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4.2 Null space, Column space, and Linear Transformations

nullspace of a matrix A is set of all solutions to \( A\vec{x} = \vec{0} \)

\[ \text{Nul } A = \{ \vec{x} : \vec{x} \text{ is in } \mathbb{R}^n \text{ and } A\vec{x} = \vec{0} \} \]

If \( A \) is \( m \times n \) then \( \text{Nul } A \) is set of all vectors that are from \( \mathbb{R}^n \) to \( \mathbb{R}^m \).

\( \vec{0} \) in \( \mathbb{R}^m \)

\( A\vec{x} = \vec{0} \) defines null space implicitly. To solve explicitly, solve \( A\vec{x} = \vec{0} \).

Is null space of A a subspace?

  1. a). \( \vec{0} \) exists? Yes, because \( A\vec{0} = \vec{0} \)
  2. b). Closed under addition? For \( \vec{u}, \vec{v} \) such that \( A\vec{u} = \vec{0}, A\vec{v} = \vec{0} \).
    \[ A(\vec{u} + \vec{v}) = A\vec{u} + A\vec{v} = \vec{0} \]
    Yes, closed under addition.
  3. c). Closed under scalar multiplication? If \( A\vec{u} = \vec{0} \), then
    \[ A(c\vec{u}) = c A\vec{u} = \vec{0} \]
    Yes, closed.
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An example of null space not related to matrix.

\[ y'' + 4y = 0 \]

linear 2nd-order homogeneous differential eq.

solution: \( y(x) = ? \)

can be viewed as a transformation of a "vector" \( y(x) \) such that it gets mapped to the zero "vector"

  • domain: all twice-differentiable functions
  • range: such that \[ y'' + 4y = 0 \]

\( y_1 = \cos 2x \quad y_2 = \sin 2x \) → these live in the null space

note \( y = 0 \) also satisfies the differential eq.

\[ (y_1 + y_2)'' + 4(y_1 + y_2) = 0 \]

\[ y_1'' + 4y_1 + y_2'' + 4y_2 = 0 \]

closed under addition

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\[ (cy_1)'' + 4(cy_1) = 0 \] \[ cy_1'' + 4cy_1 = 0 \] \[ c(y_1'' + 4y_1) = 0 \]

closed under scalar multiplication

The Null Space and Kernel

The null space of a linear transformation is also called the kernel.

Column Space

For an \( m \times n \) matrix \( A \) the column space is the linear combination of columns of \( A \).

if \( A = [\vec{a}_1, \vec{a}_2, \dots, \vec{a}_n] \)

\[ \text{Col } A = \text{span} \{ \vec{a}_1, \vec{a}_2, \dots, \vec{a}_n \} \] \[ = \{ \vec{b} : \vec{b} = A\vec{x} \text{ for some } \vec{x} \text{ in } \mathbb{R}^n \} \]
  • Nul A: associated w/ the domain of transformation
  • Col A: associated w/ the range of transformation
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We can often recover \( A \) if we have the output vector.

example

\[ \left\{ \begin{bmatrix} -3r + 2s + 3t \\ -r - 2s \\ r + 3s - 2t \\ 2r - 3s + t \end{bmatrix} \right\} \quad r, s, t \text{ are constants} \]
\[ r \begin{bmatrix} -3 \\ -1 \\ 1 \\ 2 \end{bmatrix} + s \begin{bmatrix} 2 \\ -2 \\ 3 \\ -3 \end{bmatrix} + t \begin{bmatrix} 3 \\ 0 \\ -2 \\ 1 \end{bmatrix} \]
\[ = \underbrace{\begin{bmatrix} -3 & 2 & 3 \\ -1 & -2 & 0 \\ 1 & 3 & -2 \\ 2 & -3 & 1 \end{bmatrix}}_{A} \begin{bmatrix} r \\ s \\ t \end{bmatrix} \]
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Is \(\text{Col } A\) a subspace?

From 4.1, we know if \(\{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_p \}\) span some space and \(\vec{v}_1, \vec{v}_2, \dots, \vec{v}_p\) are in a vector space, then \(\text{span } \{ \vec{v}_1, \vec{v}_2, \dots, \vec{v}_p \}\) is a subspace.

