Proof that ker(t) is a Subspace of R n

Transcription

1 Proof that ker(t) is a Subspace of R n Gene Quinn Proof that ker(t) is a Subspace of R n p.1/7

2 Suppose T : R m R n is a linear transformation with domain R m and codomain R n. Proof that ker(t) is a Subspace of R n p.2/7

3 Suppose T : R m R n is a linear transformation with domain R m and codomain R n. In this situation, the kernel of T is defined as: ker(t) = { x R m : such that T( x) = 0 R n } Proof that ker(t) is a Subspace of R n p.2/7

4 Suppose T : R m R n is a linear transformation with domain R m and codomain R n. In this situation, the kernel of T is defined as: ker(t) = { x R m : such that T( x) = 0 R n } The set-builder notation would be read as "The set of all vectors x in R m such that T( x) is the zero vector in R n ". Proof that ker(t) is a Subspace of R n p.2/7

5 Theorem: For any linear transformation T : R m R n, the kernel of T is a subspace (of R m ). Proof that ker(t) is a Subspace of R n p.3/7

6 Theorem: For any linear transformation T : R m R n, the kernel of T is a subspace (of R m ). Proof: In order to prove that ker(t) is a subspace, we must establish the following three claims: 0 m ker(t) ker(t) is closed under addition ker(t) is closed under scalar multiplication Proof that ker(t) is a Subspace of R n p.3/7

7 Claim 1: 0 m ker(t) Proof of Claim 1: By an earlier theorem, there exists an n m matrix A with the property that T( x) = A x x R m Proof that ker(t) is a Subspace of R n p.4/7

8 Claim 1: 0 m ker(t) Proof of Claim 1: By an earlier theorem, there exists an n m matrix A with the property that T( x) = A x x R m Let x be the zero vector in R m, 0 m. Then by the properties of matrix multiplication, A 0 m = 0 n for any n m matrix A. Therefore, A 0 m = T( 0 m ) = 0 n, and so by definition 0 m is in ker(t). Proof that ker(t) is a Subspace of R n p.4/7

9 Claim 2: ker(t) is closed under addition. Proof of Claim 2: Let u, v R m be arbitrary elements of ker(t). We need to show that u + v ker(t). Proof that ker(t) is a Subspace of R n p.5/7

10 Claim 2: ker(t) is closed under addition. Proof of Claim 2: Let u, v R m be arbitrary elements of ker(t). We need to show that u + v ker(t). By the definition of ker(t), T( u) = A u = 0 n and T( v) = A v = 0 n Proof that ker(t) is a Subspace of R n p.5/7

11 Claim 2: ker(t) is closed under addition. Proof of Claim 2: Let u, v R m be arbitrary elements of ker(t). We need to show that u + v ker(t). By the definition of ker(t), T( u) = A u = 0 n and T( v) = A v = 0 n By the properties of vector addition, = 0 R n. Proof that ker(t) is a Subspace of R n p.5/7

12 Claim 2: ker(t) is closed under addition. Proof of Claim 2: Let u, v R m be arbitrary elements of ker(t). We need to show that u + v ker(t). By the definition of ker(t), T( u) = A u = 0 n and T( v) = A v = 0 n By the properties of vector addition, = 0 R n. By the properties of linear transformations, T( u + v) = A( u + v) = A u + A v = 0 n + 0 n = 0 n and therefore u + v ker(t). Proof that ker(t) is a Subspace of R n p.5/7

13 Claim 3: ker(t) is closed under scalar multiplication. Proof of Claim 3: Let u R n be an arbitrary element of ker(t) and k R an arbitrary scalar. We need to show that k u ker(t). Proof that ker(t) is a Subspace of R n p.6/7

14 Claim 3: ker(t) is closed under scalar multiplication. Proof of Claim 3: Let u R n be an arbitrary element of ker(t) and k R an arbitrary scalar. We need to show that k u ker(t). By the definition of ker(t), T( u) = A u = 0 n Proof that ker(t) is a Subspace of R n p.6/7

15 Since R m is a subspace of itself, it is closed under scalar multiplication, and therefore for any real number k, k u R m. Proof that ker(t) is a Subspace of R n p.7/7

16 Since R m is a subspace of itself, it is closed under scalar multiplication, and therefore for any real number k, k u R m. By the properties of linear transformations, T(k u) = A(k u) = ka u = k 0 n = 0 n and therefore k u ker(t). Proof that ker(t) is a Subspace of R n p.7/7

17 Since R m is a subspace of itself, it is closed under scalar multiplication, and therefore for any real number k, k u R m. By the properties of linear transformations, T(k u) = A(k u) = ka u = k 0 n = 0 n and therefore k u ker(t). This completes the proof that ker(t) is a subspace. Proof that ker(t) is a Subspace of R n p.7/7