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Linear Maps & Kernel/Image

A linear map is a function between vector spaces that preserves the vector space structure. The kernel and image are fundamental subspaces that characterize the behavior of linear maps.

1. Linear Maps

Definition 4.1: Linear Map

Let VV and WW be vector spaces over the same field FF. A function T:VWT: V \to W is called a linear map (or linear transformation) if for all vectors u,vVu, v \in V and all scalars α,βF\alpha, \beta \in F:

T(αu+βv)=αT(u)+βT(v)T(\alpha u + \beta v) = \alpha T(u) + \beta T(v)
Remark 4.1: Equivalent Formulation

The linearity condition can be split into two separate properties:

  • Additivity: T(u+v)=T(u)+T(v)T(u + v) = T(u) + T(v) for all u,vVu, v \in V
  • Homogeneity: T(αv)=αT(v)T(\alpha v) = \alpha T(v) for all αF\alpha \in F and vVv \in V
Theorem 4.1: Basic Properties of Linear Maps

Let T:VWT: V \to W be a linear map. Then:

  1. T(0V)=0WT(0_V) = 0_W (T sends the zero vector to the zero vector)
  2. T(v)=T(v)T(-v) = -T(v) for all vVv \in V
  3. T(vw)=T(v)T(w)T(v - w) = T(v) - T(w) for all v,wVv, w \in V
Proof of Theorem 4.1 (Property 1):

Using homogeneity with α=0\alpha = 0:

T(0V)=T(0v)=0T(v)=0WT(0_V) = T(0 \cdot v) = 0 \cdot T(v) = 0_W
Example 4.1: Standard Examples of Linear Maps
  • Zero Map: 0:VW0: V \to W defined by 0(v)=0W0(v) = 0_W for all vVv \in V
  • Identity Map: IV:VVI_V: V \to V defined by IV(v)=vI_V(v) = v for all vVv \in V
  • Differentiation: D:Pn(R)Pn1(R)D: P_n(\mathbb{R}) \to P_{n-1}(\mathbb{R}) defined by D(p)=pD(p) = p'
Example 4.2: Non-Examples

The following are NOT linear maps:

  • Translation: T(x)=x+cT(x) = x + c for c0c \neq 0. Since T(0)=c0T(0) = c \neq 0.
  • Product map: T(x1,x2)=(x1x2,x1+x2)T(x_1, x_2) = (x_1 x_2, x_1 + x_2). Not additive.
Example 4.3: More Linear Map Examples
  • Projection: π:R3R2\pi: \mathbb{R}^3 \to \mathbb{R}^2 defined by π(x,y,z)=(x,y)\pi(x, y, z) = (x, y)
  • Rotation: In R2\mathbb{R}^2, rotation by angle θ\theta is linear
  • Integration: I:Pn(R)Pn+1(R)I: P_n(\mathbb{R}) \to P_{n+1}(\mathbb{R}) defined by I(p)(x)=0xp(t)dtI(p)(x) = \int_0^x p(t) dt
  • Matrix multiplication: For fixed AMm×n(F)A \in M_{m \times n}(F), the map T:FnFmT: F^n \to F^m defined by T(x)=AxT(x) = Ax is linear
Theorem 1.1: Linear Maps Preserve Linear Combinations

A map T:VWT: V \to W is linear if and only if for all v1,,vkVv_1, \ldots, v_k \in V and α1,,αkF\alpha_1, \ldots, \alpha_k \in F:

T(i=1kαivi)=i=1kαiT(vi)T\left(\sum_{i=1}^k \alpha_i v_i\right) = \sum_{i=1}^k \alpha_i T(v_i)

2. Kernel (Null Space)

Definition 4.2: Kernel

The kernel (or null space) of a linear map T:VWT: V \to W is:

ker(T)={vV:T(v)=0W}\ker(T) = \{v \in V : T(v) = 0_W\}
Theorem 4.2: Kernel is a Subspace

For any linear map T:VWT: V \to W, ker(T)\ker(T) is a subspace of VV.

