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43 Cards in this Set
- Front
- Back
- 3rd side (hint)
If a set V is a vector space, then it satisfies 8 axioms and 2 operations are closed.
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1. Commutativity of addition x+y=y+x
2. Associativity of addition (x+y)+z=x+(y+z) 3. Additive identity 0+x=x 4. Additive inverse x+(-x)=0 5. 1st distributive law a(x+y)=ax+ay 6. 2nd distributive law (a+b)x=ax+bx 7. Multiplicative identity 1(x)=x 8. ordinary multiplication (ab)x=a(bx) -closed under addition -closed under scalar multiplication |
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Closed Under Addition
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A set S is closed under addition if the sum of any two elements of S is in S.
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Closed Under Scalar Multiplication
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A set S is closed under scalar multiplication if the product of an arbitrary scalar and an arbitrary element of S is in S.
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A subset W of a vector space V is a subspace
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If W is closed under addition and scalar multiplication, then W is a subspace of V.
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V is a vector space, and B={b₁,b₂,..bn}. A linear combination of the elements of B
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A linear combination of the elements of B is any vector of the form
v=a₁b₁+a₂b₂....+anbn where coeffiients a₁a₂...an are arbitrary scalars. |
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Define the span of B={b₁,b₂,..bn}
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The Span(B) is the set of all linear combinations of the vectors in B.
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Define linear independence of the set B={b₁,b₂,..bn}.
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The set B is said to be linearly independent if a₁b₁+....anbn=0 implies a₁=a₂=...=an=0.
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A set B={b₁,b₂,..bn} with n>1 is linearly dependent iff
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one of the vectors bi can be written as a linear combination of the others
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If Span(B)=V, the set B={b₁,b₂,..bn} is said to
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span the vector space V.
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The set B={b₁,b₂,..bn} is a basis for the vector space V if
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the set B={b₁,b₂,..bn} is linearly independent and the set B={b₁,b₂,..bn} spans V.
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Let A be an n x m matrix. Then the following are equivalent:
1)Columns of A are linearly independent. |
2)The only solution to Ax=0 is x=0.
3)The reduced row-echelon form of the matrix A has a pivot in each column. |
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Let A be an n x m matrix. Then the following are equivalent:
1)The columns of A span Rⁿ. |
2)The equation Ax=b has a solution for every bɛRⁿ.
3)The reduced row-echelon form of the matrix A has a pivot in each row. |
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Let A be an n x n matrix. Then the following are equivalent:
1) The columns of A are linearly independent. |
2) The columns of A span Rⁿ.
3) The columns of A form a basis for Rⁿ. 4) The equation Ax=b has a unique solution for bɛRⁿ. 5) A is an invertible matrix. 6) The determinant of A is nonzero. 7) A is row equivalent to the identity matrix. |
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A matrix is invertible (or non-singular) if there exists
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an n x n matrix B s.t.
AB=BA=I |
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Let D={d₁,d₂,..dm} be a collection of vectors in Rⁿ.
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1) If m<n, then D does not span Rⁿ.
2) If m>n, then D is linearly dependent. 3) If D is a basis for Rⁿ, then m=n. |
If m<n
If m>n m=n |
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Let V and W be vector spaces. An isomorphism between V and W is
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An isomorphism between V and W is a map L:V→W that is 1-1 and onto.
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If v=a₁b₁+....+anbn, the coordinate vector is
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[v]B=(a₁...an)
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If v=a₁b₁+....+anbn, the coordinate vector is [v]B=(a₁...an). The numbers a₁...an are
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The numbers a₁...an are called the coordinates of the vector v in the B basis.
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D={d₁,d₂,..dn} and B={b₁,b₂,..bn} are two bases for vector space V. The two vectors [v]B and [v]D are related by
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[v]D=PDB[v]B
PDB(change of basis matrix)=([b₁]D....[bn]D) |
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If D={d₁,d₂,..dn} and B={b₁,b₂,..bn} are bases for a vector space V,
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PDB=PBD^-1
(PDB)PBD is the identity |
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If B, C and D are bases for the same vector space, then
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PBD=(PBC)PCD
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Given a basis B for Rⁿ and a vector v. Find PEB, PBE and [v]B.
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1. Find [v]E={[v₁]E....[vn]E}
2. Find PEB={[b₁]E....[bn]E} 3. Find PBE=PEB^-1 4. Find [v]B=PBE[v]E |
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A linear transormation from a vector space V to itself
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an operator
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L is a linear transformation from vector space V to vector space W if
L:V→W meaning ∀vɛV → L(v)ɛW |
1. L(v₁+v₂)=L(v₁)+L(v₂)
2. L(cv₁)=cL(v₁) |
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Key property of linear transformations
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-they are determined by their action on a basis
-if B is a basis for V, then L(bi) for each bi shows what L does to a vector in V. |
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Linear transformations: equation of L
-if v=a₁b₁+....+anbn |
then
L(v)=L(a₁b₁+....+anbn) =L(a₁b₁)+....+L(anbn) =a₁L(b₁)+....+anL(bn) |
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Let D={d₁,d₂,..dn} and B={b₁,b₂,..bn} be bases for vector spaces V and W and L:V→W. Then there exists [L]DB
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[Lv]D=[L]DB[v]B
where [L]DB=([Lb₁]D....[Lbn]D) |
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If L is a linear operator on a single space V, then [Lv]B=
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[Lv]B=[L]B[v]B
[Lv]B is the matrix of a linear transformation |
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Let L be an operator on a vector space with alternate bases B and D. The matrix of L relative to D is
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[L]D=PDB[L]B{PBD}
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Let V be a vector space with bases B and D, W a vector space with bases B' and D', and let L:V→W. The matrix of L relative to D and D' bases
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[L]D'D=PD'B'[L]B'B(PBD)
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Let V and W be vector spaces and L:V→W
ker(L) is a subspace of |
V
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Let V and W be vector spaces and L:V→W
Range(L) is a subspace of |
W
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Let V and W be vector spaces and L:V→W.
ker(L) is |
the set of all vectors vɛV where L(v)=0
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Let V and W be vector spaces and L:V→W.
Range(L) is |
the set of all vectors of the form L(v) where vɛV
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A linear transformation is 1-1 iff
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kerl(L)={0}
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Suppose L is a L.T. from n-dim V to m-dim W.
If n<m, then |
L is not onto
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Suppose L is a L.T. from n-dim V to m-dim W.
If n>m, then |
L is not 1-1
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Suppose L is a L.T. from n-dim V to m-dim W.
If n=m, then |
L is either isomorphism(if rank L=n) or neither 1-1 nor onto (if rank less than n)
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The dimension of V equals
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dimension of kerl(L)+dimension of Range(L)
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The rank of a L.T. L is
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the dimension of Range(L)
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If λ is eigenvalue, set of έ's s.t L(έ)=λέ is
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called the eigenspace corresponding to λ, denoted Eλ
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If A nxn matrix, then eigenvectors and eigenvalues of A are defined
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to be eigenvalues and eigenvectors of the linear operator
L(x)=Ax |
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Let V be a vec. space, L:V→V(linear operator)
If L(έ)=λέ |
έ is an eigenvector
λ is an eigenvalue of L |
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