1. A basis for a vector space is a set of linearly independent vectors that span the entire space. The standard basis for Rn is the set of n unit vectors e1, e2, ..., en.
2. The dimension of a vector space is the number of vectors in any of its bases. A vector space with dimension n has bases that contain exactly n vectors.
3. The null space of a matrix A consists of all vectors x such that Ax = 0. The dimension of the null space is called the nullity of A, and the null space always has a basis.
ppt on Vector spaces (VCLA) by dhrumil patel and harshid panchalharshid panchal
this is the ppt on vector spaces of linear algebra and vector calculus (VCLA)
contents :
Real Vector Spaces
Sub Spaces
Linear combination
Linear independence
Span Of Set Of Vectors
Basis
Dimension
Row Space, Column Space, Null Space
Rank And Nullity
Coordinate and change of basis
this is made by dhrumil patel which is in chemical branch in ld college of engineering (2014-18)
i think he is the best ppt maker,dhrumil patel,harshid panchal
ppt on Vector spaces (VCLA) by dhrumil patel and harshid panchalharshid panchal
this is the ppt on vector spaces of linear algebra and vector calculus (VCLA)
contents :
Real Vector Spaces
Sub Spaces
Linear combination
Linear independence
Span Of Set Of Vectors
Basis
Dimension
Row Space, Column Space, Null Space
Rank And Nullity
Coordinate and change of basis
this is made by dhrumil patel which is in chemical branch in ld college of engineering (2014-18)
i think he is the best ppt maker,dhrumil patel,harshid panchal
Vector Space & Sub Space Presentation
Presented By: Sufian Mehmood Soomro
Department: (BS) Computer Science
Course Title: Linear Algebra
Shah Abdul Latif University Ghotki Campus
In general, we can find the coordinates of a vector u with respect to a given basis B by solving ABuB = u, for uB, where ABis the matrix whose columns are the vectors in B. ABis called thechange of basis matrix for B.
Vector Space & Sub Space Presentation
Presented By: Sufian Mehmood Soomro
Department: (BS) Computer Science
Course Title: Linear Algebra
Shah Abdul Latif University Ghotki Campus
In general, we can find the coordinates of a vector u with respect to a given basis B by solving ABuB = u, for uB, where ABis the matrix whose columns are the vectors in B. ABis called thechange of basis matrix for B.
Row space, column space, null space And Rank, Nullity and Rank-Nullity theore...Hemin Patel
If A is an m×n matrix, then the subspace of Rn spanned by the row vectors of A is
called the row space of A, and the subspace of Rm spanned by the column vectors
is called the column space of A. The solution space of the homogeneous system of
equation Ax=0, which is a subspace of Rn, is called the nullsapce of A.
Math for Intelligent Systems - 01 Linear Algebra 01 Vector SpacesAndres Mendez-Vazquez
These are the initial notes for a class I am preparing for this summer in the Mathematics of Intelligent Systems. we will start with the vectors spaces, their basis and dimensions. The, we will look at one the basic applications the linear regression.
Chapter 12
Section 12.1: Three-Dimensional Coordinate Systems
We locate a point on a number line as one coordinate, in the plane as an ordered pair, and in
space as an ordered triple. So we call number line as one dimensional, plane as two
dimensional, and space as three dimensional co – ordinate system.
In three dimensional, there is origin (0, 0, 0) and there are three axes – x -, y - , and z – axis. X –
and y – axes are horizontal and z – axis is vertical. These three axes divide the space into eight
equal parts, called the octants. In addition, these three axes divide the space into three
coordinate planes.
– The xy-plane contains the x- and y-axes. The equation is z = 0.
– The yz-plane contains the y- and z-axes. The equation is x = 0.
– The xz-plane contains the x- and z-axes. The equation is y = 0.
If P is any point in space, let:
– a be the (directed) distance from the yz-plane to P.
– b be the distance from the xz-plane to P.
– c be the distance from the xy-plane to P.
