Overview
The kernel and range are fundamental subspaces associated with a linear transformation. They describe what gets mapped to zero and what can be reached by the transformation.
Kernel (Null Space)
The kernel of T : V → W T: V \to W T : V → W is:
ker ( T ) = { v ∈ V : T ( v ) = 0 } \ker(T) = \{\mathbf{v} \in V : T(\mathbf{v}) = \mathbf{0}\} ker ( T ) = { v ∈ V : T ( v ) = 0 }
The set of all vectors that map to zero.
Properties
ker ( T ) \ker(T) ker ( T ) is a subspace of V V V
dim ( ker ( T ) ) \dim(\ker(T)) dim ( ker ( T )) = nullity of T T T
T T T is injective (one-to-one) ⇔ ker ( T ) = { 0 } \Leftrightarrow \ker(T) = \{\mathbf{0}\} ⇔ ker ( T ) = { 0 }
Example
For T ( x , y , z ) = ( x + y , y + z ) T(x, y, z) = (x + y, y + z) T ( x , y , z ) = ( x + y , y + z ) :
T ( x , y , z ) = ( 0 , 0 ) T(x, y, z) = (0, 0) T ( x , y , z ) = ( 0 , 0 )
x + y = 0 ⇒ x = − y x + y = 0 \Rightarrow x = -y x + y = 0 ⇒ x = − y
y + z = 0 ⇒ z = − y y + z = 0 \Rightarrow z = -y y + z = 0 ⇒ z = − y
ker ( T ) = { ( − t , t , − t ) : t ∈ R } = span { ( − 1 , 1 , − 1 ) } \ker(T) = \{(-t, t, -t) : t \in \mathbb{R}\} = \text{span}\{(-1, 1, -1)\} ker ( T ) = {( − t , t , − t ) : t ∈ R } = span {( − 1 , 1 , − 1 )}
Range (Image)
The range of T : V → W T: V \to W T : V → W is:
range ( T ) = { T ( v ) : v ∈ V } = { w ∈ W : w = T ( v ) for some v } \text{range}(T) = \{T(\mathbf{v}) : \mathbf{v} \in V\} = \{\mathbf{w} \in W : \mathbf{w} = T(\mathbf{v}) \text{ for some } \mathbf{v}\} range ( T ) = { T ( v ) : v ∈ V } = { w ∈ W : w = T ( v ) for some v }
The set of all outputs of T T T .
Properties
range ( T ) \text{range}(T) range ( T ) is a subspace of W W W
dim ( range ( T ) ) \dim(\text{range}(T)) dim ( range ( T )) = rank of T T T
T T T is surjective (onto) ⇔ range ( T ) = W \Leftrightarrow \text{range}(T) = W ⇔ range ( T ) = W
Example
For T ( x , y ) = ( x + y , x − y , 2 x ) T(x, y) = (x + y, x - y, 2x) T ( x , y ) = ( x + y , x − y , 2 x ) :
T ( x , y ) = x ( 1 , 1 , 2 ) + y ( 1 , − 1 , 0 ) T(x, y) = x(1, 1, 2) + y(1, -1, 0) T ( x , y ) = x ( 1 , 1 , 2 ) + y ( 1 , − 1 , 0 )
range ( T ) = span { ( 1 , 1 , 2 ) , ( 1 , − 1 , 0 ) } \text{range}(T) = \text{span}\{(1, 1, 2), (1, -1, 0)\} range ( T ) = span {( 1 , 1 , 2 ) , ( 1 , − 1 , 0 )}
Dimension Theorem (Rank-Nullity)
For T : V → W T: V \to W T : V → W where V V V is finite-dimensional:
dim ( V ) = dim ( ker ( T ) ) + dim ( range ( T ) ) = nullity ( T ) + rank ( T ) \dim(V) = \dim(\ker(T)) + \dim(\text{range}(T)) = \text{nullity}(T) + \text{rank}(T) dim ( V ) = dim ( ker ( T )) + dim ( range ( T )) = nullity ( T ) + rank ( T )
Example
T : R 3 → R 2 T: \mathbb{R}^3 \to \mathbb{R}^2 T : R 3 → R 2 with nullity = 1
3 = 1 + rank ( T ) 3 = 1 + \text{rank}(T) 3 = 1 + rank ( T )
rank ( T ) = 2 \text{rank}(T) = 2 rank ( T ) = 2
So range ( T ) \text{range}(T) range ( T ) has dimension 2 (could be all of R 2 \mathbb{R}^2 R 2 ).
