Question

1

Answer 1

\[\large \textbf{ b}_1 = \left( \begin{array} \\ r\\0\end{array}\right) ; \textbf{ b}_2 = \left( \begin{array} \\ 0\\s\end{array}\right) \]

Answer 2

One way to figure this out is by noticing that both vectors are pointing in DIFFERENT directions - \(\textbf{b}_1\) has no component along y direction but \(\textbf{b}_2\) has some nonzero component. So they're independent and thus span entire plane

Answer 3

The standard way is to find the rank of matrix with these vectors -
the vectors are independent if you get a full column rank

Answer 4

@ganeshie8 iy intuitively makes sense but not mathematically....i m stuck with solving it

Answer 5

\[\large \left[ \begin{array} \\ r&0\\0&s\end{array}\right] \]

whats the rank of this matrix ?

Answer 6

2?

Answer 7

and you have got 2 column vectors in the matrix - full column rank => column vectors are independent

Answer 8

independent vectors => they form a basis

Answer 9

honestly, today is my first day doing linearly independence so it does not make sense to me. Sorry about that

Answer 10

we say vectors are linearly independent when there is no solution for :
\[\large c_1\mathbb{x_1} + c_3\mathbb{x_2} + c_3\mathbb{x_3} + \cdots + c_n\mathbb{x_n} = 0 \]

Answer 11

i.e. all c's are 0

Answer 12

it always has a trivial solution when all c's are 0,
so the definition excludes this trivial solution

Answer 13

there should not be any OTHER solution to above ^^

Answer 14

then only the vectors are independent

Answer 15

thats the definition of independence,
does that make sense ?

Answer 16

yes it does, thank you but how do you calculate bases, span and stuff like that

Answer 17

those are just definitions :
1) the given set of vectors "span" a space means : you can can reach every point in the space using linear combinations of the given vectors

2) basis : if the set of vectors "span" the space and if they are independent, they form basis of that space

Answer 18

consider previous example to get a feeling of these terms

Answer 19

lemme get another example to make it clear, if you don't mind.

Answer 20

okie

Answer 21

b1 = (1,1,-2)^T
b2 = (1,-2,1)^T

Answer 22

and your first goal is to figure out whether they are independent or not, right ?

Answer 23

agree

Answer 24

Sticking to the definition, you need to find out wherther there is a nontrivial solution to below :
\[\large c_1 \mathbb{b_1} + c_2\mathbb{b_2} = 0\]

Answer 25

Ofcourse as you said earlier, a trivial solution always exists when all c's are 0
but we are not interested in this, we want to see if there is any OTHER solution

Answer 26

non-zero solution

Answer 27

uhhh how do we check that

Answer 28

\[\large c_1 \mathbb{b_1} + c_2\mathbb{b_2} = 0\]

can be expressed as

\[\large \left[\begin{array}{cc}1&1\\1&-2\\-2&1 \end{array}\right] \left(\begin{array} \\ c_1 \\c_2 \end{array}\right) = 0\]

Answer 29

yes ?

Answer 30

yep

Answer 31

solve \(c_1\) and \(c_2\)

Answer 32

below is more accurate representation :

\[\large \left[\begin{array}{cc}1&1\\1&-2\\-2&1 \end{array}\right] \left(\begin{array} \\ c_1 \\c_2 \end{array}\right) = \left(\begin{array} \\ 0 \\0 \end{array}\right) \]

Answer 33

c1=c2=0?

Answer 34

no c1=c2=1?

Answer 35

the column vectors will be `independent` if the `only solution` is c1=c2=0

Answer 36

how c1=c2=1 ?
add both the columns, do u get 0 ?

Answer 37

1+1 = ?
1-2 = ?
-2+1 = ?

Answer 38

2
-1
1

Answer 39

ya i solved it in a wrong way oops

Answer 40

and why are you guessing ? you can row reduce the matrix, find out number of pivots and find out the solution right ?

Answer 41

not guessing, just confused

Answer 42

\[\large \left[\begin{array}{cc}1&1\\1&-2\\-2&1 \end{array}\right] \left(\begin{array} \\ c_1 \\c_2 \end{array}\right) = \left(\begin{array} \\ 0 \\0 \\0 \end{array}\right)\]

Answer 43

do you know how to solve above system ?

Answer 44

c1+c2=0
c1-2c2=0
-2c1+c2=0
right

Answer 45

yes thats the system,
but i want you solve it in matrix form so that we can **see** the relation between "rank" and "independednce"

Answer 46

oh then I don't, i can solve it linearly not matrox system way

Answer 47

Ohk, never did gaussian elimination before ?

Answer 48

dont thin so

Answer 49

how about row operations ?

Answer 50

yah i have done that

Answer 51

echeleon form /triangular form ?
heard these before ?

