Proving that $T:V \rightarrow W$ has a unique adjoint transformation Let $(V, \langle . , \rangle_{V})$,$(W, \langle . , \rangle_{W})$ be inner product spaces, and let $T: V \rightarrow W$ be a linear transformation. A function $T^{*}: W \rightarrow V$ is called an adjoint of $T$ if $\langle T(x),y \rangle_{W} = \langle x, T^{*}(y) \rangle_{V}$ for all $x \in V$ and $y \in W$.
Show that if $V$ and $W$ are finite dimensional, then there is a unique adjoint $T^{*}$ of $T$, and $T^{*}$ is linear.
My approach to this problem was basically one of experimentation, I feel like I need to construct $T^{*}$ in the general sense, so I began by constructing it in a specific case to gain some sort of intuition, but it doesn't seem helpful.
Let $V = \mathbb{R}^{2}$ and $W = \mathbb{R}^{3}$, both with standard inner product. Let $T: V \rightarrow W$ be defined as $T(x_{1},x_{2}) = (x_{1},x_{2},x_{1}+x_{2})$.
Then $\langle T(x_{1},x_{2}), (y_{1},y_{2},y_{3}) \rangle = \langle (x_{1},x_{2},x_{1}+x_{2}), (y_{1},y_{2},y_{3}) \rangle = x_{1}y_{1} + x_{2}y_{2} + (x_{1}+x_{2})y_{2} = x_{1}(y_{1}+y_{3}) + x_{2}(y_{2}+y_{3})$
This last expression makes it evident that I want $T^{*}(y_{1},y_{2},y_{3}) = (y_{1}+y_{3},y_{2}+y_{3})$, but from this, I'm not really getting any insight to what it looks like in general.

Hint/sketch: choose orthonormal bases $\{v_i\}$ and $\{w_j\}$ of $V$ and $W$, and represent $T$ by a matrix $A$ with respect to these bases. Then, the conjugate transpose $A^*$ will define a function $T^*:W\to V$, and you can use the orthogonality + the definition of the action of a matrix on a basis element to show that $\langle Av_i,w_j\rangle=\langle v_i,A^*w_j\rangle$ for all $i,j$. Then you can expand to all elements of $V$ and $W$ using bilinearity of the inner product, showing $T^*$ really is adjoint to $T$.