Consistency and Column Space

If we know \(A\vec{x} = \vec{v}\) and \(A\vec{x} = \vec{w}\) are both consistent, what can we say about \(A\vec{x} = \vec{v} + \vec{w}\)?

  • If \(A\vec{x} = \vec{v}\) is consistent, then \(\vec{v}\) is in \(\text{Col } A\)
  • If \(A\vec{x} = \vec{w}\) is consistent, then \(\vec{w}\) is in \(\text{Col } A\)

\(\text{Col } A\) is a subspace so \(\vec{v} + \vec{w}\) is also in \(\text{Col } A\).

Therefore \(\vec{v} + \vec{w}\) must be a linear combo of columns of \(A\).

\(\implies A\vec{x} = \vec{v} + \vec{w}\) is consistent

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Linear Transformations

A Linear transformation maps a vector space \(V\) to another vector space \(W\) such that for each \(\vec{x}\) in \(V\):

\(T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v})\)

\(T(c\vec{u}) = cT(\vec{u})\)

implies \(T(\vec{0}) = \vec{0}\)

Example

\(T: \mathbb{P}_2 \to \mathbb{R}^2\) by \(T(p) = \begin{bmatrix} p(0) \\ p(1) \end{bmatrix}\)

\(\mathbb{P}_2\): 2nd-deg polynomials

For example, if \(p(t) = 3 + 5t + 7t^2\) then \(T(p) = \begin{bmatrix} 3 \\ 15 \end{bmatrix}\)

Questions

  • Is \(T\) a linear transformation?
  • Find \(p\) in \(\mathbb{P}_2\) that spans the kernel of \(T\).
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Linear Transformations and Kernels

If \( T \) is linear, then:

\[ T(p + q) = T(p) + T(q) \]\[ T(c p) = c T(p) \]

(true, may help if think in terms of deriv. / integral or common operations w/ polynomials)

Finding the Kernel

Find \( p \) in \( \mathbb{P}_2 \) that spans kernel of \( T \):

\[ T(p) = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} p(0) \\ p(1) \end{bmatrix} \]

This means \( p(t) \) is a multiple of \( (t)(t-1) \).

\[ p(t) = c t (t - 1) \]

This polynomial ("vector") spans the kernel.

Verification of Linearity

\[ T(p + q) = \begin{bmatrix} (p + q)(0) \\ (p + q)(1) \end{bmatrix} = \begin{bmatrix} p(0) + q(0) \\ p(1) + q(1) \end{bmatrix} = \begin{bmatrix} p(0) \\ p(1) \end{bmatrix} + \begin{bmatrix} q(0) \\ q(1) \end{bmatrix} \]\[ T(c p) = \begin{bmatrix} c p(0) \\ c p(1) \end{bmatrix} = c \begin{bmatrix} p(0) \\ p(1) \end{bmatrix} \]
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Example: Subspaces under Linear Transformations

\( T: V \to W \) where \( V, W \) are vector spaces.

\( U \) is a subspace of \( V \).

Is \( T(U) \) a subspace of \( W \)?

Mapping diagram from vector space V containing subspace U to space W containing image T(U).

1. Zero Vector Property

Since \( V \) is a vector space and \( U \) is a subspace of \( V \), \( \vec{0}_V \) is in \( U \). Because \( T \) is linear, so:

\[ T(\vec{0}_V) = \vec{0}_W \]

which is in \( T(U) \), so \( T(U) \) contains the zero vector.

2. Closure Under Addition

Let \( \vec{x} \) and \( \vec{y} \) be vectors in \( U \). Then \( T(\vec{x} + \vec{y}) = T(\vec{x}) + T(\vec{y}) \) because \( T \) is linear. Since \( \vec{x} + \vec{y} \) is in \( U \), so \( T(U) \) is closed under addition.

3. Closure Under Scalar Multiplication

\( c\vec{u} \) is in \( U \) because \( U \) is a subspace and \( T \) is linear, so:

\[ T(c\vec{u}) = c T(\vec{u}) \]

This shows \( T(U) \) is closed under scalar multiplication.