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Lesson 113 — Kernel and Image

Kernel, image (range), rank, and nullity. Rank-nullity theorem. The two structures tied to every linear transformation answer 'who dies?' and 'how far do we reach?'

Used in: 3º year advanced HS (Brazil) · German Lineare Algebra Leistungskurs · Singapore H2 Mathematics · MIT 18.06 Linear Algebra

\dim V = \dim \ker T + \dim \operatorname{Im}(T)

The rank-nullity theorem says that the dimension of the domain equals the sum of the dimension of what "dies" under $T$ (the kernel) and the dimension of what is reachable (the image). It is the most central result in applied linear algebra.

Choose your door

Rigorous notation, full derivation, hypotheses

Definitions and central theorem

Kernel and image

"The null space of $T$ , denoted $\mathcal{N}(T)$ , is the set $\mathcal{N}(T) = \{v \in V : T(v) = 0\}$ ." — Beezer, A First Course in Linear Algebra, §KER

"The range of $T$ , denoted $\mathcal{R}(T)$ , is the set $\mathcal{R}(T) = \{T(v) : v \in V\}$ ." — Beezer, A First Course in Linear Algebra, §RNG

The four fundamental subspaces

For $A \in \mathbb{R}^{m \times n}$ with rank $r$ :

Strang's four fundamental subspaces. Rank r appears twice; nullity n−r and m−r fill the orthogonal complement.

Characterizations of injectivity and surjectivity

Worked examples

Example— 1· Kernel and image of a projection (basic)

Problem. Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be the projection $T(x, y, z) = (x, y, 0)$ . Find $\ker T$ , $\operatorname{Im}(T)$ , rank, and nullity. Verify the rank-nullity theorem.

Strategy. Solve $T(x,y,z) = (0,0,0)$ for the kernel. Describe $\{T(v) : v \in \mathbb{R}^3\}$ for the image. Check dimensions.

Solution.

$T(x,y,z) = (0,0,0) \iff (x,y,0) = (0,0,0) \iff x = 0$ and $y = 0$ . Thus $\ker T = \{(0, 0, z) : z \in \mathbb{R}\}$ — the $z$ -axis.

$\operatorname{Im}(T) = \{(x, y, 0) : x, y \in \mathbb{R}\}$ — the $xy$ -plane.

$\dim \ker T = 1$ ; $\dim \operatorname{Im}(T) = 2$ .

Rank-nullity check: $1 + 2 = 3 = \dim \mathbb{R}^3$ . ✓

Verification. $T$ is not injective ( $\ker T \neq \{0\}$ ). $T$ is not surjective ( $\operatorname{Im}(T) \neq \mathbb{R}^3$ ). Geometrically: $T$ "flattens" the $z$ -direction.

Source. Beezer, A First Course in Linear Algebra, §ILT, Example ILT.STLT — GNU FDL.

Example— 2· Kernel and image via row reduction (intermediate)

Problem. For $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{pmatrix}$ , find $\ker A$ , $\operatorname{Im}(A)$ , rank, and nullity.

Strategy. Row-reduce $[A \mid 0]$ for the kernel. Use pivot columns of original $A$ for the image.

Solution.

Row-reducing:

\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{pmatrix} \xrightarrow{R_2 \leftarrow R_2 - 2R_1} \begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \end{pmatrix}

Pivot in column 1. Free variables: $x_2$ and $x_3$ .

$x_1 = -2x_2 - 3x_3$ . Letting $(x_2, x_3) = (1,0)$ : $(-2, 1, 0)$ . Letting $(x_2, x_3) = (0,1)$ : $(-3, 0, 1)$ .

$\ker A = \operatorname{span}\{(-2, 1, 0), (-3, 0, 1)\}, \quad \dim = 2$

Pivot column is the 1st column of original $A$ : $\operatorname{Im}(A) = \operatorname{span}\{(1, 2)\}$ , $\dim = 1$ .

Rank-nullity: $2 + 1 = 3$ . ✓

Verification. $A \cdot (-2, 1, 0)^T = (-2 + 2, -4 + 4)^T = (0,0)^T$ . ✓

Source. Beezer, A First Course in Linear Algebra, §RNNM, Example RNNM.NSEEF — GNU FDL.

Example— 3· Injectivity, surjectivity, and rank-nullity (intermediate)

Problem. Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ with matrix $A = \begin{pmatrix} 1 & 0 & -1 & 2 \\ 0 & 1 & 2 & -1 \\ 1 & 1 & 1 & 1 \end{pmatrix}$ . Compute rank and nullity. Is $T$ injective? Surjective?

Strategy. Row-reduce $A$ . Count pivots = rank. Nullity = $4 - \operatorname{rank}$ .

Solution.

A \xrightarrow{R_3 \leftarrow R_3 - R_1 - R_2} \begin{pmatrix} 1 & 0 & -1 & 2 \\ 0 & 1 & 2 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}

Two pivots: $\operatorname{rank}(A) = 2$ . Nullity $= 4 - 2 = 2$ .

$T$ is not injective (nullity $> 0$ ; $\ker T$ has dim 2).

$T$ is not surjective (rank $= 2 < 3 = \dim \mathbb{R}^3$ ; $\operatorname{Im}(T)$ is a plane in $\mathbb{R}^3$ ).

Verification. $2 + 2 = 4 = \dim \mathbb{R}^4$ . ✓

Source. Hefferon, Linear Algebra, ch. 3, §II.2, Example 2.16 — CC-BY-SA.

Example— 4· Differentiation as a linear transformation (advanced)

Problem. Let $D: \mathcal{P}_3 \to \mathcal{P}_3$ be the differentiation operator ( $D(p) = p'$ ). Find $\ker D$ , $\operatorname{Im}(D)$ , rank, and nullity. Verify rank-nullity.

