Midterm 1 Flashcards
Linear Equation
○ a1x1 + a2x2 + … + anxn = b
organization of coefficients and variables with a solution ‘b’
System of linear equations
collection of multiple linear equations
solution of a system
list of numbers that make each equation true when values are substituted in for the corresponding ‘x’ variables
solution set
set of all possible solutions of a linear system
equivalent linear systems
systems with the same solution set
Consistent System
there is at least one solution (1 or infinite many)
Inconsistent System
there is no solution to the system for a specific input
they have EMPTY solution sets
existence and uniqueness questions
existence: does the solution/something exist?
uniqueness: if the solution exists, is there more than one solution (infinite many)?
How many solutions can a linear system have
none, one, or infinite many
size of a matrix
m x n
m: number of rows
n: number of columns
Row reduction operations
- take a multiple of one row and ADD that to another (used to eliminate entries)
- scale a row by a scalar not equal to zero (usually done to make a leading entry = 1)
- interchange rows
non-zero row/column
at least ONE of the entries has to be nonzero
leading entry
leftmost nonzero entry in a row
Conditions of Echelon Form
- leading entries of the next row are right of the leading entries in previous row above it
- everything in the column below a leading entry is all 0
- all zero rows are at the bottom
Row Reduced Echelon Form and Conditions!
represents a potential solution set for a linear system
EACH MATRIX ONLY HAS ONE RREF - row equivalent to just ONE
- all leading entries are 1s
- there are 0s ABOVE and BELOW each leading 1
if matrix is not echelon or RREF that means MORE row reductions must be done
pivot position and pivot columns
pivots are the locations in a matrix that correspond with leading 1s of RREF
pivot columns are the columns that have pivot positions
Basic/Leading Variables
variables that correspond with a pivot
basic variables have an EXACT value for a solution set
Free variables
don’t correspond to any pivots or pivot columns
can be assigned ANY value for a consistent linear system
Overdetermined System
More rows than columns
more equations than variables
- can be consistent
- can have a unique solution
– doesn’t necessarily have to be?
Underdetermined System
more columns than rows
more variables than equations
there will always be a free variable SO cannot have a unique solution
if consistent: infinite!
if inconsistent: no solution
Existence and Uniqueness Theorem
a linear system is consistent if and only if there is NOT a pivot in the last augmented column (rightmost column of augmented matrix)
● [0 0 0 0 0 | b] with b non-zero
if linear system is consistent:
1 solution (no free variables)
infinite (at least 1 free variable)
R^n
dimension we are in
n is the number of rows or entries in a vector
R is the collection of all lists if n real numbers
example:
○ R2 vector: 2 rows
○ R3 vector: 3 rows
vectors in R2 are a line to a point in 2D space
vectors in R3 are a line to a point in 3D space
Linear Combinations
Linear combinations is y = c1v1 + c2v2 + … + cpvp
where vs are a set of vectors in R^n and cs are weights or SCALARS
Y is a linear combination!
Span{v1 …vp}
Span is the set of all possible linear combinations of those sets of vectors
c1v1 + c2v2 + … + cpvp
Algebraic Properties of Rn (8 total)
○ For all u, v, w in Rn and all scalars c & d:
i. u + v = v + u v. c (u + v) = cu + cv
ii. (u + v) + w = u + (v + w) vi. (c + d) u = cu + du
iii. u + 0 = 0 + u = u vii. c (du) = (cd) u
iv. u + (-u) = -u + u = 0 viii. 1u = u
VECTOR EQUATIONS
vector equation: x1a1 + x2a2 + … + xnan = b where a and b are vectors
- has the same solution set as the linear system whose augmented matrix is [a1 a2 .. an | b]
- b can only be generated if there exists a solutions (weights ‘x’s) to the linear system with the matrix!
More about SPAN and connection to vector equation
If v1….vp are in Rn, then the set of all linear combinations of v1…vp is denoted by Span{v1…vp} this is called the subset of Rn spanned by v1….vp
is vector b in Span{v1…vp} is the same as asking x1v1+….+xpvp = b??
- solve the augmented matrix!