Proof:

We verify the subspace criterion:

  • Contains zero: T(0)=0T(0) = 0, so 0ker(T)0 \in \ker(T)
  • Closed under addition: If u,vker(T)u, v \in \ker(T), then T(u+v)=T(u)+T(v)=0+0=0T(u + v) = T(u) + T(v) = 0 + 0 = 0
  • Closed under scalar multiplication: If vker(T)v \in \ker(T) and αF\alpha \in F, then T(αv)=αT(v)=α0=0T(\alpha v) = \alpha T(v) = \alpha \cdot 0 = 0
Theorem 4.3: Injectivity via Kernel

A linear map T:VWT: V \to W is injective (one-to-one) if and only if ker(T)={0}\ker(T) = \{0\}.

Proof:

(⇒) If TT is injective and vker(T)v \in \ker(T), then T(v)=0=T(0)T(v) = 0 = T(0). By injectivity, v=0v = 0.

(⇐) If ker(T)={0}\ker(T) = \{0\} and T(u)=T(v)T(u) = T(v), then:

T(uv)=T(u)T(v)=0T(u - v) = T(u) - T(v) = 0

So uvker(T)={0}u - v \in \ker(T) = \{0\}, hence u=vu = v.

Example 4.3: Kernel Examples
  • For differentiation D:PnPn1D: P_n \to P_{n-1}, ker(D)={c:cR}\ker(D) = \{c : c \in \mathbb{R}\} (constant polynomials)
  • For T(x,y)=(x+y,x+y)T(x, y) = (x + y, x + y), ker(T)={(t,t):tR}\ker(T) = \{(t, -t) : t \in \mathbb{R}\}
  • For the identity map, ker(IV)={0}\ker(I_V) = \{0\}
Theorem 2.1: Kernel and Linear Independence

If {v1,,vk}\{v_1, \ldots, v_k\} is linearly independent and T:VWT: V \to W is injective, then {T(v1),,T(vk)}\{T(v_1), \ldots, T(v_k)\} is linearly independent.

Proof:

If αiT(vi)=0\sum \alpha_i T(v_i) = 0, then T(αivi)=0T(\sum \alpha_i v_i) = 0. Since ker(T)={0}\ker(T) = \{0\} (injectivity), αivi=0\sum \alpha_i v_i = 0. By independence, all αi=0\alpha_i = 0.

Remark 2.1: Kernel as Measure of Non-Injectivity

The kernel measures how much a linear map "collapses" the domain. A large kernel means many different vectors map to the same output, indicating non-injectivity.

3. Image (Range)

Definition 4.3: Image

The image (or range) of a linear map T:VWT: V \to W is:

im(T)={T(v):vV}\text{im}(T) = \{T(v) : v \in V\}
Theorem 4.4: Image is a Subspace

For any linear map T:VWT: V \to W, im(T)\text{im}(T) is a subspace of WW.

Proof:

We verify the subspace criterion:

  • Contains zero: T(0)=0T(0) = 0, so 0im(T)0 \in \text{im}(T)
  • Closed under addition: If w1,w2im(T)w_1, w_2 \in \text{im}(T), say wi=T(vi)w_i = T(v_i), then w1+w2=T(v1)+T(v2)=T(v1+v2)im(T)w_1 + w_2 = T(v_1) + T(v_2) = T(v_1 + v_2) \in \text{im}(T)
  • Closed under scalar multiplication: If w=T(v)im(T)w = T(v) \in \text{im}(T) and αF\alpha \in F, then αw=αT(v)=T(αv)im(T)\alpha w = \alpha T(v) = T(\alpha v) \in \text{im}(T)
Theorem 4.5: Surjectivity via Image

A linear map T:VWT: V \to W is surjective (onto) if and only if im(T)=W\text{im}(T) = W.

