1.2. Vector spaces¶

1.2.1. Subspace¶

1.2.2. Direct sum¶

Consider vector spaces \(U, V \subset W\). We say \(W\) is the direct sum of \(U\) and \(W\)

\[W = V \oplus U,\]

if \(W = V + U\), and \(\{ 0 \} = V \cap U\).

Lemma

Let \(U, V\) be subspaces of a vector space \(W\), then \(W = V \oplus U\) if and only if for every \(w \in W\) there exist unique vectors \(v \in V\) and \(u \in U\) such that \(w = v + u\).

We can sketch a visual proof for this Lemma: consider \(W = \mathbb{R}^2\), with the only non-trivial subspaces represented by lines passing through the origin, and let \(V\) be the vectors lying exclusively along the \(x\)-axis plus the zero vector, and \(U\) be the vectors lying along the \(y\)-axis plus the zero vector.

1.2.2.1. Bases under direct sum¶

Let \(b_i\) and \(k_j\) denote the bases of \(V\) and \(U\) respectively. Under a direct sum, the basis of \(W\) is given by the union \(b_i \cup k_j\).

We may also state that \(V\) and \(U\) are orthogonal, i.e. \(V \perp U\). Following from the above lemma, for every \(w \in W\), we can write the decomposition

\[w = \sum_{i=1}^{\text{dim}\ V} v^i b_i + \sum_{j=1}^{\text{dim}\ U} u^j k_j.\]

1.2.3. Dual space¶

The dual space of a vector space \(V\) is denoted \(V^\ast\), and is defined as the space of all linear functions mapping \(V\) into the reals.

Definition: Dual vector space

Given a vector space \(V\), its dual is defined

\[V^\ast := \{ \phi : V \rightarrow \mathbb{R}, \ \text{linear} \}.\]

Denoting the bases of \(V\) with \(b_i\), then the basis of \(V^\ast\) is \(\beta^i\) defined by

(1.1)¶\[\beta^i (b_j) := \delta^i_{\hphantom{i}j}.\]

We express the basis transformation for the vector space through the action of an \(n \times n\) invertible matrix \(\gamma\)

\[b'_i := \sum_{j=1}^n (\gamma^{-1})^j_{\hphantom{j}i} b_j,\]

where \(n = \text{dim} \ V\), and \((\gamma^{-1})^j_{\hphantom{j}i} \in \mathbb{R}\).

Hint

\(\gamma\) is a passive transformation, as it acts to alter the basis we are using to express our vector.

The basis transformation on the dual basis is given by the inverse transpose of \(\gamma\)

\[\beta'^i := \sum_{j=1}^n \gamma^i_{\hphantom{i}j} \beta^j. \]

1.2.4. Annihilator¶

Definition: Annihilator of a vector space

Let \(U\) be a subet of a vector space \(V\). The annihilator of \(U\) is

\[U^0 := \{ \phi : V \rightarrow \mathbb{R} \ | \ \phi(u) = 0 \ \forall \ u \in U \}.\]

As such, \(V^0 \subset V^\ast\).

Let \(b_i\) form a basis of \(V\) for \(i = 1, ..., n\) where \(n = \text{dim} \, V\), and suppose the basis of \(U\) consists of \(b_i\) for \(i = 1, ..., m < n\), with \(m = \text{dim} \, U\). We may then consider the dual basis \(\beta_i\) for \(V^\ast\), such that

\[\{ \beta_{m+1}, ..., \beta_{n} \} \subseteq U^0,\]

forms a basis for \(U^0\).

NB: from eq. (1.1), any element of \(\{ \beta_{m+1}, ..., \beta_{n} \}\) acting on any basis element of \(U\) is \(0\), asserting that \(U^0\) is indeed the annihiliator.

It follows that

(1.2)¶\[\text{dim} \, V = \text{dim} \, U + \text{dim} \, U^0.\]

1.2.5. Dual transformation¶

Let \(T\) be a linear transformation between vector spaces \(V \rightarrow W\). The dual transformation is then \(T^\ast : W^\ast \rightarrow V^\ast\) .

