Applied Linear Algebra - Test 1

Front 1

If a set V is a vector space, then it satisfies 8 axioms and 2 operations are closed.

Back 1

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

Front 2

Closed Under Addition

Back 2

A set S is closed under addition if the sum of any two elements of S is in S.

Front 3

Closed Under Scalar Multiplication

Back 3

A set S is closed under scalar multiplication if the product of an arbitrary scalar and an arbitrary element of S is in S.

Front 4

A subset W of a vector space V is a subspace

Back 4

If W is closed under addition and scalar multiplication, then W is a subspace of V.

Front 5

V is a vector space, and B={b₁,b₂,..bn}. A linear combination of the elements of B

Back 5

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.

Front 6

Define the span of B={b₁,b₂,..bn}

Back 6

The Span(B) is the set of all linear combinations of the vectors in B.

Front 7

Define linear independence of the set B={b₁,b₂,..bn}.

Back 7

The set B is said to be linearly independent if a₁b₁+....anbn=0 implies a₁=a₂=...=an=0.

Front 8

A set B={b₁,b₂,..bn} with n>1 is linearly dependent iff

Back 8

one of the vectors bi can be written as a linear combination of the others

Front 9

If Span(B)=V, the set B={b₁,b₂,..bn} is said to

Back 9

span the vector space V.

Front 10

The set B={b₁,b₂,..bn} is a basis for the vector space V if

Back 10

the set B={b₁,b₂,..bn} is linearly independent and the set B={b₁,b₂,..bn} spans V.

Front 11

Let A be an n x m matrix. Then the following are equivalent:

1)Columns of A are linearly independent.

Back 11

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.

Front 12

Let A be an n x m matrix. Then the following are equivalent:

1)The columns of A span Rⁿ.

Back 12

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.

Front 13

Let A be an n x n matrix. Then the following are equivalent:
1) The columns of A are linearly independent.

Back 13

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.

Front 14

A matrix is invertible (or non-singular) if there exists

Back 14

an n x n matrix B s.t.
AB=BA=I

Front 15

Let D={d₁,d₂,..dm} be a collection of vectors in Rⁿ.

Back 15

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.

Hint 15

If m<n
If m>n
m=n

Front 16

Let V and W be vector spaces. An isomorphism between V and W is

Back 16

An isomorphism between V and W is a map L:V→W that is 1-1 and onto.

Front 17

If v=a₁b₁+....+anbn, the coordinate vector is

Back 17

[v]B=(a₁...an)

Front 18

If v=a₁b₁+....+anbn, the coordinate vector is [v]B=(a₁...an). The numbers a₁...an are

Back 18

The numbers a₁...an are called the coordinates of the vector v in the B basis.

Front 19

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

Back 19

[v]D=PDB[v]B

PDB(change of basis matrix)=([b₁]D....[bn]D)

Front 20

If D={d₁,d₂,..dn} and B={b₁,b₂,..bn} are bases for a vector space V,

Back 20

PDB=PBD^-1

(PDB)PBD is the identity

Front 21

If B, C and D are bases for the same vector space, then

Back 21

PBD=(PBC)PCD

Front 22

Given a basis B for Rⁿ and a vector v. Find PEB, PBE and [v]B.

Back 22

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

Front 23

A linear transormation from a vector space V to itself

Back 23

an operator

Front 24

L is a linear transformation from vector space V to vector space W if

L:V→W meaning ∀vɛV → L(v)ɛW

Back 24

1. L(v₁+v₂)=L(v₁)+L(v₂)
2. L(cv₁)=cL(v₁)

Front 25

Key property of linear transformations

Back 25

-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.

Front 26

Linear transformations: equation of L
-if v=a₁b₁+....+anbn

Back 26

then
L(v)=L(a₁b₁+....+anbn)
=L(a₁b₁)+....+L(anbn)
=a₁L(b₁)+....+anL(bn)

Front 27

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

Back 27

[Lv]D=[L]DB[v]B

where
[L]DB=([Lb₁]D....[Lbn]D)

Front 28

If L is a linear operator on a single space V, then [Lv]B=

Back 28

[Lv]B=[L]B[v]B

[Lv]B is the matrix of a linear transformation

Front 29

Let L be an operator on a vector space with alternate bases B and D. The matrix of L relative to D is

Back 29

[L]D=PDB[L]B{PBD}

Front 30

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

Back 30

[L]D'D=PD'B'[L]B'B(PBD)

Front 31

Let V and W be vector spaces and L:V→W
ker(L) is a subspace of

Back 31

V

Front 32

Let V and W be vector spaces and L:V→W
Range(L) is a subspace of

Back 32

W

Front 33

Let V and W be vector spaces and L:V→W.
ker(L) is

Back 33

the set of all vectors vɛV where L(v)=0

Front 34

Let V and W be vector spaces and L:V→W.
Range(L) is

Back 34

the set of all vectors of the form L(v) where vɛV

Front 35

A linear transformation is 1-1 iff

Back 35

kerl(L)={0}

Front 36

Suppose L is a L.T. from n-dim V to m-dim W.

If n<m, then

Back 36

L is not onto

Front 37

Suppose L is a L.T. from n-dim V to m-dim W.

If n>m, then

Back 37

L is not 1-1

Front 38

Suppose L is a L.T. from n-dim V to m-dim W.

If n=m, then

Back 38

L is either isomorphism(if rank L=n) or neither 1-1 nor onto (if rank less than n)

Front 39

The dimension of V equals

Back 39

dimension of kerl(L)+dimension of Range(L)

Front 40

The rank of a L.T. L is

Back 40

the dimension of Range(L)

Front 41

If λ is eigenvalue, set of έ's s.t L(έ)=λέ is

Back 41

called the eigenspace corresponding to λ, denoted Eλ

Front 42

If A nxn matrix, then eigenvectors and eigenvalues of A are defined

Back 42

to be eigenvalues and eigenvectors of the linear operator
L(x)=Ax

Front 43

Let V be a vec. space, L:V→V(linear operator)

If L(έ)=λέ

Back 43

έ is an eigenvector
λ is an eigenvalue of L