Then the point P by the ordered triple of real numbers (a, b, c), where a, b, and c are the
coordinates of P.
– a is the x-coordinate.
– b is the y-coordinate.
– c is the z-coordinate.
– Thus, to locate a point (a, b, c) in space, start from the origin (0, 0, 0) and move a
units along the x-axis. Then, move b units parallel to the y-axis. Finally, move c
units parallel to the z-axis.
The three dimensional Cartesian co – ordinate system follows the right hand rule.
Examples:
Plot the points (2,3,4), (2, -3, 4), (-2, -3, 4), (2, -3, -4), and (-2, -3, -4).
The Cartesian product x x = {(x, y, z) | x, y, z in } is the set of all ordered triples of
real numbers and is denoted by 3 .
Note:
1. In 2 – dimension, an equation in x and y represents a curve in the plane 2 . In 3 –
dimension, an equation in x, y, and z represents a surface in space 3 .
2. When we see an equation, we must understand from the context that it is a curve in the
plane or a surface in space. For example, y = 5 is a line in 2 �but it is a plane in 3 �
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3. in space, if k, l, & m are constants, then
– x = k represents a plane parallel to the yz-plane ( a vertical plane).
– y = k is a plane parallel to the xz-plane ( a vertical plane).
– z = k is a plane parallel to the xy-plane ( a horizontal plane).
– x = k & y = l is a line.
– x = k & z = m is a line.
– y = l & z = m is a line.
– x = k, y = l and z = m is a point.
Examples: Describe and sketch y = x in 3
Example:
Solve:
Which of the points P(6, 2, 3), Q(-5, -1, 4), and R(0, 3, 8) is closest to the xz – plane? Which point
lies in the yz – plane?
Distance between two points in space:
We simply extend the formula from 2 to . 3 . The distance |p1 p2 | between the points
P1(x1,y1, z1) and P2(x2, y2, z2) is: 2 2 21 2 2 1 ...
Linear Algebra and Differential Equations by Pearson - Chapter 4Chaimae Baroudi
Chapter 4 Vector Spaces, define vectors and their properties, an introduction to vector spaces, subspaces and subsets, in addition to Linear Combinations, Bases and Dimensions for Vector Spaces. It also shows how to find Row, Column and Null Spaces for a matrix.
Linear Algebra and Differential Equations by Pearson - Chapter 3Chaimae Baroudi
Chapter 3 Linear Systems and Matrices, evolving Introduction to Matrices, Writing a System of Equations as An Augmented-coefficients matrix and converting it to echelon form by Gaussian Elimination, and to reduced-echelon form by Gauss-Jordan Elimination. It also includes Matrices Operations and Introduction to Determinants and their importance in finding Inverses of Matrices and Solving Linear Systems.
Linear Algebra and Differential Equations by Pearson - Chapter 1Chaimae Baroudi
Chapter 1 First Order Differential Equations, evolving General and Particular Solutions to ODE, Slope Fields and solving Linear, Separable, Homogeneous, Bernoulli and Exact Equations.
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Model Attribute Check Company Auto PropertyCeline George
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
Francesca Gottschalk - How can education support child empowerment.pptxEduSkills OECD
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
2. Basis
Defn - A set of vectors S = { v1, v2, , vk } in
a vector space V is called a basis for V
if and only if
S is
a) linearly independent
and
a) S spans V.
3. Standard (natural) basis for R2 is
R2=Span
Also R2=Span
Therefore, is another basis for R2
Example of a basis for R2
1 2
1 0
S , ,
0 1
e e
𝐞 , 𝐞
2 1
,
1 3
2 1
,
1 3
4. Standard (natural) basis for R3 is
R3=Span
1 2 3
1 0 0
S 0 , 1 , 0 , ,
0 0 1
e e e
Example of a basis for R3
1 2 3, ,e e e
5. The following three-dimensional vectors are linearly
independent.
Any vector in R3 can be expressed as a linear
combination of v1, v2 and v3.
v1, v2 and v3 form a basis of R3.
