Math 221: LINEAR ALGEBRA 7-2. Linear Transformations - Kernel and - PowerPoint PPT Presentation

Math 221: LINEAR ALGEBRA §7-2. Linear Transformations - Kernel and Image Le Chen 1 Emory University, 2020 Fall (last updated on 08/27/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from those by Karen Seyffarth from University of Calgary.

ker What are the kernel and the image? Definition Let V and W be vector spaces, and T : V → W a linear transformation. 1. The kernel of T (sometimes called the null space of T) is defined to be the set v ) = � ker ( T ) = { � v ∈ V | T ( � 0 } . 2. The image of T is defined to be the set im ( T ) = { T ( � v ) | � v ∈ V } .

What are the kernel and the image? Definition Let V and W be vector spaces, and T : V → W a linear transformation. 1. The kernel of T (sometimes called the null space of T) is defined to be the set v ) = � ker ( T ) = { � v ∈ V | T ( � 0 } . 2. The image of T is defined to be the set im ( T ) = { T ( � v ) | � v ∈ V } . Example If A is an m × n matrix and T A : R n → R m is the linear transformation induced by A, then ◮ ker ( T A ) = null ( A ) ; ◮ im ( T A ) = im ( A ) .

ker Example Let T : P 1 → R be the linear transformation defined by T ( p ( x )) = p (1) for all p ( x ) ∈ P 1 .

Example Let T : P 1 → R be the linear transformation defined by T ( p ( x )) = p (1) for all p ( x ) ∈ P 1 . ker ( T ) = { p ( x ) ∈ P 1 | p (1) = 0 } = { ax + b | a , b ∈ R and a + b = 0 } = { ax − a | a ∈ R } .

Example Let T : P 1 → R be the linear transformation defined by T ( p ( x )) = p (1) for all p ( x ) ∈ P 1 . ker ( T ) = { p ( x ) ∈ P 1 | p (1) = 0 } = { ax + b | a , b ∈ R and a + b = 0 } = { ax − a | a ∈ R } . im ( T ) = { p (1) | p ( x ) ∈ P 1 } = { a + b | ax + b ∈ P 1 } = { a + b | a , b ∈ R } = R .

. Thus is a subspace of , ker By the . ker Thus , and . Then and let ker Let . ker , and , . Then ker Let . ker , , respectively. Since and denote the zero vectors of and Let ker Theorem Let V and W be vector spaces and T : V → W a linear transformation. Then ker ( T ) is a subspace of V and im ( T ) is a subspace of W.

. Thus is a subspace of , ker By the . ker Thus , and . Then and let ker Let . ker , and , . Then ker Let Theorem Let V and W be vector spaces and T : V → W a linear transformation. Then ker ( T ) is a subspace of V and im ( T ) is a subspace of W. Proof that ker ( T ) is a subspace of V. 1. Let � 0 V and � 0 W denote the zero vectors of V and W , respectively. Since T ( � 0 V ) = � 0 W , � 0 V ∈ ker ( T ) .

. Let is a subspace of , ker By the . ker Thus , and . Then and let ker Theorem Let V and W be vector spaces and T : V → W a linear transformation. Then ker ( T ) is a subspace of V and im ( T ) is a subspace of W. Proof that ker ( T ) is a subspace of V. 1. Let � 0 V and � 0 W denote the zero vectors of V and W , respectively. Since T ( � 0 V ) = � 0 W , � 0 V ∈ ker ( T ) . v 1 ) = � v 2 ) = � 2. Let � v 1 ,� v 2 ∈ ker ( T ) . Then T ( � 0 , T ( � 0 , and v 2 ) = � 0 + � 0 = � T ( � v 1 + � v 2 ) = T ( � v 1 ) + T ( � 0 . Thus � v 1 + � v 2 ∈ ker ( T ) .

