$S= z^{T}z +(Ad)^{T}Ad $$
so there are $z^2$ .. therefore we have the squares of normally distributed entries -- but what about the other term ? ... d can be expressed in terms of A and z -- is this enough? 
my lecturer also told me , it should be possible to write S as a sum of $$\frac{(k-f)^2 -(k+f)}{2}$$ summands with application of the singular value decomposition to A ... but i dont see it ... hope you can help :) 
Best Regards
 A: Certainly he/she was talking about the case where $A$ has full column rank, otherwise what he/she said is false. In the extreme case where $A=0$, the minimum $S$ is always $\sum_{j=1}^{(k^2-k)/2}z_j^2$, which is the sum of $(k^2-k)/2$ independent summands, not fewer.
In general (regardless of whether $AA^T$ is invertible or not), a global minimiser of $\|z-Ad\|^2$ is given by $d=A^+z$, where $A^+$ denotes the Moore-Penrose pseudo-inverse of $A$. If $A=U\Sigma V^T$ is a singular value decomposition, then $A^+=V\Sigma^+U^T$, where $\Sigma^+$ is obtained by replacing all nonzero entries (if any) in the main diagonal of $\Sigma^T$ by their reciprocals. The minimum value of $\|z-Ad\|^2$ is thus
$$
S=\|(I-AA^+)z\|^2=\|(I-\Sigma\Sigma^+)(U^Tz)\|^2
=\left\|\pmatrix{0\\ &I_{n-r}}(U^Tz)\right\|^2,
$$
where $n=\frac{k^2-k}2$ and $r=\operatorname{rank} A$. If we put $v=U^Tz$, then $S=\sum_{j=r+1}^n v_j^2$ is a sum of $n-r$ square terms.
Now, if $A$ has full column rank, then $r=kf-\frac{f(f-1)}2$ and hence $n-r=\frac{(k-f)^2-(k+f)}2$. 
