The vector space is a collection of vectors
The vector space is a collection of vectors 𝑆 = {⃗⃗⃗ v1 , ⃗⃗⃗⃗ v2 , … } along with the following rules: Vector addition axioms 1. Closed under addition: • For any 𝑢 ⃗ and 𝑣 in 𝑆, 𝑢 ⃗ + 𝑣 is also in 𝑆 2. Commutative: • For any 𝑢 ⃗ and 𝑣 in 𝑆, 𝑢 ⃗ +𝑣 = 𝑣+𝑢 ⃗ 3. Associative: • For any 𝑢 ⃗ , 𝑣, 𝑤 ⃗⃗ in 𝑆, 𝑢 ⃗ + (𝑣 + 𝑤 ⃗⃗ ) = (𝑢 ⃗ + 𝑣) + 𝑤 ⃗⃗ 4. Zero Identity: • There exists ⃗0 in 𝑆 such that for any 𝑢 ⃗ in 𝑆, 𝑢 ⃗ + ⃗0 = 𝑢 ⃗ 5. Additive Inverse: • For any 𝑢 ⃗ in 𝑆, there exists −𝑢 ⃗ in 𝑆 such that ⃗ 𝑢 ⃗ + (−𝑢 ⃗)=0 Scalar multiplication axioms 6. Closed under scalar multiplication: • For any 𝑢 ⃗ in 𝑆 and scalar 𝑎, 𝑎𝑢 ⃗ is also in 𝑆 7. Compatible: • For any 𝑢 ⃗ in 𝑆 and scalars 𝑎, 𝑏, 𝑎 (𝑏𝑢 ⃗ ) = (𝑎𝑏)𝑢 ⃗ 8. Distributive: • for any 𝑢 ⃗ , 𝑣 in 𝑆 and scalar 𝑎, 𝑎 (𝑢 ⃗ + 𝑣 ) = 𝑎𝑢 ⃗ + 𝑎𝑣 and • for any 𝑢 ⃗ in 𝑆 and scalars 𝑎, 𝑏, (𝑎 + 𝑏)𝑢 ⃗ = 𝑎𝑢 ⃗ + 𝑏𝑢 ⃗ 9. One identity: • For any 𝑢 ⃗ in 𝑆, 1𝑢 ⃗ =𝑢 ⃗
Example: (vector space ℝ𝑛 ) ℝ2 is the set of all vectors with 𝑛-elements (“ordered n-tuple”) 𝑥1 𝑥2 ℝ𝑛 = {( ⋮ ) 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑒𝑎𝑐ℎ 𝑥𝑖 𝑖𝑠 𝑖𝑛 ℝ} 𝑥𝑛 ℝ𝑛 is a vector space! We see this by proving each of the axioms/rules listed above. I will prove 1 and 6 and leave the rest as an exercise: 𝑥1 𝑦1 𝑥2 𝑦2 1. Suppose that ( ⋮ ) and ( ⋮ ) are in ℝ𝑛 , 𝑥𝑛 𝑦𝑛 𝑥1 𝑦1 𝑥1 + 𝑦1 𝑥2 𝑦2 𝑥 +𝑦 then ( ⋮ ) + ( ⋮ ) = ( 2 ⋮ 2 ), since 𝑥𝑖 and 𝑦𝑖 are just 𝑥𝑛 𝑦𝑛 𝑥𝑛 + 𝑦𝑛 numbers in ℝ, 𝑥𝑖 + 𝑦𝑖 is also a number in ℝ, so each of the entries are just numbers in ℝ and the addition of vectors is still in ℝ𝑛 𝑥1 𝑥2 6. Suppose that ( ⋮ ) is in ℝ𝑛 and 𝑎 is a scalar in ℝ, 𝑥𝑛 𝑥1 𝑎𝑥1 𝑥2 𝑎𝑥2 Then 𝑎 ( ⋮ ) = ( ⋮ ), since 𝑎𝑥𝑖 is a product of numbers in ℝ, 𝑥𝑛 𝑎𝑥𝑛 the result is still a number in ℝ. So, the entries are in ℝ and the resulting vector is still in ℝ𝑛 .