. By the is a subspace of , ker Theorem Let V and W be vector spaces and T : V → W a linear transformation. Then ker ( T ) is a subspace of V and im ( T ) is a subspace of W. Proof that ker ( T ) is a subspace of V. 1. Let � 0 V and � 0 W denote the zero vectors of V and W , respectively. Since T ( � 0 V ) = � 0 W , � 0 V ∈ ker ( T ) . v 1 ) = � v 2 ) = � 2. Let � v 1 ,� v 2 ∈ ker ( T ) . Then T ( � 0 , T ( � 0 , and v 2 ) = � 0 + � 0 = � T ( � v 1 + � v 2 ) = T ( � v 1 ) + T ( � 0 . Thus � v 1 + � v 2 ∈ ker ( T ) . v 1 ) = � 3. Let � v 1 ∈ ker ( T ) and let k ∈ R . Then T ( � 0 , and v 1 ) = k ( � 0) = � T ( k � v 1 ) = kT ( � 0 . Thus k � v 1 ∈ ker ( T ) .

Theorem Let V and W be vector spaces and T : V → W a linear transformation. Then ker ( T ) is a subspace of V and im ( T ) is a subspace of W. Proof that ker ( T ) is a subspace of V. 1. Let � 0 V and � 0 W denote the zero vectors of V and W , respectively. Since T ( � 0 V ) = � 0 W , � 0 V ∈ ker ( T ) . v 1 ) = � v 2 ) = � 2. Let � v 1 ,� v 2 ∈ ker ( T ) . Then T ( � 0 , T ( � 0 , and v 2 ) = � 0 + � 0 = � T ( � v 1 + � v 2 ) = T ( � v 1 ) + T ( � 0 . Thus � v 1 + � v 2 ∈ ker ( T ) . v 1 ) = � 3. Let � v 1 ∈ ker ( T ) and let k ∈ R . Then T ( � 0 , and v 1 ) = k ( � 0) = � T ( k � v 1 ) = kT ( � 0 . Thus k � v 1 ∈ ker ( T ) . By the Subspace Test , ker ( T ) is a subspace of V . �

. . is a subspace of , im By the . im , Since , and such that . Then there exists and let im Let im , Since , and thus , such that . Then there exist im Let Proof that im ( T ) is a subspace of W. 1. Let � 0 V and � 0 W denote the zero vectors of V and W , respectively. Since T ( � 0 V ) = � 0 W , � 0 W ∈ im ( T ) .

. Let is a subspace of , im By the . im , Since , and such that . Then there exists and let im Proof that im ( T ) is a subspace of W. 1. Let � 0 V and � 0 W denote the zero vectors of V and W , respectively. Since T ( � 0 V ) = � 0 W , � 0 W ∈ im ( T ) . 2. Let � w 1 , � w 2 ∈ im ( T ) . Then there exist � v 1 ,� v 2 ∈ V such that T ( � v 1 ) = � w 1 , T ( � v 2 ) = � w 2 , and thus w 1 + � � w 2 = T ( � v 1 ) + T ( � v 2 ) = T ( � v 1 + � v 2 ) . Since � v 1 + � v 2 ∈ V , � w 1 + � w 2 ∈ im ( T ) .

. By the is a subspace of , im Proof that im ( T ) is a subspace of W. 1. Let � 0 V and � 0 W denote the zero vectors of V and W , respectively. Since T ( � 0 V ) = � 0 W , � 0 W ∈ im ( T ) . 2. Let � w 1 , � w 2 ∈ im ( T ) . Then there exist � v 1 ,� v 2 ∈ V such that T ( � v 1 ) = � w 1 , T ( � v 2 ) = � w 2 , and thus � w 1 + � w 2 = T ( � v 1 ) + T ( � v 2 ) = T ( � v 1 + � v 2 ) . Since � v 1 + � v 2 ∈ V , � w 1 + � w 2 ∈ im ( T ) . 3. Let � w 1 ∈ im ( V ) and let k ∈ R . Then there exists � v 1 ∈ V such that T ( � v 1 ) = � w 1 , and k � w 1 = kT ( � v 1 ) = T ( k � v 1 ) . Since k � v 1 ∈ V , k � w 1 ∈ im ( T ) .

