is that always true for two nx1 Vectors? Where x is a column vector of (x,y) and a a column vector of (a,b).
$(x-a)^T *grad(f(a) = grad(f(a))^T*(x-a)$
is that always true for two nx1 Vectors? Where x is a column vector of (x,y) and a a column vector of (a,b).
$(x-a)^T *grad(f(a) = grad(f(a))^T*(x-a)$
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Yes. (Assuming your $*$ means matrix multiplication.)
Take two row vectors $a,b \in \mathbb R ^ {n \times 1}$. Actually, they are row matrices. Denote the enties of these matrices as $\{a_i\}_{i=1,\dots,n},\{b_i\}_{i=1,\dots,n}$. Then we have
$a^T b = \sum_{i=1}^n a_i b_i = \sum_{i=1}^n b_i a_i = b^T a,$
where in the first and last $=$ the definition of matrix multiplication was used.
We just showed that the scalar product of two row vectors is commutative: $(a,b)=(b,a)$.