3.3 Kernel and Image of Linear Transformations

Kernel and image are key concepts in linear transformations. They help us understand how a transformation maps vectors between spaces. The kernel shows which vectors become zero, while the image reveals the range of possible outputs.

These ideas connect to injectivity and surjectivity. A transformation is injective if its kernel is just the zero vector, and surjective if its image is the whole codomain. The rank-nullity theorem ties it all together, linking dimensions of kernel and image.

Kernel and Image of Linear Transformations

Definition and Properties

• The kernel of a linear transformation $T:V→W$, denoted $ker(T)$ or $null(T)$, is the set of all vectors $v$ in $V$ such that $T(v)=0$, where $0$ is the zero vector in $W$
• The image of a linear transformation $T:V→W$, denoted $im(T)$ or $range(T)$, is the set of all vectors $w$ in $W$ such that $w=T(v)$ for some vector $v$ in $V$
• The kernel is a subspace of the domain $V$, while the image is a subspace of the codomain $W$
  • Example: For a linear transformation $T:\mathbb{R}^3→\mathbb{R}^2$ defined by $T(x,y,z)=(x+y,y-z)$, the kernel is the subspace of $\mathbb{R}^3$ satisfying $x+y=0$ and $y-z=0$, while the image is a subspace of $\mathbb{R}^2$

Injectivity and Surjectivity

• A linear transformation $T$ is injective (one-to-one) if and only if its kernel is the zero subspace, i.e., $ker(T)=\{0\}$
  • Example: The linear transformation $T:\mathbb{R}^2→\mathbb{R}^2$ defined by $T(x,y)=(x,y+x)$ is injective because the only solution to $T(x,y)=(0,0)$ is $(x,y)=(0,0)$
• A linear transformation $T$ is surjective (onto) if and only if its image is equal to the codomain, i.e., $im(T)=W$
  • Example: The linear transformation $T:\mathbb{R}^2→\mathbb{R}$ defined by $T(x,y)=x+y$ is surjective because for any real number $a$, there exist vectors $(x,y)$ in $\mathbb{R}^2$ such that $x+y=a$

Computing Kernel and Image

Finding the Kernel

• To find the kernel of a linear transformation $T:V→W$, solve the equation $T(v)=0$ for $v$ in $V$. The solution set is the kernel of $T$
  • Example: For a linear transformation $T:\mathbb{R}^3→\mathbb{R}^2$ defined by $T(x,y,z)=(2x-y,x+z)$, solve the system of equations $2x-y=0$ and $x+z=0$ to find the kernel
• For a linear transformation $T:\mathbb{R}^n→\mathbb{R}^m$ represented by an $m×n$ matrix $A$, the kernel is the solution set of the homogeneous system $Ax=0$
• The dimension of the kernel, denoted $dim(ker(T))$ or $nullity(T)$, is the number of free variables in the solution set of $T(v)=0$

Finding the Image

• To find the image of a linear transformation $T:V→W$, express $T(v)$ as a linear combination of the basis vectors of $W$. The span of the resulting vectors is the image of $T$
  • Example: For a linear transformation $T:\mathbb{R}^2→\mathbb{R}^3$ defined by $T(x,y)=(x+y,x-y,2y)$, express $T(x,y)$ as a linear combination of the standard basis vectors of $\mathbb{R}^3$ to find the image
• For a linear transformation $T:\mathbb{R}^n→\mathbb{R}^m$ represented by an $m×n$ matrix $A$, the image is the column space of $A$
• The dimension of the image, denoted $dim(im(T))$ or $rank(T)$, is the number of linearly independent vectors in a basis for the image

Rank-Nullity Theorem

Statement and Proof

• The Rank-Nullity Theorem states that for a linear transformation $T:V→W$ between finite-dimensional vector spaces $V$ and $W$, the dimension of the domain $V$ equals the sum of the dimensions of the kernel and the image of $T$, i.e., $dim(V)=dim(ker(T))+dim(im(T))$
• To prove the theorem, consider a basis for the kernel of $T$ and extend it to a basis for the domain $V$. Show that the images of the basis vectors not in the kernel form a basis for the image of $T$
  • Example: For a linear transformation $T:\mathbb{R}^4→\mathbb{R}^3$, if $dim(ker(T))=2$ and $dim(im(T))=2$, then the Rank-Nullity Theorem confirms that $dim(\mathbb{R}^4)=4=2+2$

Consequences and Applications

• The Rank-Nullity Theorem establishes a fundamental relationship between the nullity (dimension of the kernel) and the rank (dimension of the image) of a linear transformation
• As a consequence of the Rank-Nullity Theorem, if $T:V→W$ is a linear transformation and $dim(V)=dim(W)$, then $T$ is injective if and only if it is surjective
  • Example: For a linear transformation $T:\mathbb{R}^3→\mathbb{R}^3$, if $T$ is injective (nullity is zero), then it must also be surjective (rank is three) by the Rank-Nullity Theorem
• The Rank-Nullity Theorem can be used to determine the dimension of the kernel or image of a linear transformation when one of them is known

Injectivity and Surjectivity vs Kernel and Image

Injectivity and Kernel

• A linear transformation $T:V→W$ is injective (one-to-one) if and only if its kernel is the zero subspace, i.e., $ker(T)=\{0\}$. In other words, $T$ is injective if and only if the nullity of $T$ is zero
  • Example: The linear transformation $T:\mathbb{R}^3→\mathbb{R}^4$ defined by $T(x,y,z)=(x,y,z,0)$ is injective because the only solution to $T(x,y,z)=(0,0,0,0)$ is $(x,y,z)=(0,0,0)$
• For a linear transformation $T:\mathbb{R}^n→\mathbb{R}^m$ represented by an $m×n$ matrix $A$, the injectivity of $T$ can be determined by examining the null space of $A$ or by analyzing the rank of $A$ using Gaussian elimination

Surjectivity and Image

• A linear transformation $T:V→W$ is surjective (onto) if and only if its image is equal to the codomain, i.e., $im(T)=W$. In other words, $T$ is surjective if and only if the rank of $T$ equals the dimension of the codomain $W$
  • Example: The linear transformation $T:\mathbb{R}^3→\mathbb{R}^2$ defined by $T(x,y,z)=(x+y,y+z)$ is surjective because for any vector $(a,b)$ in $\mathbb{R}^2$, there exist vectors $(x,y,z)$ in $\mathbb{R}^3$ such that $x+y=a$ and $y+z=b$
• For a linear transformation $T:\mathbb{R}^n→\mathbb{R}^m$ represented by an $m×n$ matrix $A$, the surjectivity of $T$ can be determined by analyzing the rank of $A$ using Gaussian elimination

Injectivity, Surjectivity, and Dimension

• For a linear transformation $T:V→W$ between finite-dimensional vector spaces, the Rank-Nullity Theorem can be used to determine injectivity and surjectivity:
  • If $dim(V)<dim(W)$, then $T$ cannot be surjective
  • If $dim(V)>dim(W)$, then $T$ cannot be injective
  • If $dim(V)=dim(W)$, then $T$ is injective if and only if it is surjective (i.e., $T$ is bijective)
• Example: For a linear transformation $T:\mathbb{R}^4→\mathbb{R}^3$, $T$ cannot be surjective because $dim(\mathbb{R}^4)>dim(\mathbb{R}^3)$, but it may or may not be injective depending on its kernel