A subspace of ℝ𝑛 is a subset 𝑈 of ℝ𝑛 such that 1. Zero vector: 0 • ⃗0 = ( ⋮ ) is in 𝑈 0 2. Closed under addition: • For any 𝑢 ⃗ , 𝑣 in 𝑈, 𝑢 ⃗ + 𝑣 is also in 𝑈 3. Closed under scalar multiplication: • For any 𝑢 ⃗ in 𝑈 and scalar 𝑎, 𝑎𝑢 ⃗ is also in 𝑈 *(shortcut check): To check if a subset is a subspace, we only need to check if it contains the zero vector and if 𝑎𝑢 ⃗ + 𝑏𝑣 is in 𝑈 for any scalar 𝑎, 𝑏 and vectors 𝑢 ⃗ , 𝑣 in 𝑈 Notation: If 𝑈 is a subset of ℝ𝑛 , we write 𝑈 ⊂ ℝ𝑛 Examples Check if the following subsets are subspaces of ℝ𝑛 : 1. ℝ𝑛 itself • ℝ𝑛 contains the zero vector ⃗0 • Closed under addition: see the proof 1 above • Closed under scalar multiplication: see proof 6 above Therefore, ℝ𝑛 is a subspace of itself
⃗} 2. 𝑈 = {0 • ⃗0 is in 𝑈 ⃗ + 𝑏0 ⃗ = ⃗0 + ⃗0 = ⃗0, • Suppose that 𝑎, 𝑏 are scalars. Then 𝑎0 which is in 𝑈 Therefore, 𝑈 is a subspace of ℝ𝑛
3𝑥1 𝑥 3. 𝑈 = {( 2 ) 𝑤ℎ𝑒𝑟𝑒 𝑥𝑖 𝑎𝑟𝑒 𝑖𝑛 ℝ} ⋮ 𝑥𝑛
• ⃗0 is in 𝑈 (just let 𝑥1 , 𝑥2 , … , 𝑥𝑛 = 0) 3𝑥1 3𝑦1 𝑥 𝑦 • Suppose 𝑎, 𝑏 are scalars and ( 2 ), ( 2 ) are in 𝑈. ⋮ ⋮ 𝑥𝑛 𝑦𝑛 3𝑥1 3𝑦1 𝑥 𝑦 Then, 𝑎 ( 2 ) + 𝑏 ( 2 ) ⋮ ⋮ 𝑥𝑛 𝑦𝑛 3𝑎𝑥1 + 3𝑏𝑦1 𝑎𝑥2 + 𝑏𝑦2 =( ) ⋮ 𝑎𝑥𝑛 + 𝑏𝑦𝑛 3(𝑎𝑥1 + 𝑏𝑦1 ) 𝑎𝑥2 + 𝑏𝑦2 =( ) ⋮ 𝑎𝑥𝑛 + 𝑏𝑦𝑛 Note that this is still in 𝑈 because 𝑎𝑥𝑖 + 𝑏𝑦𝑖 is still in ℝ
Therefore, 𝑈 is a subspace of ℝ𝑛
2𝑢1 + 1 𝑢2 ) 𝑤ℎ𝑒𝑟𝑒 𝑢𝑖 𝑎𝑟𝑒 𝑖𝑛 ℝ} 4. 𝑈 = {( ⋮ 𝑢𝑛
1 • ⃗0 is in 𝑈 because if we let 𝑢1 = − 2 and the rest of the 1
2 (− 2) + 1 𝑢𝑖 = 0, we get (
0 ⋮ 0
)
0 0 =( ) ⋮ 0
2𝑥1 + 1 2𝑦1 + 1 𝑥 𝑦 • Suppose 𝑎, 𝑏 are scalars and ( ⋮2 ),( ⋮2 ) are in 𝑈. 𝑥𝑛 𝑦𝑛
2𝑥1 + 1 2𝑦1 + 1 𝑥2 𝑦2 Then 𝑎 ( )+𝑏( ) ⋮ ⋮ 𝑥𝑛 𝑦𝑛
2𝑎𝑥1 + 𝑎 + 2𝑏𝑦1 + 𝑏 2(𝑎𝑥1 + 𝑏𝑦1 ) + (𝑎 + 𝑏) 𝑎𝑥2 + 𝑏𝑦2 𝑎𝑥2 + 𝑏𝑦2 =( )=( ) ⋮ ⋮ 𝑎𝑥𝑛 + 𝑏𝑦𝑛 𝑎𝑥𝑛 + 𝑏𝑦𝑛
