Let $V$ be a finite dimensional vector space over the field $K,$ with a non-degenerate scalar product. Let $v_{0}, w_{0}$ be elements of $V .$ Let $A : V \rightarrow V$ be the linear map such that $A(v)=\left\langle v_{0}, v\right\rangle w_{0} .$ Describe $^{t} A$.
Solution. We have
$$ \langle A(v), w\rangle=\left\langle v_{0}, v\right\rangle\left\langle w_{0}, w\right\rangle=\left\langle v,\left\langle w_{0}, w\right\rangle v_{0}\right\rangle $$
So $$ ^{t} A(w)=\left\langle w_{0}, w\right\rangle v_{0} $$.
My question is: I didn't understand the notation of $\langle A(v), w\rangle$, what is this mean? Can you explain? And so that how did you get this: $ \langle A(v), w\rangle=\left\langle v_{0}, v\right\rangle\left\langle w_{0}, w\right\rangle=\left\langle v,\left\langle w_{0}, w\right\rangle v_{0}\right\rangle $ ? Can you explain? Thanks...
$A (v)=\left\langle v_{0}, v\right\rangle w_{0}$ is a vector in $V $. So it is perfect to take the scalar product between $A (v)$ and $w $, which is $\langle A(v), w\rangle$.
Note that $$\langle A(v), w\rangle=\langle\left\langle v_{0}, v\right\rangle w_{0},w\rangle=\left\langle v_{0}, v\right\rangle\left\langle w_{0}, w\right\rangle.$$ This follows directly from the property of scalar product. $$\langle cv, w\rangle=\langle v, cw\rangle=c\langle v, w\rangle.$$