Strategy. Use the basis $\{1, t, t^2, t^3\}$ of $\mathcal{P}_3$ . Apply $D$ to each basis element.

Solution.

$D(1) = 0$ , $D(t) = 1$ , $D(t^2) = 2t$ , $D(t^3) = 3t^2$ .

$\ker D = \{p \in \mathcal{P}_3 : p' = 0\} = \{p : p \text{ is constant}\} = \operatorname{span}\{1\}$ .

$\operatorname{Im}(D) = \operatorname{span}\{1, 2t, 3t^2\} = \mathcal{P}_2$ .

$\dim \ker D = 1$ , $\dim \operatorname{Im}(D) = 3$ .

Rank-nullity: $1 + 3 = 4 = \dim \mathcal{P}_3$ . ✓

Verification. $D$ is not injective (constants vanish). $D$ is not surjective (no degree-3 polynomial in image — $\dim \operatorname{Im} = 3 < 4$ ).

Source. Beezer, A First Course in Linear Algebra, §LT, Example LT.MFLT — GNU FDL.

Example— 5· Proof: injective iff kernel is trivial (proof)

Problem. Prove: $T: V \to W$ linear is injective if and only if $\ker T = \{0\}$ .

Strategy. Prove both directions. Use the definition of injectivity ( $T(u) = T(v) \Rightarrow u = v$ ) and linearity of $T$ .

Solution.

( $\Rightarrow$ ) Suppose $T$ is injective. If $v \in \ker T$ , then $T(v) = 0 = T(0)$ . By injectivity, $v = 0$ . Thus $\ker T = \{0\}$ .

( $\Leftarrow$ ) Suppose $\ker T = \{0\}$ . Let $u, v \in V$ with $T(u) = T(v)$ . Then $T(u) - T(v) = 0$ , and by linearity $T(u - v) = 0$ . Thus $u - v \in \ker T = \{0\}$ , so $u - v = 0$ , i.e., $u = v$ . Therefore $T$ is injective.

Verification. The proof is complete (both directions). The argument uses only linearity of $T$ and the definition of kernel — no extra hypotheses.

Source. Beezer, A First Course in Linear Algebra, §ILT, Theorem KILT — GNU FDL.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 25Understanding 4Modeling 6Challenge 3Proof 2