- is every b of Rn a linear combination of the vectors in {v1…vp}? == is there a pivot in every row
Identity Matrix
output the same input
● x ∈ Rn
“x is a vector with n elements”
Star Equation
matrix A of vectors times a mx1 matrix of weights
Ax =x1a1 + x2a2 + … xnan
the number of columns of A = number of entries in x
If A is an mxn matrix, with columns a1…an and if b is in Rm: CONNECT THE DIFFERENT FORMS
matrix equation = vector equation = augmented matrix for linear system
○ (Ax = b) == (x1a1 + x2a2 + … + xnan = b) == ([a1 a2 … an | b])
Span as a relation to VECTORS
Ax = b as a linear combination is basically the A vector times the x vector
Span of the columns is MULTIPLYING these two parts? (basically how do you get the linear combinations)
STAR EQUATION
span = all the possible linear combinations of vectors
x1a1 + x2a2 + … + xnan = b
Ax = b has a solution IF AND ONLY IF
b is a linear combination of the columns of A
LOGICALLY EQUIVALENT STATEMENTS (4 total)
-For each b in Rm, the equation Ax = b has a solution
- each b in Rm is a linear combination of the columns of A
- the columns of A span Rm
-A has a pivot position in every row
Properties of MATRIX equations
A(u +v) = Au +Av
A(cu) = c(Au)
Homogeneous linear system/Trvial vs nontrivial solution/Nonhomogenous linear system
Ax = 0
Trivial Solution: x = 0
NonTrivial Solution: x that is a nonzero vector that satisfies Ax = 0 == AT LEAST ONE nonzero element
nonhomogenous: Ax = b where b is not 0
Homogenous System
-ALWAYS ha at least one solution, always consistent, always has the trivial solution
-IF THERE is at least one free variable, homogeneous has a NONtrivial solution
Implicit vs Explicit description of a plane
Implicit: ○ 10x1 - 3x2 - 2x3 = 0
Explicit is PARAMETRIC VECTOR EQUATION
○ x = su + tv === ○ x = x2u + x3v
where x is the x vector and s,t are in R
free variables!
Parametric Vector FORMS
Consistent Ax = b
x = u + tv
Ax = 0
x = tv
For a consistent Ax = b, the solution set of Ax = b is the set of all vectors of the form w = p + vk where p is a solution and vk is any solution of Ax = 0
How to Write a solution set in Parametric Vector Form
- Row reduce to RREF
- express each basic variable in terms of any free variable
- write a typical solution x as a vector whose entries depend on the free variables
- decompose x into a linear combination of vectors using the free variables as parameters
Linearly Independent
Matrix A has a pivot in every column
NO free variables
x1v1 + x2v2 + … + xpvp = 0 HAS ONLY the trivial solution
Linearly dependent
Vector equation c1v1 + c2v2 + … + cpvp = 0 where not all the weights are not all zero, meaning AT LEAST ONE nonzero weight
at least ONE free variable
More columns than rows
Linearly dependent because there is going to be a free variable
likewise, IF a set of vectors is linearly independent, then number of rows will equal or be greater than the number of columns; this does not mean that if there are more rows then columns then a set of vectors is AUTOMATICALLY linearly independent
Sets with ONE vector is linearly independent
IF AND ONLY IF IT IS NOT THE ZERO VECTOR
sets with 2 or more vectors are linearly dependent IF
one of the vectors is a multiple of another OR one of the vectors is a linear combination OR in the span of at least two other vectors OR at least one of the vectors is a ZERO vector!!
Matrix Transformation
assigns a vector x in Rn to a vector T(x) in Rm
using A to turn x into T(x)
Linear Transformation
all matrix transformations are linear transformations
NOT ALL linear transformations are matrix transformations
A MATRIX TRANSFORMATION must preserve the operations of vector addition AND scalar multiplication
T(cu +dv) = cT(u) + dT(v)
Domain/CoDomain/Image of x under the action of T/Range of T
Domain: input: set Rn; number of entries in x
Codomain: output: set Rm; number of entries in T(x) - same number of entries in A or number of rows of A
Image of x under the action of T: vector T(x)
Range of T: set of all possible images T(x); span of columns of A??
Principle of superposition: ○ T(c1v1 + … + ckvk) = c1Tv1 + … + ckTvk
Solving Ax = b in the context of Transformations!
Trying to solve for all vectors x in Rn that are transformed into vector b in Rm under the action of multiplication by A
T(x) = rx
Contraction: 0 <= r < 1
Dilation: r >1
Standard Matrix for a linear transformation T
○ A = [T(e1) + … + T(en)]
e1 in R2 is (1 0)
e2 is R2 is (0 1)
ONTO
existence question!!