Answer 52

no lol don't embarass me

Answer 53

you need to know below stuff before touching basis/span :

1) row operations/gaussian elimination
2) upper triangular form/echeleon form

Answer 54

the whole linear algebra is about solving linear equations using matrices,
you should know how to solve a system in matrix form first

Answer 55

okay, I guess my teacher will go over it soon

Answer 56

\[ \large \left[\begin{array}{cc}1&1\\1&-2\\-2&1 \end{array}\right] \left(\begin{array} \\ c_1 \\c_2 \end{array}\right) = \left(\begin{array} \\ 0 \\0 \\0 \end{array}\right) \]

Answer 57

it should be taught first, basis/independence/subspaces make no sense w/o knowing how to solve above system in matrix form

Answer 58

\[\large Ax = b\]

Answer 59

you must know how to find the solution \(\large x\) before starting subspaces

Answer 60

ya i know that Ax = b

Answer 61

\[ \large \left[\begin{array}{cc}1&1\\1&-2\\-2&1 \end{array}\right] \left(\begin{array} \\ c_1 \\c_2 \end{array}\right) = \left(\begin{array} \\ 0 \\0 \\0 \end{array}\right) \]

is same as

\[\large Ax = b\]

Answer 62

\[ \large A = \left[\begin{array}{cc}1&1\\1&-2\\-2&1 \end{array}\right] \\~\\\large x = \left(\begin{array} \\ c_1 \\c_2 \end{array}\right) \\~\\\large b = \left(\begin{array} \\ 0 \\0 \\0 \end{array}\right) \]

Answer 63

ohhhh lol should have told before
i know how to do it

Answer 64

i knw you knw :)
ok so tell me how do you go about solving \(Ax = b\) ?

Answer 65

using elimination ofcourse

Answer 66

ya wait it will take me a little as i am novice

Answer 67

c1 + c2 = 0
c1 - 2c2 = 0
-2c2 + c2 =0

1 2
1 -2
-2 1

is it correct or did i mess up?

Answer 68

\[\large \left[\begin{array}{cc}1&1\\1&-2\\-2&1 \end{array}\right] \left(\begin{array} \\ c_1 \\c_2 \end{array}\right) = \left(\begin{array} \\ 0 \\0 \\0 \end{array}\right)\]

the augmented matrix is

\[\large \left[\begin{array}{|cc|c}1&1&0\\1&-2&0\\-2&1&0 \end{array}\right] \]

Answer 69

again, we want to solve it in matrix form using elimination

Answer 70

oh crap i know i know but forgot need a startup

Answer 71

my teacher already went over this

Answer 72

the augmented matrix is

\[\large {\begin{align} \\ &\left[\begin{array}{|cc|c}1&1&0\\1&-2&0\\-2&1&0 \end{array}\right] \\ ~\\ R2-R1 ~&\left[\begin{array}{|cc|c}1&1&0\\0&-3&0\\-2&1&0 \end{array}\right]
\\ ~\\ R3+2R1 ~&\left[\begin{array}{|cc|c}1&1&0\\0&-3&0\\0 &3&0 \end{array}\right]
\\ ~\\ R3+R2 ~&\left[\begin{array}{|cc|c}1&1&0\\0&-3&0\\0 &0&0 \end{array}\right]
\end{align}} \]

Answer 73

so the solution is :

-3c2 = 0
c1 + c2 = 0

solving it gives you : c1=c2=0

Answer 74

thats the only solution ^^
that proves us that the column vectors of \(\large A\) are linearly independent

Answer 75

In general, for a \(\large m \times n\) matrix :

rank = \(\large n\) means the column vectors are linearly independent

Answer 76

So you don't need to actually solve the system - you can simply find the rank of matrix with given vectors

Answer 77

rank tells you about independence

Answer 78

whoooo too much work, i will still need to watch a video how you eliminated stuff as my teacher didnt explan it well....
Thanks a lot Ganesh :)

Answer 79

row elimination works the same way as elimination method with system of equations :

you can add linear combinations of other rows to a row

Answer 80

i think its too much information for you one shot

Answer 81

i know haha, but i am in college i gettaa take all info in or ill miss

Answer 82

atleast i hope you got what "independence" means and how it is realted to "rank" of a matrix

Answer 83

oh yes i did understand it, and the credit for that goes to you. Thank you

Answer 84

thats the only important thing,
basis and span are just definitions based on independence

Answer 85

"k" independent vectors can span a "k" dimensional space

Answer 86

basis is just a "set of independent vectors" with which u can "span" a particular space.

Answer 87

\[\large \left[\begin{array}{cc} 1&0 \\ 0 & 1 \end{array}\right] \]

the column vectors in this matrix form a "basis" for "2" dimensional vector space because :

1) the column vectors are independent
2) the column vectros span the 2 dimensional space

Answer 88

as you can see they are just definitions based on independent vectors

Answer 89

can you cook up an example set of vectors that CANNOT be a basis ?

Answer 90

1 1
1 1