MA353 - Linear Algebra II

Vectorspaces and Subspaces

vectorspace

A set of vectors V is a vectorspace if:

  • closed under addition and scalar multiplication - \(\overrightarrow{v} + \overrightarrow{w} \in V\), \(c \overrightarrow{v} \in V\)
  • an identity is defined - \(1 \cdot \overrightarrow{v} = \overrightarrow{v}\)

Properties of Vectorspaces

  • commutative - \(\overrightarrow{v} + \overrightarrow{w} = \overrightarrow{w} + \overrightarrow{v}\)
  • distributive - \((a + b)\overrightarrow{v} = a\overrightarrow{v} + b \overrightarrow{v}\) and \(a(\overrightarrow{v} + \overrightarrow{w}) = a \overrightarrow{v} + b \overrightarrow{w}\)
  • \(- \overrightarrow{v} + \overrightarrow{v} = 0\)
  • \(0 + \overrightarrow{v} = \overrightarrow{v}\)

0 in a vectorspace is unique.

Proof

Subspace

subspace

Let \(W,V\) be vectorspaces. If \(W \subseteq V\), then \(W\) is a subspace.

Equivalently, \(W\) is a subspace \(W \subseteq V\) and:

  1. \(0 \in W\)
  2. \(\overrightarrow{v} + \overrightarrow{w} \in W\) for \(\overrightarrow{v},\overrightarrow{w} \in W\)
  3. \(a \overrightarrow{v} \in W\) for \(\overrightarrow{v} \in W, a \in \mathbb{R}\)

The intersection of subspaces is a subspace.

Proof

Sum of vectorspaces is also a vectorspace.

\(W_1 + W_2 \equiv \{w_1 + w_2 | w_1 \in W_1, w_2 \in W_2\}\)

Proof

Span and Linear Dependence

span

The span of a set \(S\) is all possible linear combinations of the set.

Let \(S = {v_1, ..., v_n}\)

\(\text{span} (S) = \{c_1 v_1 + ... + c_n v_n | c_1,...,c_n \in \mathbb{R}\}\)

Let \(S\) be a set, \(W\) be a vectorspace.

If \(S \subseteq W\), then \(\text{span} (S) \subseteq W\)

Proof

The span of subspaces \(W_1 \cup W_2 = W_1 + W_2\).

Proof

linear independence

A set \(S = \{s_1, ..., s_n\}\) is linearly independent if \(c_1, ..., c_n = 0\) is the only solution to \(c_1 s_1 + ... + c_n s_n = 0\)

Equivalently, a set is linearly independent if every element cannot be expressed as a combination of the other elements.

  1. Is the set \(S = \{1 + x + x^2 + x^3, x + x^2 + x^3, x^2 + x^3, x^3\}\) linearly independent?

    \begin{eqnarray*} c_1(1 + x + x^2 + x^3) + c_2 ( x + x^2 + x^3) + c_3 ( x^2 + x^3) + c_4 x^3 = 0 \\ c_1 + (c_1 + c_2)x + (c_1 + c_2 + c_3)x^2 + (c_1 + c_2 + c_3 + c_4)x^3 = 0 \\ c_1 = 0 \\ (c_1 + c_2)x = 0 \\ (c_1 + c_2 + c_3)x^2 = 0 \\ (c_1 + c_2 + c_3 + c_4)x^3 = 0 \end{eqnarray*}

Subset Dependence

If \(S_1 \subseteq S_2\) and \(S_1\) is dependent, then \(S_2\) is dependent.

Proof

If \(S_1 \subseteq S_2\) and \(S_2\) is independent, then \(S_1\) is also independent.

Proof

Basis and Dimension

A set \(B\) is called a basis of vectorspace \(V\) if

  • it is linearly independent and a generating set (\(V \subseteq \text{span} (B)\))

    or

  • \(B\) is the smallest set of \(V\) such that \(\text{span} (B) = V\)

    or

  • every vector in \(V\) can be expressed uniquely from linear combinations of \(B\)

Proof

Let \(L\) be an independent set of \(V\) with \(m\) elements. An indepenent set can be extended to create a generating set

Every basis of a finite vectorspace has the same number of elements.

Proof

For every vectorspace of dimension \(n\)

  1. \(\dim(\text{generating set}) \geq n\)

  2. \(\dim(\text{independent set}) \leq n\)

Proof

For a vectorspace \(V\) of dimension \(n\)

  1. A generating set with \(n\) elements is a basis of \(V\)

  2. An independent set with \(n\) elements is a basis of \(V\)

Proof

For a vectorspace \(V\) of dimension \(n\)

  1. A generating set with dimension greater than \(n\) can be reduced to be a basis.

  2. An independent set with dimension less than \(n\) can be extended to be a basis.

Proof

Dimension of Subspace

Let \(W\) be a subspace of \(V\). If the dimensions of \(W\) equals \(V\), then \(V = W\). If the dimension of \(W\) is zero, then \(W = \{0\}\)

Corollary

A basis \(B\) of \(W\) can be extended to be a basis of \(V\).

