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--- hilbert_space [2021/03/07 07:29] – admin
+++ hilbert_space [2021/04/02 19:03] (current) – [2.i.3 Subspaces] admin
@@ Line 90: / Line 90: @@
 is a basis for it.
-This subspace is //effectively// the same as $\mathbb{C}^2$.  If we do not bother to write down the third component then we do have a vector in $\mathbb{C}^2$ and we can always reconstruct the vector in $\mathbb{C}^3$ that it came from by just putting the zero back in the third component.  Mathematicians would say that $\mathbb{C}^2$ is //**isomorphic**// to a subspace of $\mathbb{C}^3$, which means that there exists a one-to-one map between vectors in the subspace and vectors in $\mathbb{C}^3$.
+This subspace is //effectively// the same as $\mathbb{C}^2$.  If we do not bother to write down the third component then we do have a vector in $\mathbb{C}^2$ and we can always reconstruct the vector in $\mathbb{C}^3$ that it came from by just putting the zero back in the third component.  Mathematicians would say that $\mathbb{C}^2$ is //**isomorphic**// to this subspace of $\mathbb{C}^3$, which means that there exists a one-to-one map between vectors in the subspace and vectors in $\mathbb{C}^3$.
 The distinction is somewhat important because $\mathbb{C}^2$ can be embedded in $\mathbb{C}^3$ in a variety of different ways.  For example, The set of all vectors of the form
@@ Line 166: / Line 166: @@
 As an example, in $\mathbb{R}^2$ and $\mathbb{C}^d$, the basis $\left ( \begin{array}{c} 1 \\ 0 \end{array} \right ), \left ( \begin{array}{c} 0 \\ 1 \end{array} \right )$ is orthonormal but the basis $\left ( \begin{array}{c} 1 \\ 0 \end{array} \right ), \left ( \begin{array}{c} 1 \\ 1 \end{array} \right )$ is not.
-===== 2.i.6 Orthogonal Subspaces =====
+====== 2.i.6 Orthogonal Subspaces ======
 In an inner product space, subspaces have more structure.  Suppose that $V_1 \subset V$ and $V_2 \subset V$.  $V_1$ and $V_2$ are //**orthogonal subspaces**// of $V$ if $(\psi,\phi) = 0$ for all vectors $\psi \in V_1$ and $\phi \in V_2$.  As an example, the set of all vectors of the form
@@ Line 185: / Line 185: @@
 A set of orthogonal subspaces $V_1,V_2,\cdots \subset V$ is said to //**span**// the inner product space $V$ if all vectors $\psi \in V$ can be written as
-\[\psi = \sum_j a_j \psi_j,\]
+\[\psi = \sum_j \psi_j,\]
-where the $a_j's$ are scalars and $\psi_j \in V_j$.  We sometimes write this as $V = \oplus_j V_j$
+where $\psi_j \in V_j$.  We sometimes write this as $V = \oplus_j V_j$
 As an example, let $\phi_1,\phi_2,\cdots$ be an orthonormal basis for $V$ and let $V_j$ be the one dimensional subspace consisting of all vectors of the form $a\phi_j$.  Then, $V = \oplus_j V_j$ just by the definition of a basis, i.e. all vectors $\psi \in V$ can be written as
 \[\psi = \sum_j a_j \phi_j.\]
-As a less trivial example, for any subspace $V' \subset V$, we have $V = V' \oplus V'^{\perp}$.
+As a less trivial example, for any subspace $V' \subset V$, we have $V = V' \oplus V'^{\perp}$.  To see this, note that if $\phi_1,\phi_2,\cdots$ is an orthonormal basis for $V'$ and $\chi_1,\chi_2,\cdots$ is an orthonormal basis for $V'^{\perp}$ then $\phi_1,\phi_2,\cdots,\chi_1,\chi_2,\cdots$ is a basis for $V$.  Any vector $\psi$ can be written in this basis as
+\[\psi = \sum_j a_j \phi_j + \sum_k b_k \chi_k,\]
+and then if we define
+\begin{align*}
+\psi' & = \sum_j a_j \phi_j, & \psi'^{\perp} & = \sum_k b_k \chi_k,
+\end{align*}
+we have
+\[\psi = \psi' + \psi'^{\perp},\]
+where $\psi' \in V'$ and $\psi'^{\perp} \in V'^{\perp}$.
 ====== 2.i.7 Norms ======