Adjoints and Gram Matrices

Definition Let $V$ and $W$ be finite-dimensional inner product spaces over $\mathbb{F}$ and let $T \colon V \to W$ be linear. The adjoint of $T$ is the unique map $T^{*} \colon W \to V$ that satisfies the adjoint identity

Such a map exists, is unique and is linear. Fix $w \in W$. The rule $v \mapsto \langle w, Tv \rangle_W$ is a linear functional on $V$, because $T$ and the second slot are linear, so the Riesz representation theorem gives exactly one $z \in V$ with $\langle z, v \rangle_V = \langle w, Tv \rangle_W$ for every $v$. Set $T^{*}w = z$. Conjugating both sides gives the adjoint identity, and any map that satisfies the identity must send $w$ to this same representing vector, which proves uniqueness. For linearity, conjugate-linearity of the first slot gives

so uniqueness of the representing vector gives $T^{*}(aw_1 + bw_2) = aT^{*}w_1 + bT^{*}w_2$. The adjoint depends on the inner products of both spaces. It is not the dual map $T^{\vee} \colon W^{\vee} \to V^{\vee}$, $\ell \mapsto \ell \circ T$, which needs no inner product, and in general it is not an inverse. On $\mathbb{F}^n$ and $\mathbb{F}^m$ with the standard inner products, the adjoint of $v \mapsto Av$ is $w \mapsto A^{*}w$, because $(Av)^{*}w = v^{*}A^{*}w$.

Definition Let $U$, $V$ and $W$ be finite-dimensional inner product spaces over $\mathbb{F}$, and let $T \colon U \to V$ and $S \colon V \to W$ be linear. Then

For $u \in U$ and $w \in W$, two uses of the adjoint identity give

so $T^{*}S^{*}$ satisfies the defining identity of $(ST)^{*}$, and uniqueness of the adjoint gives equality. The same uniqueness argument proves the other rules. For linear maps $T, T_1, T_2 \colon V \to W$ and a scalar $a \in \mathbb{F}$,

The first rule follows by conjugating the adjoint identity, which gives $\langle T^{*}w, v \rangle = \langle w, Tv \rangle$. The scalar rule holds because a scalar comes out of the first slot conjugated, so $\langle aTv, w \rangle = \overline{a}\,\langle Tv, w \rangle = \langle v, \overline{a}\,T^{*}w \rangle$.

Definition Let $V$ and $W$ be finite-dimensional inner product spaces over $\mathbb{F}$, let $T \colon V \to W$ be linear, let $\mathcal{B} = (b_1, \dots, b_n)$ be a basis of $V$ and $\mathcal{C} = (c_1, \dots, c_m)$ a basis of $W$, and let the Gram matrices be

In coordinates $\langle v, v' \rangle_V = [v]_{\mathcal{B}}^{*}\,G_V\,[v']_{\mathcal{B}}$, and likewise in $W$. The Gram matrix of a basis is invertible. If $G_V z = 0$, then $z^{*}G_V z = 0$, and this number is the squared norm of $\sum_i z_i b_i$, so that vector is $0$ and $z = 0$. If $A = [T]_{\mathcal{C} \leftarrow \mathcal{B}} \in \mathbb{F}^{m \times n}$, then

To derive it, write $x = [v]_{\mathcal{B}}$, $y = [w]_{\mathcal{C}}$ and $M = [T^{*}]_{\mathcal{B} \leftarrow \mathcal{C}}$. The two sides of the adjoint identity become

They agree for all columns $x$ and $y$, and standard basis columns pick out single entries, so $A^{*}G_W = G_V M$. In orthonormal bases both Gram matrices are identity matrices and the matrix of the adjoint is $A^{*}$, which is $A^{\mathsf{T}}$ when $\mathbb{F} = \mathbb{R}$. In other bases the bare conjugate transpose is generally wrong.

Definition Let $U$ be a subspace of an inner product space $V$ over $\mathbb{F}$. Its orthogonal complement is

It is a subspace, and by conjugate symmetry the order of the two slots does not matter. It meets $U$ only in $0$, because a vector orthogonal to itself has squared norm $0$. If $u_1, \dots, u_r$ span $U$, then $v \in U^{\perp}$ exactly when $\langle u_i, v \rangle = 0$ for each $i$. Now let $\dim V = n$ and $\dim U = k$. Extend a basis of $U$ to a basis of $V$ and apply Gram-Schmidt in that order. Gram-Schmidt keeps the span of each initial segment, so the result is an orthonormal basis $q_1, \dots, q_n$ of $V$ whose first $k$ vectors form a basis of $U$. Every $v$ equals $\sum_j \langle q_j, v \rangle q_j$, and it lies in $U^{\perp}$ exactly when its first $k$ coefficients vanish. Hence

Definition Let $U$ be a subspace of a finite-dimensional inner product space $V$ over $\mathbb{F}$. Then

The inclusion $U \subseteq (U^{\perp})^{\perp}$ holds in every inner product space, since each $u \in U$ is orthogonal to every vector of $U^{\perp}$. For the reverse inclusion, take the orthonormal basis $q_1, \dots, q_n$ from the previous card, with $U = \operatorname{span}\{q_1, \dots, q_k\}$ and $U^{\perp} = \operatorname{span}\{q_{k+1}, \dots, q_n\}$. A vector $v$ is orthogonal to $U^{\perp}$ exactly when $\langle q_j, v \rangle = 0$ for every $j > k$, which says that $v$ lies in $U$. Counting dimensions gives the same conclusion, because $(U^{\perp})^{\perp}$ contains $U$ and has dimension $n - (n - k) = k$. For any subset $S$ of $V$ the identity becomes $(S^{\perp})^{\perp} = \operatorname{span} S$. Finite dimension is a real hypothesis here. In infinite-dimensional inner product spaces the identity can fail.

Definition Let $V$ and $W$ be finite-dimensional inner product spaces over $\mathbb{F}$ and let $T \colon V \to W$ be linear. Then

The first equality is between subspaces of $W$, the second between subspaces of $V$. For the first, $w \in (\operatorname{im} T)^{\perp}$ means $\langle Tv, w \rangle = 0$ for every $v$, that is $\langle v, T^{*}w \rangle = 0$ for every $v$. Choosing $v = T^{*}w$ gives $\lVert T^{*}w \rVert^2 = 0$, so $T^{*}w = 0$. Conversely, $T^{*}w = 0$ makes every such pairing zero. Applying the first identity to $T^{*}$, whose adjoint is $T$, gives $(\operatorname{im} T^{*})^{\perp} = \ker T$, and taking orthogonal complements with the double complement identity gives the second. Consequently

and $\operatorname{rank} T^{*} = \operatorname{rank} T$.