Consider the map $A:\mathbb{R}^2\mapsto\mathbb{R}^n$ given by $A\left(x_{1}, x_{2}\right)=\left(2 x_{1}, x_{1}-x_{2}\right)$. I want to find the adjoint with respect to the inner product on $\mathbb{R}^2$.
My attempt:
$$\langle A x, y\rangle=\left\langle\left(2 x_{1}, x_{1}-x_{2}\right),\left(y_{1}, y_{2}\right)\right\rangle=2 x_{1} \bar{y}_{1}+\left(x_{1}-x_{2}\right) \bar{y}_{2}$$
And since we are in $\mathbb{R}^2$ and $\langle A x, y\rangle=\left\langle x, A^{*} y\right\rangle$ we can write:
$$\begin{array}{l}
2 x_{1} y_{1}+\left(x_{1}-x_{2}\right) y_{2}=\left\langle\left(x_{1}, x_{2}\right), A^{*}\left(y_{1}, y_{2}\right)\right\rangle \\
\Rightarrow A^{*}\left(y_{1}, y_{2}\right)=\left(2 y_{1}+y_{2},-y_{2}\right)
\end{array}$$
Can someone confirm if I am correct?
Best Answer
For a linear map $A:\mathbb{R}^n\to\mathbb{R}^n$, the adjoint is just the transpose.
So if
$$A=\begin{bmatrix}2&0\\1&-1\end{bmatrix}$$
you get $$A^\star = \begin{bmatrix}2&1\\0&-1\end{bmatrix}\,.$$
Your answer is correct.