Ex. 113.1Application
Definition of kernel of a linear transformation $T: V \to W$ :
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$\ker T = \{v \in V : T(v) = 0_W\}$ $\ker T = \{T(v) : v \in V\}$ $\ker T = V$ always $\ker T = W$
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Solve online Beezer, A First Course in Linear Algebra · §KER, Definition KER
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The kernel of $T: V \to W$ is the set of all domain vectors mapped to the zero vector of the codomain: $\ker T = \{v \in V : T(v) = 0_W\}$ . It is always a subspace of $V$ .
Show step-by-step (with the why)
1. Recall that $T$ is linear, so $T(0_V)=0_W$ always — the zero vector belongs to the kernel.
2. The kernel may contain vectors other than $0_V$ (when $T$ is not injective).
3. The image is the set of output vectors: $\operatorname{Im}(T)=\{T(v):v\in V\}$ — a different definition.
Ex. 113.2ApplicationAnswer key
Definition of image (range) of a linear transformation $T: V \to W$ :
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$\operatorname{Im}(T) = \{T(v) : v \in V\}$ — subspace of $W$ $\operatorname{Im}(T) = V$ $\operatorname{Im}(T) = \ker T$ $\operatorname{Im}(T) = \{0_W\}$ always
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Solve online Beezer, A First Course in Linear Algebra · §RNG, Definition RNG
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The image (range) of $T: V \to W$ is the set of all codomain vectors that are actually reached: $\operatorname{Im}(T) = \{T(v) : v \in V\}$ . It is a subspace of $W$ .
Ex. 113.3Application
If $\ker T = \{0\}$ and $\dim V = n$ , then the nullity and rank of $T$ are:
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$T$ has nullity 0 and rank $n = \dim V$ $T$ has nullity $n$ and rank 0 $T$ has nullity $n$ and rank $n$ $T$ has nullity 0 and rank 0
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Solve online Beezer, A First Course in Linear Algebra · §RNNM, Theorem RPNDD
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Nullity is $\dim \ker T$ and rank is $\dim \operatorname{Im}(T)$ . By the rank-nullity theorem: nullity + rank = $n$ . If $\ker T = \{0\}$ (nullity 0), then rank = $n$ .
Ex. 113.4ApplicationAnswer key
Nullity and rank of a linear transformation $T: V \to W$ are defined as:
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Nullity $= \dim \ker T$ ; rank $= \dim \operatorname{Im}(T)$ Nullity $= \dim \operatorname{Im}(T)$ ; rank $= \dim \ker T$ Nullity = rank alwaysNullity = $\dim W$
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Beezer, A First Course in Linear Algebra · §RNNM, Definitions NOM and ROM
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Nullity of $T$ is the dimension of the kernel: $\operatorname{nul}(T)=\dim\ker T$ . Rank of $T$ is the dimension of the image: $\operatorname{rank}(T)=\dim\operatorname{Im}(T)$ . Together they sum to $\dim V$ .
Ex. 113.5Application
The Rank-Nullity Theorem for $T: V \to W$ with $\dim V$ finite states:
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$\operatorname{nul}(T) + \operatorname{rank}(T) = \dim V$ $\operatorname{nul}(T) \cdot \operatorname{rank}(T) = \dim V$ $\operatorname{nul}(T) = \operatorname{rank}(T)$ $\operatorname{nul}(T) + \operatorname{rank}(T) = \dim W$
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Solve online Beezer, A First Course in Linear Algebra · §RNNM, Theorem RPNDD
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The Rank-Nullity Theorem: for $T: V \to W$ linear with $\dim V$ finite, $\operatorname{nul}(T) + \operatorname{rank}(T) = \dim V$ . The sum is always $\dim V$ , the domain — not the codomain.
Show step-by-step (with the why)
1. Recall: nullity counts dimensions "lost" in the kernel; rank counts dimensions "surviving" in the image.
2. Together they cover all the dimension of the domain: nothing is created, nothing is destroyed except the kernel.
3. Counter-example to fix ideas: if $T: \mathbb{R}^5 \to \mathbb{R}^3$ with rank 2, then nullity = 3, since $3+2=5$ .
Ex. 113.6Application
For $A = \begin{pmatrix}1 & 2 & 3 \\ 2 & 4 & 6\end{pmatrix}$ viewed as a transformation $T_A: \mathbb{R}^3 \to \mathbb{R}^2$ , the rank and nullity are: (Ans: rank 1, nullity 2)
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Rank 1, nullity 2Rank 2, nullity 1Rank 3, nullity 0Rank 0, nullity 3
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Solve online Beezer, A First Course in Linear Algebra · §RNNM, Example RNNM.NSEEF
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Row-reducing $A$ : the second row is twice the first, so it becomes zero. There is only 1 pivot, so rank = 1 and nullity = $3 - 1 = 2$ . Confirmation: $1 + 2 = 3 = \dim \mathbb{R}^3$ .
Show step-by-step (with the why)
1. Row-reduce $A = \begin{pmatrix}1 & 2 & 3 \\ 2 & 4 & 6\end{pmatrix}$ : $R_2 \leftarrow R_2 - 2R_1$ gives a zero row.
2. 1 pivot in column 1. Rank = 1.
3. Nullity = 3 - 1 = 2 (two free variables: $x_2$ and $x_3$ ).
Ex. 113.7Application
Projection $T: \mathbb{R}^3 \to \mathbb{R}^3$ , $T(x,y,z)=(x,y,0)$ . Kernel and nullity:
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$\ker T = \{(0, 0, z) : z \in \mathbb{R}\}$ , nullity 1 $\ker T = \{(0, 0, 0)\}$ , nullity 0 $\ker T = \mathbb{R}^3$ , nullity 3 $\ker T = \{(x, y, 0)\}$ , nullity 2
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Solve online Beezer, A First Course in Linear Algebra · §ILT, Example ILT.STLT
