Quiz Suppose u 1 , . . . , u n is a basis for U and v 1 , . . . , v k - PowerPoint PPT Presentation

Quiz Suppose u 1 , . . . , u n is a basis for U and v 1 , . . . , v k is a basis for V . Prove that u 1 , . . . , u n , v 1 , . . . , v k is a basis for U ⊕ V . Two parts to the proof: 1. Show that Span { u 1 , . . . , u n , v 1 , . . . , v k } is U ⊕ V . 2. Show that { u 1 , . . . , u n , v 1 , . . . , v k } is a linearly independent set.

Direct Sum Let U and V be two vector spaces consisting of D -vectors over a field F . Definition: If U and V share only the zero vector then we define the direct sum of U and V to be the set { u + v : u ∈ U , v ∈ V} written U ⊕ V That is, U ⊕ V is the set of all sums of a vector in U and a vector in V . In Python, [u+v for u in U for v in V] (But generally U and V are infinite so the Python is just suggestive.)

Direct Sum: Example Vectors over GF (2): Example: Let U = Span { 1000 , 0100 } and let V = Span { 0010 } . I Every nonzero vector in U has a one in the first or second position (or both) and nowhere else. I Every nonzero vector in V has a one in the third position and nowhere else. Therefore the only vector in both U and V is the zero vector. Therefore U ⊕ V is defined. U ⊕ V = { 0000 + 0000 , 1000 + 0000 , 0100 + 0000 , 1100 + 0000 , 0000 + 0010 , 1000 + 0010 , 0100 + 0010 , 1100 + 0010 } which is equal to { 0000 , 1000 , 0100 , 1100 , 0010 , 1010 , 0110 , 1110 } .

Direct Sum: Example Vectors over R : Example: Let U = Span { [1 , 2 , 1 , 2] , [3 , 0 , 0 , 4] } and let V be the null space of  0 � 1 − 1 0 . 1 0 0 − 1 I The vector [2 , − 2 , − 1 , 2] is in U because it is [3 , 0 , 0 , 4] − [1 , 2 , 1 , 2] I It is also in V because 2 3 2  0  0 � � 1 − 1 0 − 2 6 7 5 = 6 7 1 0 0 − 1 − 1 0 4 2 Therefore we cannot form U ⊕ V .

Direct Sum: Example Vectors over R : Example: I Let U = Span { [4 , − 1 , 1] } . I Let V = Span { [0 , 1 , 1] } . The only intersection is at the origin, so U ⊕ V is defined. I U ⊕ V is the set of vectors u + v where u ∈ U and v ∈ V . I This is just Span { [4 , − 1 , 1] , [0 , 1 , 1] } I Plane containing the two lines

Properties of direct sum Lemma: U ⊕ V is a vector space. (Prove using Properties V1, V2, V3.) Lemma: The union of I a set of generators of U , and I a set of generators of V is a set of generators for U ⊕ V . Proof: Suppose U = Span { u 1 , . . . , u m } and V = Span { v 1 , . . . , v n } . Then I every vector in U can be written as α 1 u 1 + · · · + α m u m , and I every vector in V can be written as β 1 v 1 + · · · + β n v n so every vector in U ⊕ V can be written as α 1 u 1 + · · · + α m u m + β 1 v 1 + · · · + β n v n QED

Properties of direct sum Direct Sum Basis Lemma: Union of a basis of U and a basis of V is a basis of U ⊕ V . Proof: Clearly I a basis of U is a set of generators for U , and I a basis of V is a set of generators for V . Therefore the previous lemma shows that I the union of a basis for U and a basis for V is a generating set for U ⊕ V . We just need to show that the union is linearly independent.

Properties of direct sum Direct Sum Basis Lemma: Union of a basis of U and a basis of V is a basis of U ⊕ V . Proof, cont’d: Let { u 1 , . . . , u m } be a basis for U . Let { v 1 , . . . , v n } be a basis for V . We need to show that { u 1 , . . . , u m , v 1 , . . . , v n } is independent. Suppose 0 = α 1 u 1 + · · · + α m u m + β 1 v 1 + · · · + β n v n . Then α 1 u 1 + · · · + α m u m = ( − β 1 ) v 1 + · · · + ( − β n ) v n | {z } | {z } in U in V Left-hand side is a vector in U , and right-hand side is a vector in V . By definition of U ⊕ V , the only vector in both U and V is the zero vector. This shows: 0 = α 1 u 1 + · · · + α m u m and 0 = ( − β 1 ) v 1 + · · · + ( − β n ) v n

Direct Sum Direct-Sum Basis Lemma: Union of a basis of U and a basis of V is a basis of U ⊕ V . Direct-Sum Dimension Corollary: dim U + dim V = dim U ⊕ V Proof: A basis for U together with a basis for V forms a basis for U ⊕ V . QED

Linear function invertibility How to tell if a linear function f : V − → W is invertible? I One-to-one? f is one-to-one if its kernel is trivial. Equivalent: if its kernel has dimension zero. I Onto? f is onto if its image equals its co-domain Recall that the image of a function f with domain V is { f ( v ) : v ∈ V} . Lemma: The image of f is a subspace of W . How can we tell if the image of f equals W ? Dimension Lemma: If U is a subspace of W then Property D1: dim U ≤ dim W , and Property D2: if dim U = dim W then U = W Use Property D2 with U = Im f . Shows that the function f is onto i ff dim Im f = dim W We conclude: f is invertible dim Ker f = 0 and dim Im f = dim W

Linear function invertibility f is one-to-one if dim Ker f = 0 and dim Im f = dim W How does this relate to dimension of the domain ? Conjecture For f to be invertible, need dim V = dim W .