For Matrix Transformations
If T ( x ) = A x T(\mathbf{x}) = A\mathbf{x} T ( x ) = A x :
Concept Matrix Form ker ( T ) \ker(T) ker ( T ) Null space of A A A range ( T ) \text{range}(T) range ( T ) Column space of A A A nullity Number of free variables rank Number of pivot columns
Injectivity, Surjectivity, Bijectivity
Property Condition Kernel Range Injective One-to-one ker ( T ) = { 0 } \ker(T) = \{\mathbf{0}\} ker ( T ) = { 0 } - Surjective Onto - range ( T ) = W \text{range}(T) = W range ( T ) = W Bijective Both ker ( T ) = { 0 } \ker(T) = \{\mathbf{0}\} ker ( T ) = { 0 } range ( T ) = W \text{range}(T) = W range ( T ) = W
Determining Properties
T T T is Injective if:
ker ( T ) = { 0 } \ker(T) = \{\mathbf{0}\} ker ( T ) = { 0 }
T ( u ) = T ( v ) ⇒ u = v T(\mathbf{u}) = T(\mathbf{v}) \Rightarrow \mathbf{u} = \mathbf{v} T ( u ) = T ( v ) ⇒ u = v
nullity ( T ) = 0 \text{nullity}(T) = 0 nullity ( T ) = 0
T T T is Surjective if:
range ( T ) = W \text{range}(T) = W range ( T ) = W
rank ( T ) = dim ( W ) \text{rank}(T) = \dim(W) rank ( T ) = dim ( W )
Every w ∈ W \mathbf{w} \in W w ∈ W has a preimage
T T T is Bijective if:
Both injective and surjective
T T T has an inverse T − 1 T^{-1} T − 1
Example: Complete Analysis
T : R 3 → R 3 T: \mathbb{R}^3 \to \mathbb{R}^3 T : R 3 → R 3 defined by T ( x , y , z ) = ( x + y , y + z , x + 2 y + z ) T(x, y, z) = (x + y, y + z, x + 2y + z) T ( x , y , z ) = ( x + y , y + z , x + 2 y + z )
Matrix form:
A = [ 1 1 0 0 1 1 1 2 1 ] A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 1 \end{bmatrix} A = 1 0 1 1 1 2 0 1 1
Row reduce:
[ 1 1 0 0 1 1 1 2 1 ] → [ 1 1 0 0 1 1 0 1 1 ] → [ 1 0 − 1 0 1 1 0 0 0 ] \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 1 \end{bmatrix} \to \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} \to \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} 1 0 1 1 1 2 0 1 1 → 1 0 0 1 1 1 0 1 1 → 1 0 0 0 1 0 − 1 1 0
rank = 2, nullity = 1
ker ( T ) = span { ( 1 , − 1 , 1 ) } \ker(T) = \text{span}\{(1, -1, 1)\} ker ( T ) = span {( 1 , − 1 , 1 )}
range ( T ) = span { ( 1 , 0 , 1 ) , ( 1 , 1 , 2 ) } \text{range}(T) = \text{span}\{(1, 0, 1), (1, 1, 2)\} range ( T ) = span {( 1 , 0 , 1 ) , ( 1 , 1 , 2 )} (2-dimensional subspace of R 3 \mathbb{R}^3 R 3 )
Not injective (ker ≠ { 0 } \ker \neq \{\mathbf{0}\} ker = { 0 } )
Not surjective (range ≠ R 3 \text{range} \neq \mathbb{R}^3 range = R 3 )