Null Space and Kernel

Definition

The null space (or kernel) of a linear transformation T:VWT: V \rightarrow W is the set of all vectors in VV that map to the zero vector in WW:

Null(T)=ker(T)={vV:T(v)=0W}\text{Null}(T) = \ker(T) = \{v \in V : T(v) = \vec{0}_W\}

where 0W\vec{0}_W is the zero vector in WW.

Properties

  1. Subspace: The null space of a linear transformation T:VWT: V \rightarrow W is a subspace of VV.

  2. Dimension: The dimension of the null space is called the nullity of TT:

nullity(T)=dim(ker(T))\text{nullity}(T) = \dim(\ker(T))
  1. Injectivity: A linear transformation TT is injective (one-to-one) if and only if ker(T)={0V}\ker(T) = {\vec{0}_V}, i.e., the null space contains only the zero vector.
    • This means distinct inputs always yield distinct outputs
    • By the Rank-Nullity Theorem, TT can only be injective if dim(V)dim(W)\dim(V) \leq \dim(W)

Null Space of a Matrix

For a matrix ARm×nA \in \mathbb{R}^{m \times n}, the null space of AA is the set of all vectors xRn\vec{x} \in \mathbb{R}^n such that Ax=0A\vec{x} = \vec{0}:

Null(A)={xRn:Ax=0}\text{Null}(A) = \{\vec{x} \in \mathbb{R}^n : A\vec{x} = \vec{0}\}

This corresponds to the null space of the linear transformation TA:RnRmT_A: \mathbb{R}^n \rightarrow \mathbb{R}^m defined by TA(x)=AxT_A(\vec{x}) = A\vec{x}.

Finding the Null Space of a Matrix

To find a basis for the null space of a matrix AA:

  1. Transform AA into row-reduced echelon form (RREF) using Gaussian elimination.
  2. Identify the free variables (variables that are not leading variables in the RREF).
  3. Express the leading variables in terms of the free variables.
  4. Write the general solution as a linear combination of vectors, each corresponding to setting one free variable to 1 and the others to 0.
  5. These vectors form a basis for the null space.

Example

Consider the matrix:

A=[123246]A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{bmatrix}

The RREF of AA is:

[123000]\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \end{bmatrix}

The equation Ax=0A\vec{x} = \vec{0} becomes:

[123000][x1x2x3]=[00]\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

This gives us:

x1+2x2+3x3=0x_1 + 2x_2 + 3x_3 = 0

Solving for x1x_1:

x1=2x23x3x_1 = -2x_2 - 3x_3

The general solution is:

x=[2x23x3x2x3]=x2[210]+x3[301]\vec{x} = \begin{bmatrix} -2x_2 - 3x_3 \\ x_2 \\ x_3 \end{bmatrix} = x_2 \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix}

Therefore, a basis for the null space of AA is:

{[210],[301]}\left\{ \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} \right\}

And the nullity of AA is 2.

Rank-Nullity Theorem

The Rank-Nullity Theorem establishes a fundamental relationship between the null space and image:

Theorem: If T:VWT: V \rightarrow W is a linear transformation and VV is finite-dimensional, then:

dim(V)=rank(T)+nullity(T)\dim(V) = \text{rank}(T) + \text{nullity}(T)

where rank(T)=dim(Im(T))\text{rank}(T) = \dim(\text{Im}(T)) and nullity(T)=dim(ker(T))\text{nullity}(T) = \dim(\ker(T)).

This theorem provides powerful insights:

  • The sum of the dimensions of the image and kernel equals the dimension of the domain
  • Understanding one space helps characterize the other
  • It directly connects to the concepts of injectivity and surjectivity

Interpretation

The Rank-Nullity Theorem tells us that the dimension of the domain equals the sum of:

  • The dimension of the image (how much of the domain is "preserved" by the transformation)
  • The dimension of the kernel (how much of the domain is "collapsed" to zero)

Matrix Form

For a matrix ARm×nA \in \mathbb{R}^{m \times n}:

n=rank(A)+dim(Null(A))n = \text{rank}(A) + \dim(\text{Null}(A))

Applications

  1. Systems of Linear Equations: The null space of a coefficient matrix represents the solution space of the homogeneous system.

  2. Linear Independence: Vectors {v1,v2,,vk}{v_1, v_2, \ldots, v_k} are linearly independent if and only if the null space of the matrix with these vectors as columns contains only the zero vector.

  3. Eigenvalues and Eigenvectors: For a square matrix AA, the eigenvectors corresponding to eigenvalue λ\lambda form the null space of (AλI)(A - \lambda I).

  4. Differential Equations: The null space of a differential operator corresponds to the solution space of the homogeneous differential equation.

Exercise

Find a basis for the null space of the matrix:

B=[121124003611]B = \begin{bmatrix} 1 & 2 & 1 & 1 \\ 2 & 4 & 0 & 0 \\ 3 & 6 & 1 & 1 \end{bmatrix}

Verify the Rank-Nullity Theorem for this matrix.