10. Adjoints#
Let’s look at orthogonality and adjoints
Orthogonality#
Let us have the Hilbert space \((H, \mathbb{F}, \langle\cdot,\cdot\rangle)\).
Definition: Orthogonal
Two vectors \(x, y \in H\) are orthogonal iff their inner product equals to zero.
In \(\mathbb{R}^2\) or \(\mathbb{R}^3\), the inner product is the dot product. In \(\mathbb{R}^2\), a vector that is orthogonal to another vector is when it is perpendicular to the other vector. In these vector space, the inner product has an intuitive geometric intepretation. However, as we are generalizing this to function space, we need to refer back to the definition, that is \(\langle x, y \rangle = 0\).
With that definition of orthogonality, let’s look at subspaces that are orthogonal to each other.
Perp#
Given \(M\) subspace of \(H\), we call \(M^\perp\) (read as M perp, “perp” as in “perpendicular”) is defined to be
Simply saying, given subspace \(M\), \(M^\perp\) is the set of all vectors in \(H\) such that it is orthogonal to every vector in \(M\). We also call \(M^\perp\) the orthogonal complement of \(M\).
From this definition only, we can show something really interesting, that is the only intersection between \(M\) and its orthogonal complement is the zero vector.
Example: Show that \(M \cap M^\perp = \{\theta\}\)
Suppose we have a vector \(x \neq \theta \in M \cap M^\perp\), this means that
But since \(x \in M \cap M^\perp \rightarrow x \in M\). From here we can see that:
This implies that \(x = \theta\) (contradiction).
Adjoints#
Let’s now only consider the field of real or complex number, \(\mathbb{F} = \mathbb{C} \text{or} \mathbb{R}\).
Definition: Adjoint
Consider 2 inner product space \(U, V\) with inner products \(\langle \cdot, \cdot \rangle _u\), \(\langle \cdot, \cdot \rangle _v\) and the map \(\mathcal{A}:U \rightarrow V\). Suppose the map is linear and continuous. Then the adjoint map of \(\mathcal{A}\), called \(\mathcal{A}^*\) is defined as: