# Column space and kernel

• beth87

## Homework Statement

If col (A) is column space of A and ker(A) null space of A
with ker(A) = {Ax = 0}
and ker(A') = {A'y = 0}

## Homework Equations

Consider the (3x2) matrix :

A = [1,2 ; 3,4 ; 5,6] (matlab syntax)

Show that

col(A) = c1 * [1,0,-1]' + c2 * [0,1,2]'

## The Attempt at a Solution

Find c1 and c2, starting from Col(A) definition.

---

We tried in this way :

Col(A) = [c1, c2, -c1+c2]'

Ax=Col(A)
Then:
[1,2 ; 3,4 ; 5,6]*[x1 ; x2]=[c1 ; c2 ; -c1+c2]

rank(A)=2
rank(A|Col(A))=2
and the system is soluble.

With Cramer I have:
x1=c2-2c1
x2=-1/2*(c2-3c1)

Ax=0 for ker(A)
and i do the system
1st eq: c2-2c1=0
2nd eq: c2-3c1=0

But here the solution is meaningless.

How can I solve this exercise?

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I didn't see any definition of Col(A), and I don't see what ker has to do with anything. The columns of your matrix are just [1,3,5] and [2,4,6]. So I would say that the definition of col(A) is a1*[1,3,5]+a2*[2,4,6]. The exercise is to show that space is the same as c1*[1,0,-1]+c2*[0,1,2]. How about finding the value of c1 and c2 corresponding to [1,3,5] and [2,4,6], would that help?

I didn't see any definition of Col(A), and I don't see what ker has to do with anything. The columns of your matrix are just [1,3,5] and [2,4,6]. So I would say that the definition of col(A) is a1*[1,3,5]+a2*[2,4,6]. The exercise is to show that space is the same as c1*[1,0,-1]+c2*[0,1,2]. How about finding the value of c1 and c2 corresponding to [1,3,5] and [2,4,6], would that help?

It is similar to my idea but how could i solve a1*[1,3,5]'+a2*[2,4,6]'=c1*[1,0,-1]'+c2*[0,1,2]'... expecially because neither c1,c2 and a1,a2 are known.

Solve [1,3,5]=c1*[1,0,-1]'+c2*[0,1,2]'. Why are you putting primes (') on things? Now do the same thing for [2,4,6].

Solve [1,3,5]=c1*[1,0,-1]'+c2*[0,1,2]'. Why are you putting primes (') on things? Now do the same thing for [2,4,6].

" ' " is the transposition from row to column vector.
So those are column vectors. We are trying.. thanks we'll write you soon

I tried this:
[1 ; 3 ; 5] + [2 ; 4 ; 6] = c1*[1 ; 0 ; -1] + c2*[0 ; 1 ; 2]
and the solutions are
c1=3 & c2=7 (confirmed by -c1+2c2=11)

could it be right?

I tried this:
[1 ; 3 ; 5] + [2 ; 4 ; 6] = c1*[1 ; 0 ; -1] + c2*[0 ; 1 ; 2]
and the solutions are
c1=3 & c2=7 (confirmed by -c1+2c2=11)

could it be right?

No. Please turn off MATLAB and think about the problem. You have two spaces A=a1*[1,3,5]+a2*[2,4,6] and C=c1*[1,0,-1]+c2*[0,1,2]. You want to show they are the same. If you could show that [1,3,5] is in C and [2,4,6] is in C, wouldn't that show that a1*[1,3,5]+a2*[2,4,6] is in C?? So that makes A a subset of C. Now can you show C is a subset of A?? If so then that makes C=A. That's a bit more work than you actually have to do, if you know something about linear independence and dimension you really don't have to go both ways.

No. Please turn off MATLAB and think about the problem. You have two spaces A=a1*[1,3,5]+a2*[2,4,6] and C=c1*[1,0,-1]+c2*[0,1,2]. You want to show they are the same. If you could show that [1,3,5] is in C and [2,4,6] is in C, wouldn't that show that a1*[1,3,5]+a2*[2,4,6] is in C?? So that makes A a subset of C. Now can you show C is a subset of A?? If so then that makes C=A. That's a bit more work than you actually have to do, if you know something about linear independence and dimension you really don't have to go both ways.

mmm... I think I give up thanks anyway for your help

mmm... I think I give up thanks anyway for your help

but we studied linear indipendence

You should still show that [1,3,5] and [2,4,6] are linear combinations of [1,0,-1] and [0,1,2]. If you know that the two pairs are linearly independent then you don't have to check the reverse.

You should still show that [1,3,5] and [2,4,6] are linear combinations of [1,0,-1] and [0,1,2]. If you know that the two pairs are linearly independent then you don't have to check the reverse.

[1,3,5] linear combination of [1,0,-1] ? ... mmmm

well [1,3,5]' = c1*[1 ; 0 ; -1]' + c2*[0 ; 1 ; 2]'
makes c1 =1 c2= 3 and 2c2-c1 = 5

but [2 ; 4 ; 6]' = c1*[1 ; 0 ; -1]' + c2*[0 ; 1 ; 2]'
makes c1=2 c2=4 and 2c2-c1=6

so which are the right values of c1 and c2??

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## Homework Statement

If col (A) is column space of A and ker(A) null space of A
with ker(A) = {Ax = 0}
and ker(A') = {A'y = 0}

## Homework Equations

Consider the (3x2) matrix :

A = [1,2 ; 3,4 ; 5,6] (matlab syntax)

Show that

col(A) = c1 * [1,0,-1]' + c2 * [0,1,2]'

Find c1 and c2, starting from Col(A) definition.

## The Attempt at a Solution

------

We tried in this way :

Col(A) = [c1, c2, -c1+c2]'

Ax=Col(A)
Then:
[1,2 ; 3,4 ; 5,6]*[x1 ; x2]=[c1 ; c2 ; -c1+c2]

rank(A)=2
rank(A|Col(A))=2
and the system is soluble.

With Cramer I have:
x1=c2-2c1
x2=-1/2*(c2-3c1)

Ax=0 for ker(A)
and i do the system
1st eq: c2-2c1=0
2nd eq: c2-3c1=0

But here the solution is meaningless.

How can I solve this exercise?

this thing
seems impossible...
consider that we aren't doing this for a maths course but econometrics and we have never done those things in maths before... so if someone could please help us more... we have been thinking about many hours and we aren't sure what we should do about this...

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well [1,3,5]' = c1*[1 ; 0 ; -1]' + c2*[0 ; 1 ; 2]'
makes c1 =1 c2= 3 and 2c2-c1 = 5

but [2 ; 4 ; 6]' = c1*[1 ; 0 ; -1]' + c2*[0 ; 1 ; 2]'
makes c1=2 c2=4 and 2c2-c1=6

so which are the right values of c1 and c2??
[1, 3, 5] and [2, 4, 6] are two different vectors. The are different linear combinations of [1, 0, -1] and [0, 1, 2]. c1= 1, c2= 3 for [1, 3, 5], while c1= 2, c2= 4 for [2, 4, 6].

That's it. Your're done! You have shown that [1, 3, 5] and [2, 4, 6] (and so any linear combination of them) can be written as linear combinations of [1, 0, -1] and [0, 1, 2].

so the question of the exercise was which are the values of c1 and c2 so the values of c1 are 1 or 2 and the values of c2 are 3 or 4?

or maybe it's like this [1,3,5]'+[2,4,6]'=c1[1,0,-1]'+c2[0,1,2]'
sum of coloumn vectors and it makes c1=3 and c2=7

You said the problem was
Show that

col(A) = c1 * [1,0,-1]' + c2 * [0,1,2]'

3. The Attempt at a Solution

Find c1 and c2, starting from Col(A) definition.
That makes no sense at all. The left hand side of your equation, "col(A)", is a vector space (the column space of A), not a vector, but the right hand side is a single vector. They are not equal for any c1 or c2!

sorry starting from space of column definition

also instead of c in the real exercise there's y upside down λ

but if you say that that before was the right solution I trust you...
Thanks a lot to everyone :-D I'll let you know

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