## Dual spaces, transposes, and adjoints

What’s the dual space of a vector space?

Let $V$ be an $n$-dimensional vector space, with basis
$$B = \{ v_1, \ldots, v_n \}.$$

Now consider an arbitrary linear function $f$ that takes a vector $v \in V$ and maps it to a scalar in, for example, $\mathbb{R}$. Then express $v$ as a linear combination of the basis vectors:
\begin{align*} f(v) &= f(c_1v_1 + \cdots + c_nv_n) \\ &= c_1 f(v_1) + \cdots + c_n f(v_n). \end{align*}

This means that if we specify $f(v_1)$ through $f(v_n)$, we’ve completely characterized $f$. In other words, if you tell me what $f(v_1)$ through $f(v_n)$ equal, I know what $f(v)$ equals for any vector $v$.

So we can specify any linear function that maps from $V$ to $\mathbb{R}$ (called a “linear functional”) as a list of $n$ scalars. This makes it clear that a linear functional is an $n$-dimensional vector itself, and so linear functions form a $n$-dimensional vector space as well:
$$V^* = \{ f_{[a_1, \ldots, a_n]} \mid a_1, \ldots, a_n \in \mathbb{R} \}.$$ This is the dual space of $V$, where I’ve used $f_{[a_1, \ldots, a_n]}$ to denote the linear functional such that $f(v_1) = a_1$, etc.

Since $V$ and its dual space $V^*$ have the same dimension, each element in $V$ has a corresponding element in $V^*$. Consider that
$$f_{[a_1, \ldots, a_n]} = a_1(c_1v_1) + \cdots + a_n(c_nv_n).$$Look at the right-hand side — if the vectors we’re dealing with are Euclidean vectors, then we could write $\vecb{a} = (a_1, \ldots, a_n)$ and express any functional as a dot product:
$$f_{\vecb{a}}(\vecb{v}) = \vecb{a} \cdot \vecb{v}.$$ We can extend this idea to any inner product space. If $V$ has an inner product, then we can very naturally associate a vector $w \in V$ with its corresponding functional in $V^*$, defined as $$f(v) = \langle v, w \rangle.$$

Now let’s throw linear transformations into the mix. Let $T$ be a linear transformation from $U$ to $W$, and let $f \in W^*$ be a linear functional on $W$.

Then if $u \in U$, observe that $f(T(u))$ gives us a scalar. Since $f$ composed with $T$ maps an element in $U$ to a scalar, the composition $f \circ T$ is a linear functional itself, an element of $U^*$!

In this way, given a functional in $W^*$ and a transformation $T$, we can generate functionals in $U^*$. So we can define another linear transformation from $W^*$ to $U^*$ that gives us this functional: $$T^\intercal(f) = f \circ T.$$

$T^\intercal$ is called the transpose of $T$, and it’s not a coincidence that the matrix of $T^\intercal$ is the transpose of the matrix of $T$ in certain bases.

Finally, let’s talk about inner product spaces again. If we have inner products for $U$ and $W$, we can, like above, associate vectors $u \in U$ and $w \in W$ with functionals in $U^*$ and $W^*$ respectively:
\begin{align*} u \quad&\longleftrightarrow\quad f_u(v) = \langle v, u \rangle \\ w \quad&\longleftrightarrow\quad f_w(v) = \langle v, w \rangle. \end{align*}

Recall that $T^\intercal$ takes a functional in $W^*$ and returns a functional in $U^*$. It’d be nice if we could deal with vectors in $W$ and $U$ instead of functionals in $W^*$ and $U^*$. Can we replace the transpose with another transformation that takes a vector in $W$, transforms it into its corresponding functional in $W^*$, apply $T^\intercal$ to it to get a functional in $U^*$, and finally transform it back into its corresponding vector in $U$?

It turns out we can: it’s called the adjoint of $T$, denoted $T^*$.

To see how $T^*$ must be defined, let’s take the definition of the transpose and substitute $f \in W^*$ with its inner product equivalent, $\langle v, w\rangle$, where $w$ is the vector in $W$ associated with $f$, and $v$ is the functional’s argument:
\begin{align*} T^\intercal(f) &= T^\intercal(\langle v, w\rangle) \\ &= (\langle v, w\rangle) \circ T \\ &= \langle T(v), w\rangle. \end{align*}

Then asking what vector in $U$ corresponds to this functional amounts to asking what vector satisfies
$$\langle T(v), w\rangle = \langle v, ?\rangle.$$

Thus, $T^*$ is the linear transformation that is defined to be the one that makes the following equality true:
$$\langle T(v), w\rangle = \langle v, T^*(w)\rangle.$$

Once we do that, we can go on to define important operators like normal, unitary, and Hermitian operators!

March 7, 2014, 1:18pm by Casey