# Finding the adjoint operator of $T(x,y) = (2x - y, x + 3y)$

Suppose $T: \mathbb{R}^2 \to \mathbb{R}^2$ is defined as $$T(x,y) = \begin{bmatrix}2x-y \\ x+3y \end{bmatrix}$$

Find the adjoint $T^{*}$ of $T$. So we can create a linear functional as follows: Choose $w = (x,y)$. Then $$T_{w}(x,y) = \langle(2x-y,x+3y), (x,y) \rangle$$

So then we need to find the vector $T^{*}w$ such that $$\langle Tv, w \rangle = \langle v, T^{*}w \rangle$$ for all $w \in W$ and $v \in V$.

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We need to? What have you done so far? Where are you stuck? What is your question? – Emily Jul 18 '12 at 17:59
$T$ is represented by the matrix \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix} – axblount Jul 18 '12 at 18:00

Method 1. From first principles.

The adjoint is completely determined by its value on $(1,0)$ and $(0,1)$ (as is any good linear transformation with domain $\mathbb{R}^2$. So let's look at it:

$$\langle T(x,y),(1,0)\rangle = \langle (2x-y,x+3y),(1,0)\rangle = 2x-y$$ so we need $$\langle (x,y),T^*(1,0)\rangle = 2x-y.$$ This means that we need $T(1,0) = (2,-1)$.

Similarly, since $$\langle T(x,y),(0,1)\rangle = \langle (2x-y,x+3y),(0,1)\rangle = x+3y$$ so what should $T^*(0,1)$ be?

Method 2. From the matrix for $T$.

The matrix of $T$ with respect to the standard basis is $$\left(\begin{array}{rr} 2& -1\\ 1&3 \end{array}\right).$$ Since $[T^*] = [T]^*$, where the left hand side indicates the adjoint and the right hand side indicates the conjugate transpose matrix, then the matrix of $T^*$ is... (And of course, from the matrix we can figure out the formula.)

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