Since 𝑎𝑥𝑖 + 𝑏𝑦𝑖 is still in ℝ, if we let 𝑣𝑖 = 𝑎𝑥𝑖 + 𝑏𝑦𝑖 , we get 2𝑣1 + 𝑎 + 𝑏 𝑣2 that the vector is ( ), which is not of the form ⋮ 𝑣𝑛 2𝑢1 + 1 𝑢2 ( ) ⋮ 𝑢𝑛 Therefore, 𝑈 is NOT a subspace of ℝ𝑛
More Examples 𝑎 𝑏 Is 𝑈 = {( 𝑐 ) 𝑤ℎ𝑒𝑟𝑒 𝑎 − 𝑏 = 0 𝑎𝑛𝑑 𝑐 2 = 𝑑 2 } a subspace of ℝ4 ? 𝑑 0 0 • ( ) is in 𝑈 since 𝑎 − 𝑏 = 0 − 0 = 0, and 𝑐 2 = 0 = 𝑑 2 0 0 • *Instead of proving the axioms directly, we could have also used a counter example: 2 1 2 1 ( ) and ( ) are in 𝑈 (check this yourself). Use scalars 𝑎 = 𝑏 = 1. 3 1 3 1 2 1 3 2 1 3 Then, 1 ( ) + 1 ( ) = ( ). If this was in 𝑈, then it must 3 1 4 −3 1 −2 2 follow the property that 𝑐 = 𝑑 2. So, 42 = (−2)2 which is not true. Therefore, the sum of vectors is not in 𝑈 and 𝑈 is not a subspace of ℝ4 .
Recall: If 𝑇: ℝ𝑛 → ℝ𝑚 is a linear transformation, then there exists an 𝑚 × 𝑛 matrix 𝐴 that induces 𝑇 The nullspace or kernel of 𝑇 or 𝐴 is 𝑥1 𝑥1 𝑥2 𝑥2 𝑛𝑢𝑙𝑙 (𝑇 ) = {( ⋮ ) 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑇 ( ⋮ ) = ⃗⃗⃗⃗⃗ 0𝑚 } 𝑥𝑛 𝑥𝑛 or 𝑥1 𝑥1 𝑥2 𝑥2 𝑛𝑢𝑙𝑙 (𝑇 ) = {( ⋮ ) 𝑠𝑜 𝑡ℎ𝑎𝑡 𝐴 ( ⋮ ) = ⃗⃗⃗⃗⃗ 0𝑚 } 𝑥𝑛 𝑥𝑛
And the image of 𝑇 or 𝐴 is 𝑦1 𝑦1 𝑥1 𝑥1 𝑦2 𝑦2 𝑥2 𝑥2 𝑖𝑚(𝑇 ) = {( ⋮ ) 𝑠𝑜 𝑡ℎ𝑎𝑡 ( ⋮ ) = 𝑇 ( ⋮ ) 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 ( ⋮ ) } 𝑦𝑚 𝑦𝑚 𝑥𝑛 𝑥𝑛 Or 𝑦1 𝑦2
𝑦1 𝑦2
𝑥1 𝑥2
𝑥1 𝑥2
𝑦𝑚
𝑦𝑚
𝑥𝑛
𝑥𝑛
𝑖𝑚(𝑇 ) = {( ⋮ ) 𝑠𝑜 𝑡ℎ𝑎𝑡 ( ⋮ ) = 𝐴 ( ⋮ ) 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 ( ⋮ )}
Example Show that 𝑛𝑢𝑙𝑙(𝑇) is a subspace of ℝ𝑛 . ⃗⃗⃗⃗𝑛 is in 𝑛𝑢𝑙𝑙(𝑇) because 𝑇(0 ⃗⃗⃗⃗𝑛 ) = ⃗⃗⃗⃗⃗ • 0 0𝑚 by property of linear transformations • Suppose that ⃗⃗⃗⃗ u1 and ⃗⃗⃗⃗ u2 are in 𝑛𝑢𝑙𝑙(𝑇), 𝑎, 𝑏 are scalars So, 𝑇(⃗⃗⃗⃗ u1 ) = ⃗⃗⃗⃗⃗ 0𝑚 and 𝑇(⃗⃗⃗⃗ u2 ) = ⃗⃗⃗⃗⃗ 0𝑚 . Then, 𝑇(𝑎u ⃗⃗⃗⃗1 + 𝑏u ⃗⃗⃗⃗2 ) = 𝑇 (𝑎u ⃗⃗⃗⃗1 ) + 𝑇(𝑏u ⃗⃗⃗⃗2 ) = 𝑎𝑇(⃗⃗⃗⃗ u1 ) + 𝑏𝑇(⃗⃗⃗⃗ u2 ) ⃗⃗⃗⃗⃗𝑚 + 𝑏0 ⃗⃗⃗⃗⃗𝑚 = 𝑎0 = ⃗⃗⃗⃗⃗ 0𝑚 + ⃗⃗⃗⃗⃗ 0𝑚 = ⃗⃗⃗⃗⃗ 0𝑚 Hence, 𝑎u ⃗⃗⃗⃗1 + 𝑏u ⃗⃗⃗⃗2 is in 𝑛𝑢𝑙𝑙(𝑇). Therefore, 𝑛𝑢𝑙𝑙(𝑇) is a subspace of ℝ𝑛
