University:
Whitman College
Course:
MATH-240 | Linear Algebra
Academic year:

2023
Views:

66

Pages:

1
Author:

charkbastardism

4.1, 32 If H, K are subspaces, show that H ∩ K is also a subspace. • Show that H ∩ K has the zero vector: Since H, K are subspaces, then ~0 ∈ H and ~0 ∈ K, so ~0 ∈ H ∩ K. • Show that H ∩ K is closed under addition: Let x1 , x2 be in H ∩ K. That means that each is in H and each is in K. Since both vectors are in H (and H is a subspace), then x1 + x2 is in H. Similarly, since both vectors are in K (and K is a subspace), then x2 + x2 is in K. Therefore, the sum is also in H ∩ K. • Show that H ∩ K is closed under scalar multiplication: Let x ∈ H ∩ K. We show that cx is in H ∩ K for all scalars c: Since x ∈ H ∩ K, then x ∈ H and x ∈ K. Since each of these are subspaces, cx ∈ H and cx ∈ K. Therefore, cx ∈ H ∩ K. The argument above does not work for unions-For example, (" H = Span 1 0 #) (" , K = Span 0 1 #) n o Then H ∩ K = ~0 , but the union (only being the two coordinate axes) would not be closed under addition- Adding the two basis vectors together, in fact, gives (1, 1), which is not in H or in K. 4.2, 30 Let T : V → W be a linear transformation from vector space V to vector space W . Show that the range of T is a subspace of W . NOTE: To show that something is in the range of T , we must show that it came from some vector in V ... • Is ~0 in the range of T ? Yes, since T (~0) = ~0, and V is a subspace (so the domain contains the zero vector). • Is the range closed under addition? Let w1 , w2 be in the range of T . Then there are vectors v1 , v2 in V such that T (v1 ) = w1 and T (v2 ) = w2 . Now, w1 + w2 = T (v1 ) + T (v2 ) = T (v1 + v2 ) And, since V is a vector space, the sum v1 + v2 is in V . Therefore, W is closed under addition. • Is the range closed under scalar multiplication? Let w be in the range, so w = T (v), for some v ∈ V . Since V is a subspace, then cv is in V for all scalars c, which tells us: cw = cT (v) = T (cv) And W is closed under scalar multiplication. 1

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