Definition: Dual transformation

For some \(T: V \rightarrow W\) the dual transformation \(T^\ast : W^\ast \rightarrow V^\ast\) is defined by

(1.3)¶\[T^\ast \omega := \omega \circ T, \ \ \forall \ \omega \in W^\ast.\]

The operation of \(T^\ast\) is commonly termed the adjoint. For all \(v \in V\) we have

\[(T^\ast \omega)\, v = \omega \, ( T v ).\]

Theorem: Matrix of dual transformation

Let \(V\) and \(W\) be vector spaces with bases defined. Let \(A\) denote the matrix of \(T: V \rightarrow W\), and \(B\) be the matrix of the dual transformation \(T^\ast: W^\ast \rightarrow V^\ast\). Then \(B = A^\text{T}\).

Proof: Starting from the definition of \(T^\ast \omega\) for any \(\omega \in W^\ast\)

\[ (T^\ast \omega) v = \omega (T v), \ \ \forall \ v \in V. \]

Substituting the matrix representations, and properly denoting the transpose

\[\begin{split}(B \omega)^\text{T} v = & \ \omega^\text{T} (A v), \\ \omega^\text{T} B^\text{T} v = & \ \omega^\text{T} A v.\end{split}\]

Since this is true for all \(\omega \in W^\ast\) and \(v \in V\), the above is equivalent to \(B^\text{T} = A\), or vice versa. \(\square\)

A direct corollary of this theorem is that

\[\text{rank}\, T^\ast = \text{rank}\, T.\]

Theorem: Annihilator of kernel and image

If \(T: V \rightarrow W\), and \(T^\ast\) is its dual transformation, then

\[\text{ker}\, T^\ast = (\text{im}\, T)^0, \ \ \ \ \ \ \text{im}\, T^\ast = (\text{ker}\, T)^0.\]

Proof: First, consider some \(\phi \in \text{im}\, T^\ast\), and \(v \in \text{ker} \, T\) – then

\[\phi (v) = T^\ast(\omega) (v),\]

for some \(\omega \in W^\ast\). It follows from eq. (1.3) that

\[\begin{split}& = w \, T(v), \\ & = 0,\end{split}\]

as \(v\) belongs to the kernel of \(T\). Therefore \(\phi \in (\text{ker} \, T)^0\) which implies \(\text{im} \, T^\ast \subseteq (\text{ker} \, T)^0\).

Second, to prove \(\text{im} \, T^\ast = (\text{ker} \, T)^0\), we may consider dimensions: using the corollary of the last theorem,

\[\text{dim} \, \text{im} \, T^\ast = \text{dim} \, \text{im} \, T.\]

Using eq. (1.2), we may write

\[\text{dim} \, (\text{ker} \, T)^0 = \text{dim} \, V - \text{dim} \, \text{ker} \, T,\]

since \(\text{ker} \, T \subset V\). From rank-nullity (see Rank and nullity) theorem

\[\text{dim} \, V = \text{dim} \, \text{im} \, T + \text{dim} \, \text{ker} \, T,\]

and so

\[\begin{split}\text{dim} \, (\text{ker} \, T)^0 = & \, \text{dim} \, \text{im} \, T, \\ = & \, \text{dim} \, \text{im} \, T^\ast. \\\end{split}\]

As both have the same dimension, they must be equal. The exact same line of thinking follows for the other equation. \(\square\)

To rephrase the above theorem, we can say that the subset of \(W^\ast\) that is mapped to \(0\) by \(T^\ast\) is the same subset of \(W^\ast\) that would map vectors in the image of \(T\) to \(0\). Likewise, the subset of \(V^\ast\) covered by the image of \(T^\ast\) take vectors in \(V\), that are mapped to \(0\) by \(T\), also to \(0\).

1.2.6. Projections¶

Consider \(W = V \oplus U\); we construct an operator \(P: W \rightarrow V\), element wise \(w \mapsto P(w)\). For some \(w = v + u\), with \(v \in V\), \(u \in U\), we then define \(P(w) := u\).

That is to say \(P\) is a linear transformation, with \(\text{im}\, P = V\), \(\text{ker}\, P = U\), and \(P^2 = P\).

\(P\) is called the projection of \(W\) onto \(V\) along \(U\).

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