Example 4.4: Image Examples
  • For T(x,y,z)=(x,y,0)T(x, y, z) = (x, y, 0), im(T)={(x,y,0):x,yR}\text{im}(T) = \{(x, y, 0) : x, y \in \mathbb{R}\} (the xy-plane)
  • For the identity map, im(IV)=V\text{im}(I_V) = V
  • For the zero map, im(0)={0}\text{im}(0) = \{0\}
Theorem 3.1: Image and Spanning Sets

If {v1,,vk}\{v_1, \ldots, v_k\} spans VV, then {T(v1),,T(vk)}\{T(v_1), \ldots, T(v_k)\} spans im(T)\text{im}(T). If TT is surjective, then {T(v1),,T(vk)}\{T(v_1), \ldots, T(v_k)\} spans WW.

Remark 3.1: Image as Measure of Surjectivity

The image measures how much of the codomain is "covered" by the map. If im(T)=W\text{im}(T) = W, the map is surjective; otherwise, it misses some vectors in WW.

4. Computing Kernel and Image

Theorem 4.6: Image from Basis

If {v1,,vn}\{v_1, \ldots, v_n\} spans VV, then {T(v1),,T(vn)}\{T(v_1), \ldots, T(v_n)\} spans im(T)\text{im}(T).

Proof:

Any wim(T)w \in \text{im}(T) equals T(v)T(v) for some v=αiviv = \sum \alpha_i v_i. Then:

w=T(v)=T(αivi)=αiT(vi)w = T(v) = T\left(\sum \alpha_i v_i\right) = \sum \alpha_i T(v_i)

So {T(v1),,T(vn)}\{T(v_1), \ldots, T(v_n)\} spans im(T)\text{im}(T).

Example 4.5: Finding Kernel and Image

Let T:R3R2T: \mathbb{R}^3 \to \mathbb{R}^2 be defined by T(x,y,z)=(x+y,y+z)T(x, y, z) = (x + y, y + z).

Kernel: T(x,y,z)=(0,0)T(x, y, z) = (0, 0) gives:

x+y=0,y+z=0x + y = 0, \quad y + z = 0

So y=xy = -x and z=y=xz = -y = x. Thus:

ker(T)={(t,t,t):tR}\ker(T) = \{(t, -t, t) : t \in \mathbb{R}\}

Image: Apply TT to the standard basis:

T(1,0,0)=(1,0),T(0,1,0)=(1,1),T(0,0,1)=(0,1)T(1,0,0) = (1,0), \quad T(0,1,0) = (1,1), \quad T(0,0,1) = (0,1)

So im(T)=span{(1,0),(1,1),(0,1)}=R2\text{im}(T) = \text{span}\{(1,0), (1,1), (0,1)\} = \mathbb{R}^2 (since these span R2\mathbb{R}^2).

Example 4.6: Computing via Matrix Representation

For a linear map T:RnRmT: \mathbb{R}^n \to \mathbb{R}^m given by T(x)=AxT(x) = Ax for matrix AA:

  • ker(T)={x:Ax=0}\ker(T) = \{x : Ax = 0\} (null space of AA)
  • im(T)={Ax:xRn}\text{im}(T) = \{Ax : x \in \mathbb{R}^n\} (column space of AA)

Row reduce AA to find kernel (free variables) and image (pivot columns).

5. Composition of Linear Maps

Composing linear maps gives another linear map. Understanding composition is crucial for understanding how linear transformations interact and combine.

Theorem 5.1: Composition is Linear

If T:UVT: U \to V and S:VWS: V \to W are linear maps, then their composition ST:UWS \circ T: U \to W defined by (ST)(u)=S(T(u))(S \circ T)(u) = S(T(u)) is also linear.