Show that 𝑖𝑚(𝑇) is a subspace of ℝ𝑚 ⃗⃗⃗⃗𝑛 ) • ⃗⃗⃗⃗⃗ 0𝑚 is in 𝑖𝑚(𝑇) because ⃗⃗⃗⃗⃗ 0𝑚 = 𝑇(0 • Suppose that ⃗⃗⃗ v1 and 𝑣 ⃗⃗⃗⃗2 are in 𝑖𝑚(𝑇), 𝑎, 𝑏 are scalars So, ⃗⃗⃗ v1 = 𝑇(⃗⃗⃗⃗ u1 ) and ⃗⃗⃗⃗ v2 = 𝑇(⃗⃗⃗⃗ u2 ) for some ⃗⃗⃗⃗ u1 𝑎𝑛𝑑 ⃗⃗⃗⃗ u2 in 𝑛𝑢𝑙𝑙(𝑇) Then, 𝑎v⃗⃗⃗1 + 𝑏v ⃗⃗⃗⃗2 = 𝑎𝑇(⃗⃗⃗⃗ u1 ) + 𝑏𝑇 (⃗⃗⃗⃗ u2 ) = 𝑇(𝑎u ⃗⃗⃗⃗1 ) + 𝑇 (𝑏u ⃗⃗⃗⃗2 ) ⃗⃗⃗⃗2 ) = 𝑇(𝑎u ⃗⃗⃗⃗1 + 𝑏b Hence, 𝑎v ⃗⃗⃗1 + 𝑏v ⃗⃗⃗⃗2 is in 𝑖𝑚(𝑇). Therefore, 𝑖𝑚(𝑇) is a subspace of ℝ𝑚
More Examples Find the nullspace and image of the linear transformation −2 2 ]. induced by the matrix 𝐴 = [ −1 1 𝑥1 𝑥1 0 The nullspace is 𝑛𝑢𝑙𝑙 (𝑇) = {(𝑥 ) 𝑠𝑜 𝑡ℎ𝑎𝑡 𝐴 (𝑥 ) = ( )} 2 2 0 −2𝑥1 + 2𝑥2 −2 2 𝑥1 0 0 So, we need [ ] [𝑥 ] = [ ] [ ]=[ ] −𝑥1 + 𝑥2 −1 1 2 0 0 Equating each entry: −2𝑥1 + 2𝑥2 = 0 → 𝑥1 = 𝑥2 −𝑥1 + 𝑥2 = 0 𝑥 Therefore, the nullsapce is 𝑛𝑢𝑙𝑙 (𝑇 ) = {( ) 𝑤ℎ𝑒𝑟𝑒 𝑥 𝑖𝑛 ℝ} 𝑥 𝑦1 𝑦1 𝑥1 𝑥1 The image is 𝑖𝑚(𝑇) = {(𝑦 ) 𝑠𝑜 𝑡ℎ𝑎𝑡 (𝑦 ) = 𝐴 (𝑥 ) 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 (𝑥 )} 2 2 2 2 𝑦1 2(𝑥 − 𝑥1 ) −2𝑥1 + 2𝑥2 −2 2 𝑥1 So, [𝑦 ] = [ ] [𝑥 ] = [ ]=[ 2 ] −𝑥1 + 𝑥2 (𝑥2 − 𝑥1 ) 2 −1 1 2 𝑦1 2𝑢 If we let 𝑢 = 𝑥2 − 𝑥1 , then 𝑢 is in ℝ and [𝑦 ] = [ ] 2 𝑢 2𝑢 Therefore, the image is 𝑖𝑚(𝑇 ) = {[ ] 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑢 𝑖𝑛 ℝ} 𝑢
Suppose that 𝐴 is an 𝑛 × 𝑛 matrix and 𝜆 is a number. Recall: 𝜆 is an eigenvalue of 𝐴 if 𝐴x⃗ = 𝜆x⃗ for some non-zero ℝ𝑛 vector x⃗. In this case, the vector x⃗ is an eigenvector of 𝐴 corresponding to 𝜆 Let’s define 𝐸𝐴 (𝜆) = {x⃗ 𝑖𝑛 ℝ𝑛 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝐴x⃗ = 𝜆x⃗} = {x⃗ 𝑖𝑛 ℝ𝑛 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 (𝐴 − 𝜆𝐼 )x⃗ = ⃗0} = 𝑛𝑢𝑙𝑙 (𝐴 − 𝜆𝐼 ) Therefore, if 𝑛𝑢𝑙𝑙(𝐴 − 𝜆𝐼) contains more than just the zero vector ⃗0, then 𝜆 is an eigenvalue of 𝐴 and every vector in 𝑛𝑢𝑙𝑙(𝐴 − 𝜆𝐼 ) is a corresponding eigenvector. We say that 𝐸𝐴 (𝜆) is the eigenspace corresponding to 𝜆
