Positive semi-definite matrix Suppose a square symmetric matrix $V$ is given 
$V=\left(\begin{array}{ccccc}
 \sum w_{1s} &   &    &  & \\
   & \ddots &   & -w_{ij} \\
   &  & \ddots & & \\
   & -w_{ij} &   & \ddots & \\
   & & & & \sum w_{ns}
 \end{array}\right) \in\mathbb{R}^{n\times n},$
with values $w_{ij}> 0$, hence with only positive diagonal entries.
Since the above matrix is diagonally dominant, it is positive semi-definite. However, I wonder if it can be proved that
$a\cdot diag(V)-V~~~~~a\in[1, 2]$
is also positive semi-definite. ($diag(V)$ denotes a diagonal matrix whose entries are those of $V$, hence all positive) In case of $a=2$, the resulting
$2\cdot diag(V)-V$ 
is also diagonally dominant (positive semi-definite), but is it possible to prove for $a\in[1,2]$?
.........................................
Note that the above proof would facilitate my actual problem; is it possible to prove
$tr[(X-Y)^T[a\cdot diag(V)-V](X-Y)]\geq 0$, 
where $tr(\cdot)$ denotes matrix trace, for $X, Y\in\mathbb{R}^{n\times 2}$ and $a\in[1,2]$ ?
Also note that
$tr(Y^TVY)\geq tr(X^TVX)$ and $tr(Y^Tdiag(V)Y)\geq tr(X^Tdiag(V)X)$.
(if that facilitates the quest., assume $a=1$)
.....................................................
Since the positive semi-definiteness could not generally be guaranteed for $a<2$, the problem casts to: for which restrictions on a does the positive semi-definiteness of a⋅diag(V)−V still hold? 
Note the comment from DavideGiraudo, and his claim for case $w_{ij}=1$, for all $i,j$. Could something similar be derived for general $w_{ij}$≥0? 

In the case $w_{ij}=1$ we have $V=\pmatrix{n&-1&-1&\cdots &-1 \\\ 
-1&n&-1&\ldots &-1 \\\
\vdots&\vdots&\ddots& &-1\\\
-1&-1&-1&\ldots&n}$ and $$M_a:=a\operatorname{diag}(V)-V=\pmatrix{n(a-1)&1&1&\cdots &1 \\\ 
1&n(a-1)&1&\ldots &1 \\\
\vdots&\vdots&\ddots& &1\\\
1&1&1&\ldots&n(a-1)}.$$
We can compute the determinant $\det(M_a-XI_n)$ adding to the first line all the other one. We get 
$$\det(M_A-XI_N)=(n(a-1)-(n-1)X)(n(a-1)-1-X)^{n-1},$$
and if we want $M_a$ positive semi-definite we should have $n(a-1)-1\geq 0$ so $a-1\geq\frac 1n$.