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The projection $T(x,y,z)=(x,y,0)$ maps to zero exactly when $x=0$ and $y=0$ : $\ker T = \{(0,0,z): z\in\mathbb{R}\}$ — the $z$ -axis. Nullity = 1.
Show step-by-step (with the why)
1. Solve $T(x,y,z)=(0,0,0)$ : requires $x=0$ , $y=0$ , $z$ free.
2. $\ker T = \operatorname{span}\{(0,0,1)\}$ . Nullity = 1.
3. Rank = $3 - 1 = 2$ . Image = $xy$ -plane. Confirms: $1+2=3$ .
Ex. 113.8ApplicationAnswer key
Continuing the previous exercise: image and rank of the projection $T(x,y,z)=(x,y,0)$ :
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$\operatorname{Im}(T) = \{(x,y,0) : x, y \in \mathbb{R}\}$ , rank 2 $\operatorname{Im}(T) = \mathbb{R}^3$ , rank 3 $\operatorname{Im}(T) = \{(0,0,0)\}$ , rank 0 $\operatorname{Im}(T) = \{(x,0,0)\}$ , rank 1
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Solve online Beezer, A First Course in Linear Algebra · §RNG, Example RNG.NRFO
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The projection $T(x,y,z)=(x,y,0)$ reaches exactly vectors of the form $(x,y,0)$ : the $xy$ -plane. A basis is $\{(1,0,0),(0,1,0)\}$ . Rank = 2.
Ex. 113.9Application
$T: \mathbb{R}^3 \to \mathbb{R}$ , $T(x,y,z)=x+y+z$ . Kernel and its dimension: (Ans: plane $x+y+z=0$ , dim 2)
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$\ker T = \{(x,y,z): x+y+z=0\}$ , dimension 2 $\ker T = \{(0,0,0)\}$ , dimension 0 $\ker T = \{(x,y,z): x=y=z\}$ , dimension 1 $\ker T = \mathbb{R}^3$ , dimension 3
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Solve online Hefferon, Linear Algebra · Ch. 3 §II.2, Example 2.4
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$T(x,y,z)=x+y+z=0$ defines a plane in $\mathbb{R}^3$ . Kernel basis: $\{(-1,1,0),(-1,0,1)\}$ . Nullity = 2. Rank = 1 (image = all of $\mathbb{R}$ ). Confirms: $2+1=3$ .
Show step-by-step (with the why)
1. Solve $x+y+z=0$ : free variables $y,z$ ; $x=-y-z$ .
2. Setting $(y,z)=(1,0)$ : vector $(-1,1,0)$ . Setting $(y,z)=(0,1)$ : vector $(-1,0,1)$ .
3. Kernel basis: $\{(-1,1,0),(-1,0,1)\}$ , dimension 2.
Ex. 113.10Application
$T: \mathbb{R}^4 \to \mathbb{R}^3$ with matrix $\begin{pmatrix}1&0&-1&2\\0&1&2&-1\\1&1&1&1\end{pmatrix}$ . Rank, nullity, injectivity, surjectivity:
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Rank 2, nullity 2; $T$ is neither injective nor surjectiveRank 3, nullity 1; $T$ is surjectiveRank 4, nullity 0; $T$ is injectiveRank 1, nullity 3; $T$ is not injective
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Solve online Hefferon, Linear Algebra · Ch. 3 §II.2, Example 2.16
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Row-reducing: $R_3 \leftarrow R_3 - R_1 - R_2$ zeros the third row. 2 pivots: rank = 2, nullity = $4 - 2 = 2$ . Not injective (nullity greater than 0); not surjective (rank 2 less than 3 = codomain dimension).
Ex. 113.11Understanding
A linear transformation $T: V \to W$ is injective if and only if:
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$T$ injective $\Leftrightarrow$ $\ker T = \{0\}$ $T$ injective $\Leftrightarrow$ $\ker T = V$ $T$ injective $\Leftrightarrow$ $\operatorname{Im}(T) = \{0\}$ $T$ injective $\Leftrightarrow$ rank = $\dim W$
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Solve online Beezer, A First Course in Linear Algebra · §ILT, Theorem KILT
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$T$ is injective when $T(u)=T(v)\Rightarrow u=v$ . Equivalently: $\ker T=\{0\}$ . If any nonzero vector goes to zero, there exist two distinct vectors with the same image.
Show step-by-step (with the why)
1. Direction 1. Suppose $T$ is injective. If $T(v)=0=T(0)$ , then $v=0$ . Thus $\ker T=\{0\}$ .
2. Direction 2. Suppose $\ker T=\{0\}$ . If $T(u)=T(v)$ , then $T(u-v)=0$ , so $u-v\in\ker T=\{0\}$ , hence $u=v$ .
Ex. 113.12UnderstandingAnswer key
A linear transformation $T: V \to W$ is surjective if and only if:
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$T$ surjective $\Leftrightarrow$ $\operatorname{Im}(T) = W$ $\Leftrightarrow$ rank $(T) = \dim W$ $T$ surjective $\Leftrightarrow$ $\ker T = \{0\}$ $T$ surjective $\Leftrightarrow$ $\operatorname{Im}(T) \subsetneq W$ $T$ surjective $\Leftrightarrow$ $\dim V = \dim W$
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Solve online Beezer, A First Course in Linear Algebra · §SLT, Definition SLT
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$T$ is surjective when every $w \in W$ has a preimage: $\exists v \in V$ with $T(v)=w$ . It is equivalent to $\operatorname{Im}(T)=W$ , i.e., rank $= \dim W$ .
Ex. 113.13Application
$T: \mathbb{R}^2 \to \mathbb{R}^2$ , $T(x,y)=(x+y, x-y)$ . Is $T$ injective?
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$T$ injective, since $\ker T = \{0\}$ $T$ is not injective $T$ is surjective but not injective $T$ is the zero transformation
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Solve online Beezer, A First Course in Linear Algebra · §ILT, Example ILT.NKAO
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$T(x,y)=(x+y,x-y)$ . Solve $T(x,y)=(0,0)$ : system $x+y=0$ , $x-y=0$ . Adding: $2x=0$ , so $x=0$ ; then $y=0$ . Trivial kernel — injective.
Show step-by-step (with the why)
1. Solve $x+y=0$ and $x-y=0$ : adding gives $2x=0\Rightarrow x=0$ ; subtracting gives $2y=0\Rightarrow y=0$ .
2. $\ker T = \{(0,0)\}$ — trivial. $T$ is injective.
3. Since $T:\mathbb{R}^2\to\mathbb{R}^2$ is injective with domain and codomain of the same dimension, it is also surjective (bijective).
Ex. 113.14Application
$T: \mathbb{R}^3 \to \mathbb{R}^2$ linear arbitrary. Can $T$ be injective?
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$T$ is not injective: $\ker T \neq \{0\}$ $T$ is injective $T$ is bijectiveCannot decide without more information
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Solve online Beezer, A First Course in Linear Algebra · §ILT, Theorem ILTD
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$T: \mathbb{R}^3 \to \mathbb{R}^2$ . By the rank-nullity theorem: nullity + rank = 3, but rank $\leq \dim \mathbb{R}^2 = 2$ . Thus nullity $\geq 1 > 0$ . Therefore $\ker T \neq \{0\}$ and $T$ is not injective.
Ex. 113.15Application
$T: \mathbb{R}^2 \to \mathbb{R}^3$ linear arbitrary. Can $T$ be surjective?
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$T$ is not surjective: rank $(T) \leq \dim V = 2 < 3 = \dim W$ $T$ is surjectiveDepends on the specific matrix of $T$ $T$ is bijective if $\ker T = \{0\}$
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Solve online Beezer, A First Course in Linear Algebra · §SLT, Theorem SLTD
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$T: \mathbb{R}^2 \to \mathbb{R}^3$ . Rank $\leq \dim \mathbb{R}^2 = 2 < 3 = \dim \mathbb{R}^3$ . Thus $\operatorname{Im}(T)$ has dimension at most 2, which cannot cover $\mathbb{R}^3$ . $T$ can never be surjective.
Ex. 113.16Understanding
If $\dim V = \dim W$ (equal finite), then for $T: V \to W$ linear:
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For $\dim V = \dim W$ finite: injective $\Leftrightarrow$ surjective $\Leftrightarrow$ bijectiveInjective always implies surjective independent of dimensionsSurjective always implies injectiveBijective only exists when $\dim V = 0$
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Solve online Beezer, A First Course in Linear Algebra · §IVLT, Theorem ILTIS
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When $\dim V = \dim W = n$ : nullity + rank = $n$ and rank $\leq n$ . Injective means nullity = 0, hence rank = $n = \dim W$ — surjective. The equivalence holds only with equal finite dimensions.
Ex. 113.17Application
$T: \mathbb{R}^2 \to \mathbb{R}^2$ given by the identity matrix $I_2$ . Classification:
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$T$ injective and surjective (bijective) $T$ injective but not surjective $T$ surjective but not injective $T$ is neither injective nor surjective
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Solve online Hefferon, Linear Algebra · Ch. 3 §II.2, Example 2.3
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Matrix $A = \begin{pmatrix}1&0\\0&1\end{pmatrix}$ (identity): $T(x,y)=(x,y)$ . Kernel = $\{(0,0)\}$ (injective). Image = $\mathbb{R}^2$ (surjective). Bijective.
Ex. 113.18ApplicationAnswer key
$T: \mathbb{R}^3 \to \mathbb{R}^2$ , $T(v) = 0$ for all $v$ (zero transformation). Classification:
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$T$ is neither injective nor surjective $T$ is injective and surjective $T$ is injective $T$ is surjective
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Solve online Hefferon, Linear Algebra · Ch. 3 §II.1, Example 1.5
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Zero transformation $T(v)=0$ for all $v$ : $\ker T = V$ (nullity = $\dim V$ ) and $\operatorname{Im}(T) = \{0\}$ (rank = 0). Not injective (except if $\dim V = 0$ ) and not surjective (except if $\dim W = 0$ ).
Ex. 113.19Application
$T: \mathbb{R}^3 \to \mathbb{R}^2$ with matrix $\begin{pmatrix}1&0&1\\0&1&1\end{pmatrix}$ . Rank and surjectivity:
Select the correct option
Rank 2; $T$ surjective (but not injective since nullity 1)Rank 1; $T$ surjectiveRank 3; $T$ is not surjectiveRank 2; $T$ is injective
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Solve online Beezer, A First Course in Linear Algebra · §SLT, Example SLT.NSAQ
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Matrix $A = \begin{pmatrix}1&0&1\\0&1&1\end{pmatrix}$ : row-reducing, 2 pivots. Rank = 2 = $\dim\mathbb{R}^2$ . Surjective. Nullity = $3-2=1$ : not injective.
Show step-by-step (with the why)
1. Row-reduce $A$ : already in row-echelon form, 2 pivots (columns 1 and 2).
2. Rank = 2 = codomain dimension $\mathbb{R}^2$ : surjective.
3. Nullity = $3-2=1$ : $\ker T$ has dimension 1, so not injective.
Ex. 113.20Understanding
The kernel $\ker T$ of any linear transformation is always:
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$\{0\}$ is always a subspace of $V$ , therefore $\ker T$ is always nonempty $\ker T$ can be empty $\ker T$ is empty when $T$ is injective $\ker T$ is nonempty only when $T$ is not injective
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Solve online Beezer, A First Course in Linear Algebra · §KER, Theorem ZLTLT
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$T$ is linear, so $T(0_V)=0_W$ . Therefore $0_V \in \ker T$ always — the kernel is never empty. When $T$ is injective, $\ker T = \{0_V\}$ (trivial, but not empty).
Ex. 113.21Application
For $A = \begin{pmatrix}1&2&3\\2&4&6\end{pmatrix}$ , find a basis for $\ker A$ : (Ans: dim 2)
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Basis of $\ker A = \{(-2,1,0),(-3,0,1)\}$ , dimension 2Basis of $\ker A = \{(1,2,3)\}$ , dimension 1Basis of $\ker A = \{(0,0,0)\}$ , dimension 0Basis of $\ker A = \{(-2,1,0)\}$ , dimension 1
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Solve online Beezer, A First Course in Linear Algebra · §RNNM, Example RNNM.NSEEF
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Row-reducing $A$ : pivot in column 1. Free variables: $x_2, x_3$ . From the row-reduced form: $x_1 = -2x_2 - 3x_3$ . Setting $(x_2,x_3)=(1,0)$ : $(-2,1,0)$ . Setting $(x_2,x_3)=(0,1)$ : $(-3,0,1)$ .