T: Rn -> Rm is ONTO Rm if each b in Rm is the image of AT LEAST one x in Rn
Ax = b is ALWAYS consistent
at least one solution
- standard matrix has a pivot in EVERY ROW
- columns of A SPANS Rm = means every b in Rm is going to have a solution, every b is a linear combination of the columns of A
Geometric: get to any vector with an image
ONE-TO-ONE
UNIQUENESS QUESTION
T: Rn -> Rm is ONE TO ONE if each b in Rm is the image of AT MOST one (possible 0) x in Rn
T(x) = Ax = b has either ONE solution or NO solution
pivot in EVERY column : no free variables!!
columns of A are LINEARLY INDEPENDENT
ONE TO ONE if and only if solution to T(x) = 0 is the trivial solution
if and only if EVERY COLUMN of A is pivotal
Cannot have multiple vectors that have the same image
Reflections through x1 axis, x2 axis, line x2 = x1, line x2 = -x1; the origin
1 0
0 -1
——-
-1 0
0 1
——-
0 1
1 0
———
0 -1
-1 0
———-
-1 0
0 -1
Contractions and Expansions
if k is between 0 and 1: contraction
k > 1: expansion
Horizontal contraction & expansion:
k 0
0 1
————
Vertical contraction & expansion:
1 0
0 k
Shears
Horizontal Shear:
1 k
0 1
Verticle Shear
1 0
k 1
if k is positive: right or up shear
if k is negative: left or down shear
Projections
on the x1 axis:
1 0
0 0
on the x2 axis:
0 0
0 1
Rotations COUNTERCLOCKWISE!!!!!!!!
cos(a) -sin(a)
sin (a) cos(a)
Properties of Matrix Addition (6 total)
a. A + B = B + A d. r (A + B) = rA + rB
b. (A + B) + C = A + (B + C) e. (r + s) A = rA + sA
c. A + 0 = A f. r (sA) = (rs) A
Matrix Multiplication
CAN ONLY MULTIPLE if the left matrix has the same number of columns as the number of rows of the RIGHT matrix
Properties of Matrix Multiplication (5 total)
a. A (BC) = (AB) C
b. A (B + C) = AB + AC
c. (B + C) A = BA + CA
d. r (AB) = (rA) B = A (rB)
e. ImA = A = AIn
matrices that commute
Matrices A and B commute if AB = BA
Transposing Matrices
take the first row of the matrix and make it the first column
and continue like that
WARNINGS when multiplying matrices
○ Order when multiplying matrices matters
■ In general, AB ≠ BA
○ AB = AC does not suggest B = C
○ If AB is the zero matrix, cannot conclude in general that either A = 0 or B = 0
Properties of Matrix Transposing (4 total)
a. (AT)T = A
b. (A + B)T = AT + BT
c. For any scalar r, (rA)T = rAT
d. (AB)T = BTAT
○ The transpose of a product of matrices equals the product of their transposes in the reverse order
Power of Matrices
only SQUARE matrices!!!!
Uniqueness of RREF Theorem
each matrix is row equivalent to one and only RREF
Theorem: Parametric Vector Form of a Nonhomogeneous System
● Suppose the equation Ax = b is consistent for some given b, and let p be a solution. Then the solution set of Ax = b is the set of all vectors of the form w = p + vh, where vh is any solution of the homogeneous equation Ax = 0.
Theorem: Characterization of Linearly Dependent Sets
indexed set S= {v1…vp} of 2 or more vectors are linearly dependent if and only if at least one of the vectors in S is a linear combination of the others; if S is linearly dependent and v1 != 0, then some vj is a linear combination of the preceding vectors
Theorem 8: Linear Dependence based on Matrix Size
More columns than rows means linear dependence
Theorem : Using the Standard Matrix to find Columns of A
For T: Rn -> Rm (linear transformation), there exists a unique matrix A such that T(x) = ax for all x in Rn
T(ej), where ej is the jth column of the identity matrix in Rn
A = [T(e1) … T(en)]
One-To-One use the Homogenous Equation Theorem
Let T: Rn -> Rm be a linear transformation. Then T is one-to-one if and only if the equation T(x) = 0 has only the trivial solution!!