Example: (vector space ℝ𝑛 ) ℝ2 is the set of all vectors with 𝑛-elements (“ordered n-tuple”) 𝑥1 𝑥2 ℝ𝑛 = {( ⋮ ) 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑒𝑎𝑐ℎ 𝑥𝑖 𝑖𝑠 𝑖𝑛 ℝ} 𝑥𝑛 ℝ𝑛 is a vector space! We see this by proving each of the axioms/rules listed above. I will prove 1 and 6 and leave the rest as an exercise: 𝑥1 𝑦1 𝑥2 𝑦2 1. Suppose that ( ⋮ ) and ( ⋮ ) are in ℝ𝑛 , 𝑥𝑛 𝑦𝑛 𝑥1 𝑦1 𝑥1 + 𝑦1 𝑥2 𝑦2 𝑥 +𝑦 then ( ⋮ ) + ( ⋮ ) = ( 2 ⋮ 2 ), since 𝑥𝑖 and 𝑦𝑖 are just 𝑥𝑛 𝑦𝑛 𝑥𝑛 + 𝑦𝑛 numbers in ℝ, 𝑥𝑖 + 𝑦𝑖 is also a number in ℝ, so each of the entries are just numbers in ℝ and the addition of vectors is still in ℝ𝑛 𝑥1 𝑥2 6. Suppose that ( ⋮ ) is in ℝ𝑛 and 𝑎 is a scalar in ℝ, 𝑥𝑛 𝑥1 𝑎𝑥1 𝑥2 𝑎𝑥2 Then 𝑎 ( ⋮ ) = ( ⋮ ), since 𝑎𝑥𝑖 is a product of numbers in ℝ, 𝑥𝑛 𝑎𝑥𝑛 the result is still a number in ℝ. So, the entries are in ℝ and the resulting vector is still in ℝ𝑛 .
A subspace of ℝ𝑛 is a subset 𝑈 of ℝ𝑛 such that 1. Zero vector: 0 • ⃗0 = ( ⋮ ) is in 𝑈 0 2. Closed under addition: • For any 𝑢 ⃗ , 𝑣 in 𝑈, 𝑢 ⃗ + 𝑣 is also in 𝑈 3. Closed under scalar multiplication: • For any 𝑢 ⃗ in 𝑈 and scalar 𝑎, 𝑎𝑢 ⃗ is also in 𝑈 *(shortcut check): To check if a subset is a subspace, we only need to check if it contains the zero vector and if 𝑎𝑢 ⃗ + 𝑏𝑣 is in 𝑈 for any scalar 𝑎, 𝑏 and vectors 𝑢 ⃗ , 𝑣 in 𝑈 Notation: If 𝑈 is a subset of ℝ𝑛 , we write 𝑈 ⊂ ℝ𝑛 Examples Check if the following subsets are subspaces of ℝ𝑛 : 1. ℝ𝑛 itself • ℝ𝑛 contains the zero vector ⃗0 • Closed under addition: see the proof 1 above • Closed under scalar multiplication: see proof 6 above Therefore, ℝ𝑛 is a subspace of itself