Show step-by-step (with the why)
1. Row-reduce $\begin{pmatrix}1&2&3\\2&4&6\end{pmatrix}$ : $R_2\leftarrow R_2-2R_1$ gives a zero row.
2. Row-echelon form: $\begin{pmatrix}1&2&3\\0&0&0\end{pmatrix}$ . 1 pivot, free variables $x_2,x_3$ .
3. $x_1=-2x_2-3x_3$ . For each free variable set to 1 (others 0): basis vectors of the kernel.
Ex. 113.22Application
Continuing: for $A = \begin{pmatrix}1&2&3\\2&4&6\end{pmatrix}$ , a basis for $\operatorname{Im}(A)$ is:
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Column 1 of $A$ original: $\{(1,2)\}$ — basis of the image, rank 1Columns 1 and 2 of $A$ original: rank 2Row 1 of $A$ : $\{(1,2,3)\}$ Image is all of $\mathbb{R}^2$ , rank 2
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Solve online Beezer, A First Course in Linear Algebra · §RNNM, Example RNNM.NSEEF
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The pivot column (column 1 in row-echelon form) corresponds to column 1 of original $A$ : $(1,2)^T$ . Basis of image: $\{(1,2)\}$ , dimension 1. Verification: $1+1=2$ (nullity + rank = columns of $A$ ). But $A$ has 3 columns: $2+1=3$ . ✓
Ex. 113.23Application
Projection $T: \mathbb{R}^3 \to \mathbb{R}^3$ , $T(x,y,z)=(x,y,0)$ . Bases of kernel and image:
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Basis of $\ker T$ : $\{(0,0,1)\}$ ; basis of $\operatorname{Im}(T)$ : $\{(1,0,0),(0,1,0)\}$ Basis of $\ker T$ : $\{(1,0,0)\}$ ; basis of $\operatorname{Im}(T)$ : $\{(0,0,1)\}$ Basis of $\ker T$ : $\{(1,0,0),(0,1,0)\}$ ; basis of $\operatorname{Im}(T)$ : $\{(0,0,1)\}$ Kernel trivial; image is all of $\mathbb{R}^3$
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Solve online Beezer, A First Course in Linear Algebra · §RNG, Example RNG.NRFO
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Projection $T(x,y,z)=(x,y,0)$ . Kernel: $x=y=0$ , $z$ free — basis $\{(0,0,1)\}$ . Image: vectors of the form $(x,y,0)$ — basis $\{(1,0,0),(0,1,0)\}$ .
Ex. 113.24Application
Integration $I: \mathcal{P}_2 \to \mathcal{P}_3$ , $I(p)(t)=\int_0^t p(s)\,ds$ . Kernel and image:
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Kernel: $\{0\}$ ; image: $\mathcal{P}_2 = \operatorname{span}\{1, t, t^2\}$ ; nullity 0, rank 3Kernel: $\{$ constants $\}$ ; image: $\mathcal{P}_3$ Kernel: $\mathcal{P}_3$ ; image: $\{0\}$ Kernel: $\{0\}$ ; image: $\mathcal{P}_3$ ; bijective
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Solve online Hefferon, Linear Algebra · Ch. 3 §II.1, Example 1.9
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Integration $I(p)(t)=\int_0^t p(s)\,ds$ from $\mathcal{P}_2$ to $\mathcal{P}_3$ . If $I(p)=0$ then the integral polynomial is identically zero, so $p=0$ . Trivial kernel. Image: polynomials with zero constant term, span $\{t,t^2/2,t^3/3\} \cong \mathcal{P}_2$ , rank 3.
Ex. 113.25ApplicationAnswer key
Differentiation $D: \mathcal{P}_3 \to \mathcal{P}_3$ , $D(p)=p'$ . Kernel, image, nullity, rank:
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Kernel: $\operatorname{span}\{1\}$ (constants), dim 1; image: $\mathcal{P}_2$ , dim 3Kernel: $\{0\}$ , dim 0; image: $\mathcal{P}_3$ , dim 4Kernel: $\mathcal{P}_3$ , dim 4; image: $\{0\}$ Kernel: even polynomials, dim 2; image: odd polynomials
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Solve online Beezer, A First Course in Linear Algebra · §LT, Example LT.MFLT
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Differentiation $D: \mathcal{P}_3 \to \mathcal{P}_3$ . Polynomials whose derivative is zero are constants: $\ker D = \operatorname{span}\{1\}$ , nullity 1. $D(1)=0, D(t)=1, D(t^2)=2t, D(t^3)=3t^2$ : image = $\operatorname{span}\{1,2t,3t^2\}=\mathcal{P}_2$ , rank 3. Verifies: $1+3=4=\dim\mathcal{P}_3$ . ✓
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1. $D(p)=0 \Rightarrow p'=0 \Rightarrow p$ is constant. Kernel basis: $\{1\}$ .
2. Apply $D$ to basis $\{1,t,t^2,t^3\}$ : $0,1,2t,3t^2$ .
3. Image = $\operatorname{span}\{1,2t,3t^2\}=\mathcal{P}_2$ , dim 3.
Ex. 113.26Application
$B = \begin{pmatrix}1&0&1\\2&1&4\\1&1&3\end{pmatrix}$ . Rank, nullity, and basis of $\ker B$ :
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Rank 2, nullity 1; basis of $\ker$ : $\{(-1,-2,1)\}$ Rank 3, nullity 0Rank 1, nullity 2Rank 2, nullity 1; basis of $\ker$ : $\{(1,2,0)\}$
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Solve online Hefferon, Linear Algebra · Ch. 3 §II.2, Example 2.10
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Row-reducing $B$ : $R_2\leftarrow R_2-2R_1$ , $R_3\leftarrow R_3-R_1$ give 2 pivots in columns 1 and 2. Rank = 2, nullity = 1. Free variable: $x_3$ . From RREF: $x_1=-(-1)x_3=x_3$ ... solve correctly: $x_1+x_3=0$ , $x_2+2x_3=0$ . Kernel basis: $(-1,-2,1)$ .
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1. Row-reduce $\begin{pmatrix}1&0&1\\2&1&4\\1&1&3\end{pmatrix}$ .
2. $R_2\leftarrow R_2-2R_1$ : second row becomes $(0,1,2)$ .
3. $R_3\leftarrow R_3-R_1$ : third row becomes $(0,1,2)$ . Then $R_3\leftarrow R_3-R_2$ : zero row.
4. 2 pivots: rank 2. Free variable $x_3$ . $x_2=-2x_3$ , $x_1=-x_3$ . Basis vector: $(-1,-2,1)$ .
Ex. 113.27ApplicationAnswer key
Continuing: basis of $\operatorname{Im}(B)$ for $B = \begin{pmatrix}1&0&1\\2&1&4\\1&1&3\end{pmatrix}$ :
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Basis of $\operatorname{Im}(B)$ : columns 1 and 2 of $B$ original, $\{(1,2,1)^T,(0,1,1)^T\}$ Basis of $\operatorname{Im}(B)$ : columns 1, 2 and 3 of $B$ originalBasis of $\operatorname{Im}(B)$ : rows 1 and 2 of $B$ Basis of $\operatorname{Im}(B)$ : column 3 of $B$ original