Proof:

For u1,u2Uu_1, u_2 \in U and αF\alpha \in F:

(ST)(αu1+u2)=S(T(αu1+u2))=S(αT(u1)+T(u2))=αS(T(u1))+S(T(u2))=α(ST)(u1)+(ST)(u2)(S \circ T)(\alpha u_1 + u_2) = S(T(\alpha u_1 + u_2)) = S(\alpha T(u_1) + T(u_2)) = \alpha S(T(u_1)) + S(T(u_2)) = \alpha (S \circ T)(u_1) + (S \circ T)(u_2)
Theorem 5.2: Kernel and Image of Composition

For linear maps T:UVT: U \to V and S:VWS: V \to W:

  1. ker(T)ker(ST)\ker(T) \subseteq \ker(S \circ T)
  2. im(ST)im(S)\text{im}(S \circ T) \subseteq \text{im}(S)
  3. If SS is injective, then ker(ST)=ker(T)\ker(S \circ T) = \ker(T)
  4. If TT is surjective, then im(ST)=im(S)\text{im}(S \circ T) = \text{im}(S)
Proof:

(1) If uker(T)u \in \ker(T), then T(u)=0T(u) = 0, so (ST)(u)=S(0)=0(S \circ T)(u) = S(0) = 0. Thus uker(ST)u \in \ker(S \circ T).

(2) Any wim(ST)w \in \text{im}(S \circ T) equals S(T(u))S(T(u)) for some uu, so wim(S)w \in \text{im}(S).

(3) If uker(ST)u \in \ker(S \circ T), then S(T(u))=0S(T(u)) = 0. Since SS is injective, T(u)=0T(u) = 0, so uker(T)u \in \ker(T).

(4) If wim(S)w \in \text{im}(S), say w=S(v)w = S(v), and TT is surjective, then v=T(u)v = T(u) for some uu, so w=S(T(u))im(ST)w = S(T(u)) \in \text{im}(S \circ T).

Example 5.1: Composition Example

Let T:R2R3T: \mathbb{R}^2 \to \mathbb{R}^3 be T(x,y)=(x,y,0)T(x, y) = (x, y, 0) and S:R3R2S: \mathbb{R}^3 \to \mathbb{R}^2 be S(x,y,z)=(x,y)S(x, y, z) = (x, y).

Then (ST)(x,y)=S(x,y,0)=(x,y)(S \circ T)(x, y) = S(x, y, 0) = (x, y), so ST=IR2S \circ T = I_{\mathbb{R}^2} (identity).

Note: ker(T)={0}\ker(T) = \{0\} and ker(ST)={0}\ker(S \circ T) = \{0\} (consistent with (3)).

Theorem 5.3: Associativity of Composition

For linear maps T:UVT: U \to V, S:VWS: V \to W, R:WXR: W \to X:

(RS)T=R(ST)(R \circ S) \circ T = R \circ (S \circ T)

6. Linear Maps and Subspaces

Linear maps interact naturally with subspaces through preimages and restrictions. These constructions are fundamental for understanding the structure of linear transformations.

Definition 6.1: Preimage

For a linear map T:VWT: V \to W and a subspace UWU \subseteq W, the preimage (or inverse image) is:

T1(U)={vV:T(v)U}T^{-1}(U) = \{v \in V : T(v) \in U\}
Theorem 6.1: Preimage of Subspace is Subspace

If T:VWT: V \to W is linear and UWU \subseteq W is a subspace, then T1(U)T^{-1}(U) is a subspace of VV.

Proof:

We verify the subspace criterion:

  • Contains zero: T(0)=0UT(0) = 0 \in U, so 0T1(U)0 \in T^{-1}(U)
  • Closed under addition: If v1,v2T1(U)v_1, v_2 \in T^{-1}(U), then T(v1),T(v2)UT(v_1), T(v_2) \in U, so T(v1+v2)=T(v1)+T(v2)UT(v_1 + v_2) = T(v_1) + T(v_2) \in U
  • Closed under scalar multiplication: If vT1(U)v \in T^{-1}(U) and αF\alpha \in F, then T(αv)=αT(v)UT(\alpha v) = \alpha T(v) \in U
Definition 6.2: Restriction

For a linear map T:VWT: V \to W and a subspace UVU \subseteq V, the restriction of TT to UU is the map TU:UWT|_U: U \to W defined by TU(u)=T(u)T|_U(u) = T(u) for all uUu \in U.

