Question

Let V be a vector space and ℓ : V → R be a linear map. If z ∈ V is not in the
nullspace of ℓ, show that every x ∈ V can be decomposed uniquely as x = v + cz,
where v is in the nullspace of ℓ and c is a scalar.

Answer 1

its some crazy stuff though.

Answer 2

is it interesting stuff

Answer 3

yea i like it more than calculus

Answer 4

ohh really!! i want to learn then

Answer 5

why do u like it more than calculus?

Answer 6

there exists weird things in algebra...but calculus from my experience is mostly alot of methodology...more ambiguity in linear algebra.

Answer 7

like why is there called an empty set which contains nothing..

Answer 8

I don't see questions like these often :D

Answer 9

does that excite you?

Answer 10

it excites me

Answer 11

a little :)

Answer 12

Well, @dan815 after you :)

Answer 13

i am excited for different reasons than u

Answer 14

iam excited because i dont understand these words haHahaha

Answer 15

hahaha. dan atm has more of an idea how cows eat grass than this question right now

Answer 16

Oh, make no mistake, I DO NOT know the answer here... (yet) but I'm working on it :)

Answer 17

I suppose we start with the essentials. Could I use f instead of ℓ ?
It's easier to type it that way :)

Answer 18

yea k.

Answer 19

okay, the usuals...
Let V be a vector space, and f: V -> R be a linear map.
Let z be an element of v such that f(v) is not equal to 0

Answer 20

And now, let x be an element of V.

Answer 21

yepyepyep

Answer 22

That's a HARD one!

Answer 23

AND UR HERE TO HELP :) haha

Answer 24

Haha Yes , Yes I am :D

Answer 25

goodie !

Answer 26

Thankfully, the co-domain is just the set of reals. I think we can do this now... :D

Answer 27

Im still tryna figure it out here.. HM.......

Answer 28

Why don't we get rid of the annoying trivial case of when x itself is in the kernel (nullspace) of f. Then it follows trivially, just set v = x and c = 0

Answer 29

alright...mmm. i've got till 11:30pm so im 1hour

Answer 30

Great!

Answer 31

relax :P
Now what if x is not in the (bloody) nullspace ?

Well then, we can set
\[\Large c = \frac{f(x)}{f(z)}\]

Answer 32

haha

Answer 33

Which exists, because f(z) is not in the nullspace (f(z) is not zero)

Answer 34

\[\Large\qquad \ \ f(x) = cf(z)\]\[\Large \implies f(x)= f(cz)\]

Answer 35

oH I WAS AT THE WROND dESTINATION^

Answer 36

(remembering of course that f is a linear transformation)

Answer 37

Okay, that said, we set
\[\Large v = x - cz\]

Answer 38

Let's TRANSFORM both sides...
\[\Large f(v) = f(x-cz) = f(x)-f(cz) = 0\]

Answer 39

i'm with ya

Answer 40

me too

Answer 41

And you have
\[\Large f(v) = 0\]
implying, of course, that v is in the nullspace of f.

Answer 42

haha zofia. let terenz do all the talking :)

Answer 43

okk

Answer 44

-.-
Anyway,
\[\Large v = x - cz\]
Then
\[\Large x = v + cz\]

Successfully decomposing x as desired.
Now, maybe you can show uniqueness?

Answer 45

oK cHRIS, sowwie!

Answer 46

nono. i didn't mean it like that! like ur learning too:)

Answer 47

so we are both students:) haha yea ok.

Answer 48

riiigghht! :D

Answer 49

uhh, guys? focus? :D
Showing uniqueness... any idea guys? :)

Answer 50

um. let me think

Answer 51

Here, let me start you out :)
Suppose we have two such decompositions of x...
\[\Large x = v_1 + c_1z\]\[\Large x = v_2 + c_2z\]

You have to show that in fact,
\[\Large v_1 =v_2\]and\[\Large c_1=c_2\]

Answer 52

equality of both sides?

Answer 53

Equality, specifically, of v1 and v2
as well as c1 and c2.

Answer 54

yea

Answer 55

Where both \(v_1\) and \(v_2\) are in the nullspace of f
and \(c_1\) and \(c_2\) are scalars

Answer 56

Stuck? :)

Answer 57

u read mi mind

Answer 58

You could start with this... remembering that since both expressions are equal to x, then you have
\[\Large v_1 + c_1z = v_2 + c_2z\]
Could you work from there?

Answer 59

that wat i meant by equality of both sides...

Answer 60

so c1=c2 from z:

Answer 61

I get it !

Answer 62

I'm afraid it's not that simple, as you haven't established equality between c1 and c2 yet :P

Answer 63

o poo

Answer 64

i forgot abotu that

Answer 65

@Zofia1 yes? :)

Answer 66

Yes.

Answer 67

I meant, you know how to do it?

Answer 68

Im doing on paper and here wait.

Answer 69

No rush.

Answer 70

you group c1z an c2z on one side?

Answer 71

Yup :)
\[\Large v_1 - v_2 = c_2z - c_1z\]

Answer 72

and factor?

Answer 73

Very intuitive :)
\[\Large v_1 - v_2 = (c_2-c_1)z\]

Answer 74

then we estalibished f(v)=0?

Answer 75

hmm... almost, not quite :)

Answer 76

Remember, we have a linear transformation f, where both v1 and v2 are elements of its nullspace and z specifically isn't...

Answer 77

does this mean v1=0 and v2=0? because they are in the nullspace

Answer 78

Don't be in such a rush :D
Let's apply the linear transformation f on both sides, shall we?
\[\Large f(v_1 - v_2 )= f[(c_2-c_1)z]\]

Answer 79

Now, on the left, f being a linear transformation, we can have...
\[\Large f(v_1) - f(v_2 )= f[(c_2-c_1)z]\]
With me so far?

Answer 80

yea. i get that.

Answer 81

On the right, since c2 - c1 is just a scalar, likewise, we can simplify into...
\[\Large f(v_1) - f(v_2 )= (c_2-c_1)f(z)\]

Answer 82

cant we do that to the right hand side?

Answer 83

o isee

Answer 84

c2 - c1 is a scalar...

Answer 85

OH!

Answer 86

Okay, ready to proceed?
Or can you demonstrate it from here?

Answer 87

I MEAN OH!

Answer 88

hmm?

Answer 89

left hand side is 0 because we said f(v)=0 right?

Answer 90

Left hand side is zero because BOTH \(\large f(v_1)\) and \(\large f(v_2)\) are zero, as both v1 and v2 are in the nullspace

Answer 91

now f(z) is not in the nullspace so that means c2-c1 must only equal zero.

Answer 92

for this to occur c2=c1

Answer 93

Yes continue... (you haven't shown equality between v1 and v2 quite yet...)

Answer 94

ok. so we showed equality with the scalars.

Answer 95

Right...

Answer 96

we can still use this yea? f(v1)−f(v2)=(c2−c1)f(z)

Answer 97

f(v1−v2)=f[(c2−c1)z]

Answer 98

that one