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Solve online Hefferon, Linear Algebra · Ch. 3 §II.2, Example 2.10
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The pivot columns of $B$ are in positions 1 and 2 (from the row-echelon form). The image basis uses these columns from original $B$ : $(1,2,1)^T$ and $(0,1,1)^T$ .
Ex. 113.28ModelingAnswer key
$T: \mathbb{R}^4 \to \mathbb{R}^2$ , $T(x,y,z,w)=(x+z,y+w)$ . Basis of $\ker T$ : (Ans: nullity 2)
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$\ker T = \{(x,y,z,w) : x+z=0,\; y+w=0\}$ , basis $\{(-1,0,1,0),(0,-1,0,1)\}$ $\ker T = \{(0,0,0,0)\}$ $\ker T = \{(x,y,z,w): x=y\}$ $\ker T = \mathbb{R}^4$
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Solve online Beezer, A First Course in Linear Algebra · §KER, Example KER.NKAO
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$T(x,y,z,w)=(x+z,y+w)$ . Solve $x+z=0$ and $y+w=0$ : $x=-z, y=-w$ . Free variables: $z,w$ . Basis vectors: $(-1,0,1,0)$ and $(0,-1,0,1)$ . Nullity = 2, rank = 2.
Ex. 113.29ApplicationAnswer key
For an invertible matrix $A$ of order $n$ , the kernel of the transformation $T_A(x)=Ax$ has:
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Nullity 0, rank $n$ ; basis of $\ker$ : empty (only zero)Nullity $n$ , rank 0Nullity 1, rank $n-1$ Nullity $n$ , rank $n$
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Solve online Beezer, A First Course in Linear Algebra · §RNNM, Corollary RPNDD
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Invertible matrix $A$ ( $n\times n$ ): $Ax=0$ implies $x=A^{-1}0=0$ . Thus $\ker A=\{0\}$ . Nullity = 0. By rank-nullity: rank = $n$ . $T_A$ is bijective.
Ex. 113.30Application
$C = \begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$ . Nullity, rank, and image of $T_C$ :
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Nullity 2, rank 1; $T$ not injective; image: $\operatorname{span}\{(1,1,1)\}$ Nullity 1, rank 2; $T$ is injectiveNullity 0, rank 3Nullity 3, rank 0
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Solve online Hefferon, Linear Algebra · Ch. 3 §II.2, Example 2.12
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Row-reducing: $R_2\leftarrow R_2-R_1$ zeros second row; $R_3\leftarrow R_3-R_1$ zeros third row. 1 pivot: rank = 1. Nullity = $3-1=2$ . Image: pivot column of original $C$ = $(1,1,1)^T$ .
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1. $C = \begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$ : all rows equal.
2. Row reduction eliminates rows 2 and 3. 1 pivot, rank = 1.
3. Nullity = $3-1=2$ . Image = $\operatorname{span}\{(1,1,1)^T\}$ .
Ex. 113.31Modeling
The system $Ax=b$ has a particular solution $x_p$ . The general solution is:
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General solution: $x = x_p + h$ , $h \in \ker A$ arbitraryGeneral solution: $x = x_p$ (unique particular solution)General solution: $x = A^{-1}b$ No general solution exists when $\ker A \neq \{0\}$
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Solve online Beezer, A First Course in Linear Algebra · §PSPHS, Theorem PSPHS
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When $Ax=b$ has solution $x_p$ and $\ker A \neq \{0\}$ , the general solution is $x = x_p + h$ with $h \in \ker A$ . Since $A(x_p+h)=Ax_p+Ah=b+0=b$ . The kernel parameterizes the infinitely many solutions.
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1. If $x_p$ satisfies $Ax_p=b$ and $h\in\ker A$ , then $A(x_p+h)=b+0=b$ : any $x_p+h$ is a solution.
2. Conversely, if $Ax=b$ , then $A(x-x_p)=0$ , so $x-x_p\in\ker A$ .
3. Conclusion: the solution set is the affine subspace $x_p + \ker A$ .
Ex. 113.32Modeling
The linear system $Ax=b$ has a solution if and only if:
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$b \in \operatorname{Im}(A)$ (consistency condition) $b = 0$ $b \notin \operatorname{Im}(A)$ $\ker A = \{0\}$
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Solve online Hefferon, Linear Algebra · Ch. 3 §II.1, Lemma 1.10
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$Ax=b$ has a solution $\Leftrightarrow$ $b$ is a linear combination of columns of $A$ $\Leftrightarrow$ $b \in \operatorname{Im}(A) = \operatorname{Col}(A)$ . Uniqueness depends on $\ker A$ ; existence depends on $\operatorname{Im}(A)$ .
Ex. 113.33Modeling
$A = \begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$ , $b = (1,0,0)^T$ . The system $Ax=b$ :
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Rank $(A) = 2 < 3 = \dim \mathbb{R}^3$ ; $\operatorname{Im}(A)$ is a plane, $b=(1,0,0)^T \notin \operatorname{Im}(A)$ ; no solutionThe system has a unique solutionThe system has infinitely many solutionsCannot decide
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Solve online Hefferon, Linear Algebra · Ch. 3 §II.1, Example 1.12
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Row-reducing $A = \begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$ : 1 pivot, rank = 1. Image = $\operatorname{span}\{(1,1,1)^T\}$ . Vector $b=(1,0,0)^T$ is not a multiple of $(1,1,1)^T$ . No solution.
Ex. 113.34Modeling
Evaluation $T: \mathcal{P} \to \mathbb{R}$ , $T(p) = p(1)$ . Kernel of $T$ :
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$\ker T = \{p \in \mathcal{P} : p(1)=0\}$ — ideal of polynomials vanishing at 1, infinite dimension $\ker T = \{0\}$ $\ker T = \mathcal{P}$ entire $\ker T = \{$ constants $\}$
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Solve online Hefferon, Linear Algebra · Ch. 3 §II.1, Exercise 1.17