Properties

Sum

\(\dim(W_1 + W_2) = \dim (W_1) + \dim (W_2) - \dim (W_1 \cap W_2)\)

  1. Line and coincident plane

    \(\dim(\text{line + plane}) = 2 = \dim(\text{line}) + \dim(\text{plane}) - \text{dim(line} \cap \text{plane)}\)

    where \(\text{dim(plane)} - \text{dim(line} \cap \text{plane)} = 1\)

  2. Line and noncoincident plane

    \(\dim(\text{line + plane}) = 3 = \dim(\text{line}) + \dim(\text{plane}) - \dim(\text{line} \cap \text{plane)}\)

    where \(\dim(\text{plane}) - \dim(\text{line} \cap \text{plane)} = 0\)

Direct Sum

\(W_1 + W_2 = W_1 \oplus W_2\) iff \(\text{dim} (W_1 + W_2) = \text{dim} (W_1) + \text{dim} (W_2)\)

Proof

Linear Transformations

A linear transformation is a mapping \(T:V \rightarrow W\) from one vectorspace to another such that

  • addition is preserved

    • Let \(a,b \in V\). \(T(a) + T(b) = T(a + b)\)
  • scalar multiplication is preserved

    • Let \(a \in \mathbb{R}, x \in V\), \(T(ax) = aT(x)\)

Properties

  • \(T(\overrightarrow{0}) = \overrightarrow{0}\)
    • proof: \(T(\overrightarrow{0}) = T(0*\over{v}) = 0T(\overrightarrow{v}) = \overrightarrow{0}\)
  • \(T(\overrightarrow{u} - \overrightarrow{v}) = T(\overrightarrow{u}) - T(\overrightarrow{v})\)
    • proof: \(T(\overrightarrow{u} - \overrightarrow{v}) = T(\overrightarrow{u}) + T(-1 \overrightarrow{v}) = T(\overrightarrow{u}) - T(\overrightarrow{v})\)
  1. For \(T:P_n(\mathbb{R}) \rightarrow P_{n-1}(\mathbb{R})\) where \(T(f) = f'\) for \(f \in P_n(\mathbb{R})\). Show \(T\) is linear.

    Let \(a,b \in \mathbb{R}\), \(f,g \in P_n(\mathbb{R})\).

    \(T(af + bg) = (af + bg)' = af' + bg' = aT(f) + bT(g)\)

Identity Transformation

A mapping \(T:v \rightarrow W\) such that \(T(\overrightarrow{v} = \overrightarrow{v}\) for \(\overrightarrow{v} \in V\) #enddefinition

#+begindefinition Zero Transformation

A mapping \(T:V \rightarrow W\) such that \(T(\overrightarrow{v}) = \overrightarrow{0}\) for \(\overrightarrow{v} in V\)

Kernel and Image

Kernel

The kernel (\(\ker(T)\) or \(N(T)\)) of a transformation \(T:V \rightarrow W\) is the set of vectors in \(V\) that map to \(\overrightarrow{0}\) in \(W\)

Image

The image (\(Im(T)\) or \(R(T)\)) of a transformation \(T:V \rightarrow W\) is the set of vectors in \(W\) mapped to by vecotrs in \(V\).

  1. Prove the kernel of \(T\) is a subspace of \(V\). (show \(0 \in \ker(T)\), \(\ker(T)\) closed for addition, \(\ker(T)\) closed for scalar multiplication

    1. By definition \(T(0) = 0\), so \(0 \in \ker(T)\)

    2. For \(\overrightarrow{u}, \overrightarrow{v} \in \ker(T)\), we have \(T(\overrightarrow{u}) = T(\overrightarrow{v}) = 0\)

      \(T(\overrightarrow{u} + \overrightarrow{v}) = T(\overrightarrow{u}) + T(\overrightarrow{v}) = 0\)

      so \(\overrightarrow{u} + \overrightarrow{v} \in \ker(T)\)

    3. For \(a \in R\), \(v \in \ker(T)\) we have \(T(\overrightarrow{v}) = 0\)

      \(T(a\overrightarrow{v}) = aT(\overrightarrow{v}) = 0\), so \(a \overrightarrow{v} \in \ker(T)\)
  2. Let \(T:\mathbb{R}_3 \to \mathbb{R}_2\), where \[T \left( \begin{matrix} a\\b\\c \end{matrix} \right) = \left( \begin{matrix} a-b \\ 0 \\ 2c \end{matrix} \right)\]. Find the kernel of \(T\)

    \[\ker(T) = \left\{ \left( \begin{matrix} a\\b\\c \end{matrix} \right) \middle| T \left( \begin{matrix} a\\b\\c \end{matrix} \right) = 0 \right\}\]

    \[\left( \begin{matrix} a-b \\ 0 \\ 2c \end{matrix} \right) = 0\], so \(a = b\) and \(c = 0\)

    \[\ker(T) = \left\{ \left( \begin{matrix} a\\a\\0 \end{matrix} \right) \middle| a \in \mathbb{R} \right\}\]

  3. Find the image of \(T\).

    \[Im(T) = \left\{ T(\overrightarrow{v}) \middle| \overrightarrow{v} \in \mathbb{R}_3 \right\} = \left\{ \left( \begin{matrix} a-b \\ 0 \\ 2c \end{matrix} \right) \middle| a,b,c \in \mathbb{R} \right\} = \left\{ \left( \begin{matrix} a\\ 0 \\ b \end{matrix} \right) \middle| a,b \in \mathbb{R} \right\}\]