⃗} 2. 𝑈 = {0 • ⃗0 is in 𝑈 ⃗ + 𝑏0 ⃗ = ⃗0 + ⃗0 = ⃗0, • Suppose that 𝑎, 𝑏 are scalars. Then 𝑎0 which is in 𝑈 Therefore, 𝑈 is a subspace of ℝ𝑛
3𝑥1 𝑥 3. 𝑈 = {( 2 ) 𝑤ℎ𝑒𝑟𝑒 𝑥𝑖 𝑎𝑟𝑒 𝑖𝑛 ℝ} ⋮ 𝑥𝑛
• ⃗0 is in 𝑈 (just let 𝑥1 , 𝑥2 , … , 𝑥𝑛 = 0) 3𝑥1 3𝑦1 𝑥 𝑦 • Suppose 𝑎, 𝑏 are scalars and ( 2 ), ( 2 ) are in 𝑈. ⋮ ⋮ 𝑥𝑛 𝑦𝑛 3𝑥1 3𝑦1 𝑥 𝑦 Then, 𝑎 ( 2 ) + 𝑏 ( 2 ) ⋮ ⋮ 𝑥𝑛 𝑦𝑛 3𝑎𝑥1 + 3𝑏𝑦1 𝑎𝑥2 + 𝑏𝑦2 =( ) ⋮ 𝑎𝑥𝑛 + 𝑏𝑦𝑛 3(𝑎𝑥1 + 𝑏𝑦1 ) 𝑎𝑥2 + 𝑏𝑦2 =( ) ⋮ 𝑎𝑥𝑛 + 𝑏𝑦𝑛 Note that this is still in 𝑈 because 𝑎𝑥𝑖 + 𝑏𝑦𝑖 is still in ℝ
Therefore, 𝑈 is a subspace of ℝ𝑛
2𝑢1 + 1 𝑢2 ) 𝑤ℎ𝑒𝑟𝑒 𝑢𝑖 𝑎𝑟𝑒 𝑖𝑛 ℝ} 4. 𝑈 = {( ⋮ 𝑢𝑛
1 • ⃗0 is in 𝑈 because if we let 𝑢1 = − 2 and the rest of the 1
2 (− 2) + 1 𝑢𝑖 = 0, we get (
0 ⋮ 0
)
0 0 =( ) ⋮ 0
2𝑥1 + 1 2𝑦1 + 1 𝑥 𝑦 • Suppose 𝑎, 𝑏 are scalars and ( ⋮2 ),( ⋮2 ) are in 𝑈. 𝑥𝑛 𝑦𝑛
2𝑥1 + 1 2𝑦1 + 1 𝑥2 𝑦2 Then 𝑎 ( )+𝑏( ) ⋮ ⋮ 𝑥𝑛 𝑦𝑛
2𝑎𝑥1 + 𝑎 + 2𝑏𝑦1 + 𝑏 2(𝑎𝑥1 + 𝑏𝑦1 ) + (𝑎 + 𝑏) 𝑎𝑥2 + 𝑏𝑦2 𝑎𝑥2 + 𝑏𝑦2 =( )=( ) ⋮ ⋮ 𝑎𝑥𝑛 + 𝑏𝑦𝑛 𝑎𝑥𝑛 + 𝑏𝑦𝑛
Since 𝑎𝑥𝑖 + 𝑏𝑦𝑖 is still in ℝ, if we let 𝑣𝑖 = 𝑎𝑥𝑖 + 𝑏𝑦𝑖 , we get 2𝑣1 + 𝑎 + 𝑏 𝑣2 that the vector is ( ), which is not of the form ⋮ 𝑣𝑛 2𝑢1 + 1 𝑢2 ( ) ⋮ 𝑢𝑛 Therefore, 𝑈 is NOT a subspace of ℝ𝑛
More Examples 𝑎 𝑏 Is 𝑈 = {( 𝑐 ) 𝑤ℎ𝑒𝑟𝑒 𝑎 − 𝑏 = 0 𝑎𝑛𝑑 𝑐 2 = 𝑑 2 } a subspace of ℝ4 ? 𝑑 0 0 • ( ) is in 𝑈 since 𝑎 − 𝑏 = 0 − 0 = 0, and 𝑐 2 = 0 = 𝑑 2 0 0 • *Instead of proving the axioms directly, we could have also used a counter example: 2 1 2 1 ( ) and ( ) are in 𝑈 (check this yourself). Use scalars 𝑎 = 𝑏 = 1. 3 1 3 1 2 1 3 2 1 3 Then, 1 ( ) + 1 ( ) = ( ). If this was in 𝑈, then it must 3 1 4 −3 1 −2 2 follow the property that 𝑐 = 𝑑 2. So, 42 = (−2)2 which is not true. Therefore, the sum of vectors is not in 𝑈 and 𝑈 is not a subspace of ℝ4 .
