The Adjoint of a Linear Map

# The Adjoint of a Linear Map

Let $V$ be a vector space over the field $\mathbb{F}$. Recall from the Linear Functionals page that a linear functional is a linear map $\varphi : V \to \mathbb{F}$ from the vector space $V$ into the underlying field, in this case, $\mathbb{F}$. Also recall that if $V$ is a finite-dimensional inner product space and if $\varphi$ is a linear functional on $V$ then there exists a unique vector $v \in V$ such that for every $u \in V$ we have that

(1)
\begin{align} \quad \varphi (u) = <u, v> \end{align}

We will now look at another type of linear map. Let $V$ and $W$ be finite-dimensional nonzero inner product spaces, and let $T$ be a linear map from $V$ to $W$, that is, $T \in \mathcal L(V, W)$.

 Definition: If $V$ and $W$ are finite-dimensional nonzero inner product spaces and $T \in \mathcal L (V, W)$ then the adjoint of $T$ denoted $T^*$ is the linear map $T^* : W \to V$ is defined by considering the linear function $\varphi : V \to \mathbb{F}$ defined by $\varphi (v) =$ and for a fixed $w \in W$ we define $T^* (w)$ to be the unique vector in $V$ such that $=$.

The subscripts on the inner product brackets denote which inner product (on $V$ or on $W$) we're applying.

Note that we have defined the adjoint to be a function $T^* : W \to V$. In fact, the following proposition tells us that $T^*$ is more than just a function and is actually also a linear map from $W$ to $V$.

 Proposition 1: Let $V$ and $W$ be finite-dimensional nonzero inner product spaces over the field $\mathbb{F}$. Then if $T \in \mathcal L(V, W)$, then $T^* \in \mathcal L(W, V)$.
• Proof: Let $T \in \mathcal L(V, W)$. Let $w_1, w_2 \in W$ and $a, b \in \mathbb{F}$. Then we have that:
(2)
\begin{align} \quad <T(v), aw_1 + bw_2> = <T(v), aw_1> + <T(v), bw_1> \\ \quad <T(v), aw_1 + bw_2> = \bar{a}<T(v), w_1> + \bar{b}<T(v), w_2)> \\ \quad <T(v), aw_1 + bw_2> = \bar{a}<v, T^*(w_1)> + \bar{b}<v, T^*(w_2)> \\ \quad <T(v), aw_1 + bw_2> = <v, aT^*(w_1)> + <v, bT^*(w_2)> \\ \quad <T(v), aw_1 + bw_2> = <v, aT^*(w_1) + bT^*(w_2)> \end{align}
• Thus we have that $T^*(aw_1 + bw_2) = aT^*(w_1) + bT^*(w_2)$ since $<T(v), aw_1 + bw_2> = <v, T^*(aw_1 + bw_2)>$. Thus the additivity and homogeneity properties hold and so $T^* \in \mathcal L (W, V)$. $\blacksquare$

We will now look at some examples of finding the adjoints of some linear maps.

## Example 1

Let $T : \mathbb{R}^4 \to \mathbb{R}^2$ be defined by $T(x, y, z, w) = (x + 2y, 3z + 4z)$ for each $(x, y, z, w) \in \mathbb{R}^4$. Define the inner product on both $\mathbb{R}^4$ and $\mathbb{R}^2$ to be the standard dot product. Find $T^*$.

Since $T : \mathbb{R}^4 \to \mathbb{R}^2$ then $T^* : \mathbb{R}^2 \to \mathbb{R}^4$. Fix $(m, n) \in \mathbb{R}^2$. Then:

(3)
\begin{align} <(x, y, z, w), T^*(m, n)> \\ = <T(x, y, z, w), (m, n)> \\ = < (x + 2y, 3z + 4w), (m, n)> \\ = mx + 2my + 3nz + 4nw \\ = <(x, y, z, w), (m, 2m, 3n, 4n)> \end{align}

Therefore we have that $T^*(m, n) = (m, 2m, 3n, 4n)$. Fix $r(x) = b_0 + b_1x + b_2x^2 \in \wp_2 (\mathbb{R})$.