Theorem 6.2: Properties of Restriction

For T:VWT: V \to W and subspace UVU \subseteq V:

  1. ker(TU)=ker(T)U\ker(T|_U) = \ker(T) \cap U
  2. im(TU)=T(U)\text{im}(T|_U) = T(U) (the image of the subspace UU under TT)
Example 6.1: Preimage Example

Let T:R3R2T: \mathbb{R}^3 \to \mathbb{R}^2 be T(x,y,z)=(x,y)T(x, y, z) = (x, y) and U={(0,t):tR}U = \{(0, t) : t \in \mathbb{R}\} (y-axis in R2\mathbb{R}^2).

Then T1(U)={(0,y,z):y,zR}T^{-1}(U) = \{(0, y, z) : y, z \in \mathbb{R}\} (the yz-plane in R3\mathbb{R}^3).

Example 6.2: Restriction Example

Let T:R3R3T: \mathbb{R}^3 \to \mathbb{R}^3 be T(x,y,z)=(x,y,0)T(x, y, z) = (x, y, 0) (projection onto xy-plane).

Restrict to U={(x,y,0):x,yR}U = \{(x, y, 0) : x, y \in \mathbb{R}\} (xy-plane). Then TU=IUT|_U = I_U (identity on UU).

Theorem 6.3: Image of Subspace

If T:VWT: V \to W is linear and UVU \subseteq V is a subspace, then T(U)={T(u):uU}T(U) = \{T(u) : u \in U\} is a subspace of WW.

Proof:

This is essentially the same as showing im(TU)\text{im}(T|_U) is a subspace, which follows from TUT|_U being linear.

Frequently Asked Questions

What's the difference between 'linear map', 'linear transformation', and 'linear operator'?

These terms are often used interchangeably, but with slight distinctions: 'Linear map' is the most general term for any linear function T: V → W. 'Linear transformation' is synonymous, though some authors prefer it when V = W. 'Linear operator' typically refers to linear maps from a space to itself (T: V → V), also called endomorphisms.

How do I check if a function is linear?

Method 1: Verify both properties separately: (a) T(u + v) = T(u) + T(v) for all u, v (additivity), and (b) T(αv) = αT(v) for all scalars α and vectors v (homogeneity). Method 2: Verify the combined condition: T(αu + βv) = αT(u) + βT(v) for all scalars α, β and vectors u, v. Quick check: If T(0) ≠ 0, then T is NOT linear.

Why must linear maps preserve the zero vector?

This follows from homogeneity: T(0) = T(0·v) = 0·T(v) = 0. This is a useful quick test: if T(0) ≠ 0, the map is definitely NOT linear.

How do kernel and image relate to matrices?

For T represented by matrix A: ker(T) = null space of A (solution space of Ax = 0), im(T) = column space of A. This connects abstract linear algebra to matrix computations and row reduction.

What's the geometric meaning of the kernel?

The kernel is what T 'collapses' to zero. For a projection onto a plane, the kernel is the line perpendicular to that plane. For differentiation, the kernel is constant functions. It measures the 'dimension loss' of the map.

Linear Maps & Kernel/Image Practice
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Is T(x,y)=(x+y,xy)T(x, y) = (x + y, x - y) linear from R2\mathbb{R}^2 to R2\mathbb{R}^2?
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Is T(x)=x+1T(x) = x + 1 linear from R\mathbb{R} to R\mathbb{R}?
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What is ker(T)\ker(T) where T(x,y)=(x+y,x+y)T(x,y) = (x+y, x+y)?
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If TT is linear and injective, what is dim(kerT)\dim(\ker T)?
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Is the image of a linear map always a subspace?
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T:R3R3T: \mathbb{R}^3 \to \mathbb{R}^3, T(x,y,z)=(x,y,0)T(x,y,z) = (x,y,0). What is im(T)\text{im}(T)?
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If im(T)=W\text{im}(T) = W, then TT is:
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What is ker(IV)\ker(I_V) where IVI_V is the identity on VV?
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If ker(T){0}\ker(T) \neq \{0\}, then TT is:
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If dimV=5\dim V = 5 and dim(kerT)=2\dim(\ker T) = 2, what is dim(im T)\dim(\text{im } T)?
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