Recall: If 𝑇: ℝ𝑛 → ℝ𝑚 is a linear transformation, then there exists an 𝑚 × 𝑛 matrix 𝐴 that induces 𝑇 The nullspace or kernel of 𝑇 or 𝐴 is 𝑥1 𝑥1 𝑥2 𝑥2 𝑛𝑢𝑙𝑙 (𝑇 ) = {( ⋮ ) 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑇 ( ⋮ ) = ⃗⃗⃗⃗⃗ 0𝑚 } 𝑥𝑛 𝑥𝑛 or 𝑥1 𝑥1 𝑥2 𝑥2 𝑛𝑢𝑙𝑙 (𝑇 ) = {( ⋮ ) 𝑠𝑜 𝑡ℎ𝑎𝑡 𝐴 ( ⋮ ) = ⃗⃗⃗⃗⃗ 0𝑚 } 𝑥𝑛 𝑥𝑛
And the image of 𝑇 or 𝐴 is 𝑦1 𝑦1 𝑥1 𝑥1 𝑦2 𝑦2 𝑥2 𝑥2 𝑖𝑚(𝑇 ) = {( ⋮ ) 𝑠𝑜 𝑡ℎ𝑎𝑡 ( ⋮ ) = 𝑇 ( ⋮ ) 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 ( ⋮ ) } 𝑦𝑚 𝑦𝑚 𝑥𝑛 𝑥𝑛 Or 𝑦1 𝑦2
𝑦1 𝑦2
𝑥1 𝑥2
𝑥1 𝑥2
𝑦𝑚
𝑦𝑚
𝑥𝑛
𝑥𝑛
𝑖𝑚(𝑇 ) = {( ⋮ ) 𝑠𝑜 𝑡ℎ𝑎𝑡 ( ⋮ ) = 𝐴 ( ⋮ ) 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 ( ⋮ )}
Example Show that 𝑛𝑢𝑙𝑙(𝑇) is a subspace of ℝ𝑛 . ⃗⃗⃗⃗𝑛 is in 𝑛𝑢𝑙𝑙(𝑇) because 𝑇(0 ⃗⃗⃗⃗𝑛 ) = ⃗⃗⃗⃗⃗ • 0 0𝑚 by property of linear transformations • Suppose that ⃗⃗⃗⃗ u1 and ⃗⃗⃗⃗ u2 are in 𝑛𝑢𝑙𝑙(𝑇), 𝑎, 𝑏 are scalars So, 𝑇(⃗⃗⃗⃗ u1 ) = ⃗⃗⃗⃗⃗ 0𝑚 and 𝑇(⃗⃗⃗⃗ u2 ) = ⃗⃗⃗⃗⃗ 0𝑚 . Then, 𝑇(𝑎u ⃗⃗⃗⃗1 + 𝑏u ⃗⃗⃗⃗2 ) = 𝑇 (𝑎u ⃗⃗⃗⃗1 ) + 𝑇(𝑏u ⃗⃗⃗⃗2 ) = 𝑎𝑇(⃗⃗⃗⃗ u1 ) + 𝑏𝑇(⃗⃗⃗⃗ u2 ) ⃗⃗⃗⃗⃗𝑚 + 𝑏0 ⃗⃗⃗⃗⃗𝑚 = 𝑎0 = ⃗⃗⃗⃗⃗ 0𝑚 + ⃗⃗⃗⃗⃗ 0𝑚 = ⃗⃗⃗⃗⃗ 0𝑚 Hence, 𝑎u ⃗⃗⃗⃗1 + 𝑏u ⃗⃗⃗⃗2 is in 𝑛𝑢𝑙𝑙(𝑇). Therefore, 𝑛𝑢𝑙𝑙(𝑇) is a subspace of ℝ𝑛
Show that 𝑖𝑚(𝑇) is a subspace of ℝ𝑚 ⃗⃗⃗⃗𝑛 ) • ⃗⃗⃗⃗⃗ 0𝑚 is in 𝑖𝑚(𝑇) because ⃗⃗⃗⃗⃗ 0𝑚 = 𝑇(0 • Suppose that ⃗⃗⃗ v1 and 𝑣 ⃗⃗⃗⃗2 are in 𝑖𝑚(𝑇), 𝑎, 𝑏 are scalars So, ⃗⃗⃗ v1 = 𝑇(⃗⃗⃗⃗ u1 ) and ⃗⃗⃗⃗ v2 = 𝑇(⃗⃗⃗⃗ u2 ) for some ⃗⃗⃗⃗ u1 𝑎𝑛𝑑 ⃗⃗⃗⃗ u2 in 𝑛𝑢𝑙𝑙(𝑇) Then, 𝑎v⃗⃗⃗1 + 𝑏v ⃗⃗⃗⃗2 = 𝑎𝑇(⃗⃗⃗⃗ u1 ) + 𝑏𝑇 (⃗⃗⃗⃗ u2 ) = 𝑇(𝑎u ⃗⃗⃗⃗1 ) + 𝑇 (𝑏u ⃗⃗⃗⃗2 ) ⃗⃗⃗⃗2 ) = 𝑇(𝑎u ⃗⃗⃗⃗1 + 𝑏b Hence, 𝑎v ⃗⃗⃗1 + 𝑏v ⃗⃗⃗⃗2 is in 𝑖𝑚(𝑇). Therefore, 𝑖𝑚(𝑇) is a subspace of ℝ𝑚
