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Part A

On Euclidean $\mathbb{R}^3$, let $L(x) = (3x_1 + x_2, x_2 - x_3, 5x_1 + x_3)^T$. Compute $L^{\dagger}(x)$.

\begin{align*} L(x) &= (3x_1 + x_2, x_2 - x_3, 5x_1 + x_3)^T \\ [L(x)]_{\mathcal{E}} &= \begin{pmatrix} 3 & 1 & 0 \\ 0 & 1 & -1 \\ 5 & 0 & 1 \end{pmatrix} \\ [L(x)^{\dagger}]_{\mathcal{E}} &= \begin{pmatrix} 3 & 0 & 5 \\ 1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix} \\ L(x)^{\dagger} &= (3x_1 + 5x_3, x_1 + x_2, -x_2 + x_3)^T \\ \end{align*}

Part B

Repeat the above, only with $\mathbb{R}^3$ having the nonstandard inner product $\langle x | y \rangle = x_1 y_1 + 2x_2 y_2 + 3x_3 y_3$

The book lists this as the answer but I don't see how to get this:

\begin{align*} [L^{\dagger}] = G^{-1} [L] G = \begin{pmatrix} 3 & 0 & 15 \\ 1/2 & 1 & 0 \\ 0 & -2/3 & 1 \end{pmatrix} \\ \end{align*}

I can calculate the metric for the given non-standard inner product: \begin{align*} G_{ij} &= \langle e_i | e_j \rangle \\ G &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \\ \end{align*}

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When in doubt, go back to the definitions. The adjoint has the property that $\langle x|L^\dagger(y)\rangle = \langle L(x)|y\rangle$. If you write this out in matrix form this is $$\mathbf x^TG[L^\dagger]_{\mathcal E}\mathbf y = \mathbf x^T[L]_{\mathcal E}^TG\mathbf y.$$ This holds for all $\mathbf x$ and $\mathbf y$, so $G[L^\dagger]_{\mathcal E} = [L]_{\mathcal E}^TG$, from which $[L^\dagger]_{\mathcal E} = G^{-1}[L]_{\mathcal E}^TG$.

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