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Evaluation $T: \mathcal{P} \to \mathbb{R}$ , $T(p)=p(1)$ . Kernel: polynomials with $p(1)=0$ . Example: $(t-1), (t-1)^2, (t-1)^3, \ldots$ all are in the kernel. Infinite dimension.
Ex. 113.35Modeling
A matrix $A$ of order $m \times n$ with rank $r < n$ satisfies:
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Rank $(A) < n \Rightarrow \ker A \neq \{0\} \Rightarrow Ax=0$ has nontrivial solutionRank $(A) < n$ does not affect the kernelRank $(A) = n$ implies $\ker A \neq \{0\}$ The kernel of $A$ depends only on $m$
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Solve online Beezer, A First Course in Linear Algebra · §RNNM, Corollary RNNM
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If $A$ is $m \times n$ with rank $r < n$ , by rank-nullity: nullity = $n - r > 0$ . Thus $\ker A$ contains nonzero vectors. The homogeneous system has nontrivial solutions.
Ex. 113.36Challenge
For $S: U \to V$ and $T: V \to W$ linear, the rank of the composition satisfies:
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rank $(T \circ S) \leq \min($ rank $(T)$ , rank $(S))$ rank $(T \circ S) =$ rank $(T) +$ rank $(S)$ rank $(T \circ S) =$ rank $(T) \cdot$ rank $(S)$ rank $(T \circ S) \geq \max($ rank $(T)$ , rank $(S))$
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Solve online Hefferon, Linear Algebra · Ch. 3 §II.2, Exercise 2.28
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$\operatorname{Im}(T\circ S) = T(\operatorname{Im}(S)) \subseteq \operatorname{Im}(T)$ , so rank $\leq$ rank $T$ . Also $\operatorname{Im}(T\circ S) \subseteq T(W)$ where $W = \operatorname{Im}(S)$ has dim = rank $S$ , so rank $\leq$ rank $S$ . Hence the min.
Ex. 113.37Challenge
For composition $T \circ S$ , nullity satisfies:
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nullity $(T \circ S) \geq$ nullity $(S)$ nullity $(T \circ S) =$ nullity $(S)$ nullity $(T \circ S) \leq$ nullity $(S)$ nullity $(T \circ S) = 0$ always
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Solve online Hefferon, Linear Algebra · Ch. 3 §II.2, Exercise 2.29
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$\ker S \subseteq \ker(T \circ S)$ : if $S(u) = 0$ , then $T(S(u)) = T(0) = 0$ . Thus nullity $(T \circ S) \geq$ nullity $(S)$ .
Ex. 113.38Challenge
For $A$ square $n \times n$ , rank $< n$ implies that $A$ is not invertible?
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Yes: rank $(A) < n \Rightarrow$ nullity $> 0 \Rightarrow Ax=0$ has nontrivial solution; thus $A$ is not invertibleNo: matrices with low rank can be invertibleDepends on the number of rows of $A$ This is true only for symmetric matrices
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Solve online Beezer, A First Course in Linear Algebra · §RNNM, Corollary RPNDD
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For $A$ square $n \times n$ : invertible $\Leftrightarrow$ rank = $n$ $\Leftrightarrow$ nullity = 0 $\Leftrightarrow$ $\ker A = \{0\}$ . If rank $< n$ : nullity $> 0$ , $Ax=0$ has nontrivial solution $v$ . If $A$ were invertible: $v = A^{-1}Av = A^{-1}0 = 0$ — contradiction.
Ex. 113.39ProofAnswer key
Prove that $\ker T$ is a subspace of $V$ for any linear transformation $T: V \to W$ :
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$\ker T$ is a subspace of $V$ (closed under addition and scaling; contains $0_V$ ) $\ker T$ is a subspace of $W$ $\ker T$ may not contain $0_V$ $\ker T$ is a subspace only when $T$ is injective
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Beezer, A First Course in Linear Algebra · §KER, Theorem KLTA
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Verification of three axioms: (1) $T(0_V)=0_W$ , so $0_V\in\ker T$ ; (2) if $u,v\in\ker T$ , $T(u+v)=T(u)+T(v)=0+0=0$ , so $u+v\in\ker T$ ; (3) if $u\in\ker T$ and $\alpha\in\mathbb{R}$ , $T(\alpha u)=\alpha T(u)=\alpha\cdot 0=0$ .
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1. Zero. $T(0_V)=0_W$ (linearity) $\Rightarrow 0_V\in\ker T$ .
2. Sum. $u,v\in\ker T\Rightarrow T(u+v)=0+0=0\Rightarrow u+v\in\ker T$ .
3. Scalar. $u\in\ker T,\,\alpha\in\mathbb{R}\Rightarrow T(\alpha u)=\alpha\cdot0=0\Rightarrow\alpha u\in\ker T$ .
Ex. 113.40Proof
Prove that $\operatorname{Im}(T)$ is a subspace of $W$ for any linear transformation $T: V \to W$ :
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$\operatorname{Im}(T)$ is a subspace of $W$ (contains $0_W$ ; closed under addition and scaling) $\operatorname{Im}(T)$ is a subspace of $V$ $\operatorname{Im}(T)$ can be $\emptyset$ $\operatorname{Im}(T)$ is a subspace only when $T$ is surjective
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Beezer, A First Course in Linear Algebra · §RNG, Theorem RLTA
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(1) $T(0_V)=0_W\in\operatorname{Im}(T)$ . (2) $w_1=T(v_1),w_2=T(v_2)\Rightarrow w_1+w_2=T(v_1+v_2)\in\operatorname{Im}(T)$ . (3) $w=T(v)\Rightarrow\alpha w=T(\alpha v)\in\operatorname{Im}(T)$ . All three axioms are satisfied.

Sources for this lesson

A First Course in Linear Algebra — Robert A. Beezer · 3rd ed. · EN · GNU FDL · §KER, §RNG, §ILT, §SLT, §RNNM, §PSPHS, §IVLT, §LT. Primary source for exercises.
Linear Algebra — Jim Hefferon · 4th ed. · EN · CC-BY-SA · ch. 3, §II. Secondary source for modeling exercises, composition, and examples of polynomial spaces.
Linear Algebra Done Right — Sheldon Axler · 4th ed. · EN · CC-BY-NC · §3D. Abstract perspective without determinants (reference for the 7 doors).