More Examples Find the nullspace and image of the linear transformation −2 2 ]. induced by the matrix 𝐴 = [ −1 1 𝑥1 𝑥1 0 The nullspace is 𝑛𝑢𝑙𝑙 (𝑇) = {(𝑥 ) 𝑠𝑜 𝑡ℎ𝑎𝑡 𝐴 (𝑥 ) = ( )} 2 2 0 −2𝑥1 + 2𝑥2 −2 2 𝑥1 0 0 So, we need [ ] [𝑥 ] = [ ] [ ]=[ ] −𝑥1 + 𝑥2 −1 1 2 0 0 Equating each entry: −2𝑥1 + 2𝑥2 = 0 → 𝑥1 = 𝑥2 −𝑥1 + 𝑥2 = 0 𝑥 Therefore, the nullsapce is 𝑛𝑢𝑙𝑙 (𝑇 ) = {( ) 𝑤ℎ𝑒𝑟𝑒 𝑥 𝑖𝑛 ℝ} 𝑥 𝑦1 𝑦1 𝑥1 𝑥1 The image is 𝑖𝑚(𝑇) = {(𝑦 ) 𝑠𝑜 𝑡ℎ𝑎𝑡 (𝑦 ) = 𝐴 (𝑥 ) 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 (𝑥 )} 2 2 2 2 𝑦1 2(𝑥 − 𝑥1 ) −2𝑥1 + 2𝑥2 −2 2 𝑥1 So, [𝑦 ] = [ ] [𝑥 ] = [ ]=[ 2 ] −𝑥1 + 𝑥2 (𝑥2 − 𝑥1 ) 2 −1 1 2 𝑦1 2𝑢 If we let 𝑢 = 𝑥2 − 𝑥1 , then 𝑢 is in ℝ and [𝑦 ] = [ ] 2 𝑢 2𝑢 Therefore, the image is 𝑖𝑚(𝑇 ) = {[ ] 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑢 𝑖𝑛 ℝ} 𝑢
Suppose that 𝐴 is an 𝑛 × 𝑛 matrix and 𝜆 is a number. Recall: 𝜆 is an eigenvalue of 𝐴 if 𝐴x⃗ = 𝜆x⃗ for some non-zero ℝ𝑛 vector x⃗. In this case, the vector x⃗ is an eigenvector of 𝐴 corresponding to 𝜆 Let’s define 𝐸𝐴 (𝜆) = {x⃗ 𝑖𝑛 ℝ𝑛 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝐴x⃗ = 𝜆x⃗} = {x⃗ 𝑖𝑛 ℝ𝑛 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 (𝐴 − 𝜆𝐼 )x⃗ = ⃗0} = 𝑛𝑢𝑙𝑙 (𝐴 − 𝜆𝐼 ) Therefore, if 𝑛𝑢𝑙𝑙(𝐴 − 𝜆𝐼) contains more than just the zero vector ⃗0, then 𝜆 is an eigenvalue of 𝐴 and every vector in 𝑛𝑢𝑙𝑙(𝐴 − 𝜆𝐼 ) is a corresponding eigenvector. We say that 𝐸𝐴 (𝜆) is the eigenspace corresponding to 𝜆
