I don't think anyone knows what you mean by monotonicity of a vector-valued function, or why you are mixing together linear transformations and quadratic forms. In particular your matrix $A$ has the property you describe if and only if
$(A + A^T) / 2$ has the property. Take any square matrix $B,$ take its skew-symmetric part $C = (B - B^T)/2,$ then for any column vector $w$ we have $w^T C w = 0.$ Put another way, your condition is far more sensible for the (symmetric) Hessian matrix of second partials for a function taking $\mathbf R^n$ to $\mathbf R.$

Define a matrix $Q_n$ with orthogonal columns given by this pattern (example for $n=6$):
$$ Q_n \; \; = \; \;
\left( \begin{array}{cccccc}
1 & -1 & -1 & -1 & -1 & -1\\\
1 & 1 & -1 & -1 & -1 & -1 \\\
1 & 0 & 2 & -1 & -1 & -1 \\\
1 & 0 & 0 & 3 & -1 & -1 \\\
1 & 0 & 0 & 0 & 4 & -1 \\\
1 & 0 & 0 & 0 & 0 & 5
\end{array}
\right) . $$
Note that, if desired, $Q_n$ can be made into a genuine orthogonal matrix by dividing the column entries by
$\sqrt n, \; \sqrt 2, \; \sqrt 6, \; \sqrt {12}, \; \sqrt {20}, \; \sqrt {30} $ and generally
dividing column $j$ by $\sqrt {j^2 - j} $ when $j \geq 2.$

The correct change of basis for a linear transformation matrix $E$ is $P^{-1} E P.$ The correct change of basis for a quadratic form symmetric (Gram) matrix $G$ is $U^T G U.$ The overlap of the two concepts is when we insist on an orthogonal matrix $W^T = W^{-1}$ and take $W^T G W.$

Anyway, take $$ A_S = (A + A^T) / 2. $$ Then look at
$$ Q_n^T A_S Q_n, $$ ignore row 1 and column 1, and check the lower right $n-1$ by $n-1$ block for positive semidefiniteness. This is exactly the condition you have asked about, but I have built in a little flexibility.

The lower right $n-1$ by $n-1$ block is exactly $$ R_n^T A_S R_n, $$ with the rectangular matrix:
$$ R_n \; \; = \; \;
\left( \begin{array}{ccccc}
-1 & -1 & -1 & -1 & -1\\\
1 & -1 & -1 & -1 & -1 \\\
0 & 2 & -1 & -1 & -1 \\\
0 & 0 & 3 & -1 & -1 \\\
0 & 0 & 0 & 4 & -1 \\\
0 & 0 & 0 & 0 & 5
\end{array}
\right) . $$

Finally, Suvrit gave the same answer but with rectangular matrix $S_n$ given by:
$$ S_n \; \; = \; \;
\left( \begin{array}{ccccc}
1 & 0 & 0 & 0 & 0\\\
-1 & 1 & 0 & 0 & 0 \\\
0 & -1 & 1 & 0 & 0 \\\
0 & 0 & -1 & 1 & 0 \\\
0 & 0 & 0 & -1 & 1 \\\
0 & 0 & 0 & 0 & -1
\end{array}
\right) . $$
Look at the entries of $ S_n^T A_S S_n.$