Proof that im ( T ) is a subspace of W. 1. Let � 0 V and � 0 W denote the zero vectors of V and W , respectively. Since T ( � 0 V ) = � 0 W , � 0 W ∈ im ( T ) . 2. Let � w 1 , � w 2 ∈ im ( T ) . Then there exist � v 1 ,� v 2 ∈ V such that T ( � v 1 ) = � w 1 , T ( � v 2 ) = � w 2 , and thus � w 1 + � w 2 = T ( � v 1 ) + T ( � v 2 ) = T ( � v 1 + � v 2 ) . Since � v 1 + � v 2 ∈ V , � w 1 + � w 2 ∈ im ( T ) . 3. Let � w 1 ∈ im ( V ) and let k ∈ R . Then there exists � v 1 ∈ V such that T ( � v 1 ) = � w 1 , and k � w 1 = kT ( � v 1 ) = T ( k � v 1 ) . Since k � v 1 ∈ V , k � w 1 ∈ im ( T ) . By the Subspace Test , im ( T ) is a subspace of W . �

Definition Let V and W be vector spaces and T : V → W a linear transformation. Then the dimension of ker ( T ) , dim ( ker ( T )) is called the nullity of T and is denoted nullity ( T ) ; the dimension of im ( T ) , dim ( im ( T )) is called the rank of T and is denoted rank ( T ) .

dim ker Example If A is an m × n matrix, then im ( T A ) = im ( A ) = col ( A ) . It follows that rank ( T A ) = dim ( im ( T A )) = dim ( col ( A )) = rank ( A ) .

Example If A is an m × n matrix, then im ( T A ) = im ( A ) = col ( A ) . It follows that rank ( T A ) = dim ( im ( T A )) = dim ( col ( A )) = rank ( A ) . Also, ker ( T A ) = null ( A ) , so nullity ( T A ) = dim ( null ( A )) = n − rank ( A ) .

rank ; since dim im , Since im nullity dim ker . Thus is a basis of ker , independent subset of is an span From this, we see that ker and we found that Finding bases of the kernel and the image Example (continued) For the linear transformation T defjned by T : P 1 → R T ( p ( x )) = p (1) for all p ( x ) ∈ P 1 , ker ( T ) = { ax − a | a ∈ R } im ( T ) = R .

rank and dim im , we found that Since im Finding bases of the kernel and the image Example (continued) For the linear transformation T defjned by T : P 1 → R T ( p ( x )) = p (1) for all p ( x ) ∈ P 1 , ker ( T ) = { ax − a | a ∈ R } im ( T ) = R . From this, we see that ker ( T ) = span { ( x − 1) } ; since { ( x − 1) } is an independent subset of P 1 , { ( x − 1) } is a basis of ker ( T ) . Thus dim ( ker ( T )) = 1 = nullity ( T ) .

and we found that Finding bases of the kernel and the image Example (continued) For the linear transformation T defjned by T : P 1 → R T ( p ( x )) = p (1) for all p ( x ) ∈ P 1 , ker ( T ) = { ax − a | a ∈ R } im ( T ) = R . From this, we see that ker ( T ) = span { ( x − 1) } ; since { ( x − 1) } is an independent subset of P 1 , { ( x − 1) } is a basis of ker ( T ) . Thus dim ( ker ( T )) = 1 = nullity ( T ) . Since im ( T ) = R , dim ( im ( T )) = 1 = rank ( T ) .

: Suppose This gives us a system of four equations in the four variables . Then ker Problem Let T : � M 22 → � M 22 be defined by � a � a + b � a b � b + c � b � ∈ � T = for all M 22 . c d c + d d + a c d Then T is a linear transformation (you should be able to prove this). Find a basis of ker ( T ) and a basis of im ( T ) .