Span {v1, v2 , v3} = R3.
1 2 3
1 1 1
2 , 0 , 1
1 2 0
v v v
Example of a basis for R3
6. The standard (natural) basis for Rn is
denoted by e1, e2, , en where
0
0
th row1
0
0
i ie
Example of a basis for Rn
7. Theorem - If S = { v1, v2, , vn } is a basis
for a vector space V, then every vector in V
can be written in one and only one way as a
linear combination of the vectors in S
Theorem - If S = { v1, v2, , vn } is a set of
vectors spanning a vector space V, then S
contains a basis T for V
Basis
8. Theorem - If S = { v1, v2, , vn } is a basis for
a vector space V and T = { w1, w2, , wr } is
a linearly independent set of vectors in V, then
r ≤ n.
Corollary - If S = { v1, v2, , vn } and
T = { w1, w2, , wm } are bases for a
vector space V, then n = m,
i.e.
every basis for V contains the same number of
vectors.
Basis
9. Defn - The dimension of a nonzero vector
space V is the number of vectors in a basis
for V.
Notation is dim V.
The dimension of the trivial vector space
{ 0 } is defined as 0.
Dimension of a Vector Space
10. Corollary - If a vector space V has dimension
n, then a largest linearly independent subset of
vectors in V contains n vectors and is a basis
for V
Corollary - If a vector space V has dimension
n, the smallest set of vectors that spans V
contains n vectors and is a basis for V.
Basis and Dimension
11. Corollary - If vector space V has dimension n,
then any subset of m > n vectors must be
linearly dependent.
Corollary - If vector space V has dimension n,
then any subset of m < n vectors cannot span
V.
Theorem - If S is a linearly independent set of
vectors in a finite dimensional vector space V,
then there is a basis T for V, which contains S.
Basis and Dimension
12. Theorem - Let V be an n-dimensional vector
space
a) If S = { v1, v2, , vn } is a linearly
independent set of vectors in V, then S is a
basis for V.
b) If S = { v1, v2, , vn } spans V, then S is a
basis for V.
Basis and Dimension
13. Consider the homogeneous system Ax = 0 where
A reduces to
Basis for Solution Sets : Homogeneous
1 2 2 2 3 0
4 6 4 6 6 4
1 3 4 2 4 0
2 4 4 3 4 2
A
1 0 2 0 3 4
0 1 2 0 1 0
0 0 0 1 2 2
0 0 0 0 0 0
14. Corresponding system of equations is
51 3 6
52 3
54 6
2 3 4 0
2 0
2 2 0
x x x x
x x x
x x x
1
2
3
4
5
6
2 3 4 2 3 4
2 2 1 0
1 0 0
2 2 0 2 2
0 1 0
0 0 1
x r s t
x r s
x r
r s t
x s t
x s
x t
Let x6 = t, x5 = s, x3 = r
Then x4 = –2s + 2t,
x2 = –2r – s,
x1 = 2r + 3s – 4t
Basis for Solution Sets : Homogeneous
15. The null space of A is spanned by the independent
set of 3 vectors (below) and thus has dimension 3.
Dimension of null space is called the nullity of A.
2 3 4
2 1 0
1 0 0
, ,
0 2 2
0 1 0
0 0 1
Basis for Solution Sets : Homogeneous
16. Consider the linear system
The solution consists of all vectors of the form
for arbitrary r and s
Basis for Solution Sets : Nonhomogeneous
1
2
3
1 2 1 3
2 4 2 6
3 6 3 9
x
x
x
1
2
3
1 2 1 2 1
1 1 1 0
2 2 0 1
x r s
x r r s
x s
17. The set of linear combinations, ,
forms a plane
passing through the origin of R3.
This plan is a linear space.
Basis for Solution Sets
2 1
1 0
0 1
r s
18. The solution vectors, ,
form a plane parallel to the plane above,
displaced from the origin by the vector
This plan is not a linear space.
Basis for Solution Sets
1 2 1
1 1 0
2 0 1
r s
1
1
2