Compute the adjoint map for complex number

Question

Consider the vector space $V = \mathbb{C}^2$ over $\mathbb{C}$ and the linear operator $φ: V \to V$ such that$φ(z_1, z_2)=(2z_1 + iz_2, (1 − i)z_1)$

Let $φ^*$ be the adjoint map of $φ$. Give the matrix representation of $φ^∗$ with respect to the standard basis of $(\mathbb{C}^2, \mathbb{C})$ and compute $φ^∗(u)$ for the vector $$u =(3 − i, 1 + 2i)$$

For this question what I tried is the following

I first get the matrix representation for $φ$ with respect to these matrices then I take the transpose of it since $φ^*=φ^T$. But I'm not sure if that's the correct way to solve it or either how to make the matrix representation for the complex number. What I did is to have a $(2\times4)$ matrix of each real and imaginary component of $z$ but I still think it's wrong. Can someone show me how to describe complex numbers in matrices?

Answer 1

You don't have to represent complex numbers as matrices since you are not considering them as group actions. Also, you need to write $\varphi^*$ with respect to the standard basis of $\mathbb{C}^2$, which is $\{(1,0),(0,1),(i,0),(0,i)\}$. So there is no need to write $i=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$.

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Apparently, $\varphi$ could be represented by $$\begin{pmatrix} i & 2\\ 1-i & 0 \end{pmatrix}$$ and thus according to Richard Jensen's answer that

For complex vector spaces, the adjoint of a matrix is computed by transposing and complex conjugating all entries.

The adjoint matrix, $\varphi^*$, of $\varphi$ is $$\varphi^*=\begin{pmatrix} -i &1+i\\ 2 &0 \end{pmatrix}\sim (z_1,z_2)\to((1+i)z_2-iz_1,2z_1)$$ Thus, $$\varphi^*(3−i,1+2i)=(-2,6-2i)$$

Answer 2

For complex vector spaces, the adjoint of a matrix is computed by transposing and complex conjugating all entries. For example

$\begin{pmatrix}1 & i \\ 2-i & i \end{pmatrix}^* = \begin{pmatrix}1 & 2 + i \\ -i & -i \end{pmatrix}$