# Vector Subspaces

## Recommended Posts

We consider a linear vector space V of dimension n. W is a proper subspace of V.We take a vector 'e' belonging to V-W and N vectors y_i belonging to W;i=1,2,3…N;N>>n the dimension of V. All y_i cannot obviously be independent the number N being greater than the dimension of V the parent vector space; k of the  y_i vectors are considered to be linearly independent where k is the dimension of W. The rest of the y_i are linear combinations of these k, basic y_i vectors of W.
We consider sums

αi=e+yi;i=1,2N
(1)
Now each alpha_i=e+y_i belongs to V-W. We prove it as follows
If possible let alpha_i belong to W. We have

e=αiyi=αi+(yi)
(2)

Both alpha_i and  -y_i belong to W. Therefore their sum ‘e’ should belong to W . This contradicts our postulate that   e belongs to V-W. Therefore each alpha_i belongs to V-W.
Next we consider the equation

Σi(ciαi)=0
(3)

Σi=Ni=1(ci(e+yi))=0

eΣi=Ni=1ci=Σi=Ni=1ciyi
(4)
The right side of (4) belongs to W  while the left side belongs to V-W. If the left  side belonged to W them (1/Sigma c_i)(Sigma c_i)e=e would belong to W which is not the case. |The right side being a linear combination of vectors from W belongs to W.The only solution to avoid this predicament would be to assume Sigma  c_i on the left side of (4) to be zero: that each side of (4) represents the null vector. We cannot have all c_i=0 [individually]in an exclusive manner  since that would make the space N dimensional , in view of (3), [N is  much greater than n, the dimension of the parent vector space V].
Equations

Σi=Ni=1ci=0
(3.1)

Σi=Ni=1ciyi=0
(3.2)
From (3.1)

cN=c1c2c3..cN1
(4)
Considering (3.2) with (4) we have,

yN=c1c1+c2+c3..+cN1y1+c2c1+c2+c3..+cN1y2+..+cN1c1+c2+c3..+cN1yN1
(5.1)

yN=a1y1+a2y2+aN1yN1
(5.2)
Where,

ai=cic1+c2++cN1
(5.3)
From (5.3) we have the identity

a1+a2++aN1=1
(6)
But the N(>>N) vectors were chosen arbitrarily. Equation (5.2) should not come under the constraint of equation (6).We could have chosen y_N in the form of (5.2) in a manner  that (6) is violated.
##### Share on other sites
• 3 weeks later...
Posted (edited)
On 3/12/2021 at 11:47 PM, Anamitra Palit said:

Now each alpha_i=e+y_i belongs to V-W.

I saw your similar post on another site where it didn't get any traction. May I suggest a couple of minor notational changes that will improve clarity? Since $e \in V \setminus W$, I'd call it $v$. Likewise I'd call the $y_i$'s $w_i$. These minor changes would decrease the cognitive burden on the reader; and (if you don't mind my saying) your exposition here and on the other site are already a little convoluted and the reader can use all the help they can get.

On 3/12/2021 at 11:47 PM, Anamitra Palit said:

Σi=Ni=1ci=0

This notation's hard to figure out. You did convince me on the other site that $\displaystyle \sum_{i = 1}^N c_i = 0$, but here I don't know what you're trying to say.

On 3/12/2021 at 11:47 PM, Anamitra Palit said:

yN=c1c1+c2+c3..+cN1y1+c2c1+c2+c3..+cN1y2+..+cN1c1+c2+c3..+cN1yN1

Are there some parens missing? I think so but who can be sure?

I don't mean to be making only picky stylistic complaints, but I did make an honest attempt to work through your exposition on the other site and gave up due to lack of clarity. I suspect others might have done the same.

You will get better responses if you clarify the exposition and notation.

At least here you are no longer trying to divide through by $c_N$, which for all we know might be 0. So there is some incremental improvement over the version on the other site.

Also where do you use the fact that N >> n? Your argument (whatever it is) seems like it would go  through (or not) just as well for N = n + 1.

In fact if you could give a concrete example with, say, $V = \mathbb R^3$ and $W = \mathbb R^2$ you might either figure out where you're making a mistake, or at the very least others would see what you're trying to do.

Also do you happen to know Mathjax markup? For example if I write "e^{i \pi} + 1 = 0" between two "math" tags on this site, I get this beautifully rendered

$e^{i \pi} + 1 = 0$

It's a small learning curve at first but greatly improves readability. As someone with terrible handwriting I wish they'd had this when I rode a dinosaur to school.

Edited by wtf
##### Share on other sites
Posted (edited)
2 hours ago, wtf said:

yN=c1c1+c2+c3..+cN1y1+c2c1+c2+c3..+cN1y2+..+cN1c1+c2+c3..+cN1yN1

ps -- I follow everything up to this line. If you can please put in proper parens and show exactly how you got this I'd find it very helpful.

Also please note that the $y_i$'s are presumably taken to be all distinct from each other, else you can't be sure they contain a linearly independent subset.

And also note that when ask us to consider the equation

On 3/12/2021 at 11:47 PM, Anamitra Palit said:

Σi(ciαi)=0

you need to specify that at least one of the $c_i$'s is not zero. There's no reason they couldn't all be. When you say, "We cannot have all c_i=0 [individually]in an exclusive manner  since that would make the space N dimensional," that is not true. With all the c_i's equal to zero, that equation is true in any vector space no matter what the dimensions of the space and its subspaces.

I think you are confusing this with the definition of linear independence, which says that if $\displaystyle \sum c_i v_i = 0$ implies that all the c_i's must be zero, then the v_i's are linearly independent. But just because you have $\displaystyle \sum c_i v_i = 0$, that doesn't mean all the c_i's can't all happen to be 0 regardless of whether the v_i's are linearly independent.

Finally, you keep using the notation

Σi=Ni=1 ...

which I imagine is supposed to mean $\displaystyle \sum_{i= 1}^N$ but is incredibly confusing in context. Can you please fix that throughout?

Edited by wtf
##### Share on other sites

pps -- I looked at your handle and found this.

You most definitely do know how to do math markup. Can you please do us a favor and mark up this linear algebra post?

##### Share on other sites

OP never came back? After crossposting this to two different discussion forums?

## Create an account

Register a new account