Extracting an invertible function V W Starting with a linear function f we will extract a largest possible subfunction that is invertible. Make it onto by setting co-domain to be image of f . Make it one-to-one by getting rid of extra domain elements sharing same image.

Extracting an invertible function V W Starting with a linear function f we will extract a largest possible subfunction that is invertible. Make it onto by setting co-domain to be image of f . Make it one-to-one by getting rid of extra domain elements sharing same image.

Extracting an invertible function V W Start with linear function f : V − → W Step 1: Choose smaller co-domain W ∗ Step 2: Choose smaller domain V ∗ Step 3: Define function f ∗ : V ∗ − → W ∗ by f ∗ ( x ) = f ( x ) In fact, we will end up selecting a basis of W ∗ and a basis of V ∗ .

Extracting an invertible function V W Start with linear function f : V − → W Step 1: Choose smaller co-domain W ∗ Step 2: Choose smaller domain V ∗ Step 3: Define function f ∗ : V ∗ − → W ∗ by f ∗ ( x ) = f ( x ) In fact, we will end up selecting a basis of W ∗ and a basis of V ∗ .

Extracting an invertible function V W Start with linear function f : V − → W Step 1: Choose smaller co-domain W ∗ Step 2: Choose smaller domain V ∗ Step 3: Define function f ∗ : V ∗ − → W ∗ by f ∗ ( x ) = f ( x ) In fact, we will end up selecting a basis of W ∗ and a basis of V ∗ .

Extracting an invertible function from linear function f : V − → W V W I Choose smaller co-domain W ∗ Let W ∗ be image of f Let w 1 , . . . , w r be a basis of W ∗ I Choose smaller domain V ∗ Let v 1 , . . . , v r be pre-images of w 1 , . . . , w r That is, f ( v 1 ) = w 1 , . . . , f ( v r ) = w r Let V ∗ = Span { v 1 , . . . , v r } I Define function f ∗ : V ∗ − → W ∗ by f ∗ ( x ) = f ( x ) We will show: I f ∗ is onto I f ∗ is one-to-one (kernel is trivial) I Bonus: v 1 , . . . , v r form a basis for V ∗

Extracting an invertible function from linear function f : V − → W Onto: I Choose smaller co-domain W ∗ Let w be any vector in co-domain W ∗ . There Let W ∗ be image of f are scalars α 1 , . . . , α r such that Let w 1 , . . . , w r be a basis of W ∗ w = α 1 w 1 + · · · + α r w r Because f is linear, I Choose smaller domain V ∗ Let v 1 , . . . , v r be pre-images of w 1 , . . . , w r f ( α 1 v 1 + · · · + α r v r ) That is, f ( v 1 ) = w 1 , . . . , f ( v r ) = w r = α 1 f ( v 1 ) + · · · + α r f ( v r ) Let V ∗ = Span { v 1 , . . . , v r } = α 1 w 1 + · · · + α r w r I Define function f ∗ : V ∗ − so w is image of α 1 v 1 + · · · + α r v r ∈ V ∗ . → W ∗ QED by f ∗ ( x ) = f ( x ) We will show: I f ∗ is onto I f ∗ is one-to-one (kernel is trivial) I Bonus: v 1 , . . . , v r form a basis for V ∗

Extracting an invertible function from linear function f : V − → W One-to-one: I Choose smaller co-domain W ∗ By One-to-One Lemma, need only show kernel Let W ∗ be image of f is trivial. Let w 1 , . . . , w r be a basis of W ∗ Suppose v ∗ is in V ∗ and f ( v ∗ ) = 0 I Choose smaller domain V ∗ Because V ∗ = Span { v 1 , . . . , v r } , there are Let v 1 , . . . , v r be pre-images of w 1 , . . . , w r scalars α 1 , . . . , α r such that That is, f ( v 1 ) = w 1 , . . . , f ( v r ) = w r v ∗ = α 1 v 1 + · · · + α r v r Let V ∗ = Span { v 1 , . . . , v r } Applying f to both sides, I Define function f ∗ : V ∗ − 0 = f ( α 1 v 1 + · · · + α r v r ) → W ∗ by f ∗ ( x ) = f ( x ) = α 1 w 1 + · · · + α r w r We will show: Because w 1 , . . . , w r are linearly independent, I f ∗ is onto α 1 = · · · = α r = 0 I f ∗ is one-to-one (kernel is trivial) so v ∗ = 0 QED I Bonus: v 1 , . . . , v r form a basis for V ∗

Extracting an invertible function from linear function f : V − → W Bonus: v 1 , . . . , v r form a basis for V ∗ I Choose smaller co-domain W ∗ Need only show linear independence Let W ∗ be image of f Suppose 0 = α 1 v 1 + · · · + α r v r Let w 1 , . . . , w r be a basis of W ∗ Applying f to both sides, I Choose smaller domain V ∗ 0 = f ( α 1 v 1 + · · · + α r v r ) Let v 1 , . . . , v r be pre-images of w 1 , . . . , w r = α 1 w 1 + · · · + α r w r That is, f ( v 1 ) = w 1 , . . . , f ( v r ) = w r Let V ∗ = Span { v 1 , . . . , v r } Because w 1 , . . . , w r are linearly independent, I Define function f ∗ : V ∗ − α 1 = · · · = α r = 0. QED → W ∗ by f ∗ ( x ) = f ( x ) We will show: I f ∗ is onto I f ∗ is one-to-one (kernel is trivial) I Bonus: v 1 , . . . , v r form a basis for V ∗