Question

Suppose that x and y are vectors and M is a subspace in a vector space V. Let H be the subspace spanned by M and x, and let K be the subspace spanned by M and y.
Prove that if y is in H but not in M, then x is in K
Please, help

Answer 1

I have to use dimension to prove. I am not allowed using explicitly way

Answer 2

If you're supposed to use the dimension formula to prove this then you must be given some dimensions though, in your question it doesn't say that the dimensions are finite. So you must take this as a premise.

Introduce a dimension, lets say \(n\) for the Vector Space \(V\). Both \(x,y\) are in \(V\). Introduce a dimension for \(M\) and also a Basis for \(M\), you can complete this Basis to a Basis of \(H\) because \(M\) is a subspace of \(V\). Do the same for the Subspace \(K\).

Answer 3

dim H = dim (Span ( M + x))= dim M + dim x
dim K = dim (Span (M + y))= dim M + dim y
if y \(\in\)H \(\rightarrow y = (x_0+x_1+....+xi)\)
am I Ok so far?

Answer 4

The dimension of \(H\) and the dimension of \(K\) should be identical right?, because they are both spanned by the same subset and one vector, so their dimension is lets say \(p\leq n\) where \(n\) is the dimension of the Vector Space \(P\), which Basis did you introduce? Are those complex vector spaces?

Answer 5

For basis: no indication, most of the time, we have both C and R

Answer 6

for dim H and dim K, I don't think they are equal.:(

Answer 7

I would have started of with a Basis for \(M\). Mustn't necessary be the first step, but it's what I first jumped into my mind, it must be true that \(x \notin M\) such that \(x\) is linearly independent from \(M\)

Introduce a Basis for \(M\) (using the theorem that every Vector Space has a Basis) so that \(\mathcal{M}= \lbrace v_1, v_2 , \dots , v_{p-1} \rbrace\) is a Basis of \(M\), if we include \(x \in H\) to that Basis we have completed it to a Basis of \(H\) with dimension \(p\)

\(\mathcal{H}= \lbrace v_1, v_2 , \dots , v_{p-1}, x \rbrace \) with dimension \(p\)

Answer 8

yes, got this part. :)

Answer 9

The way I read the problem, and it might be that I misunderstand it, but the same procedure should be possible to create a basis for \(K\) lets call it \(\mathcal{K}\), that's why I said their dimensions must be equal. They are both spanned by the same subspace and one vector (for which we must make some premise of course)

Answer 10

In that manner \(\mathcal{K}:=\lbrace \mathcal{M}, y\rbrace \)

Answer 11

It doesn't say K = H
let say m is dimension of M
3 is dim (x)
7 is dim (y)
then dim H \(\neq\)dimK

Answer 12

No it does not say that the vector Spaces are equal, and I don't say that myself, I am only talking about the dimensions.

Lets say, my house has 3 levels, and your house has 3 levels, our houses have the same dimensions, do we live together in the same house?

Answer 13

lol If you want , we can!!! ( joke)

Answer 14

\( \dim A = \dim B (\neg \implies) A=B \)

Answer 15

got you now

Answer 16

give me time to try. I will post my work to get check,

Answer 17

sure thing, the rest should be considering the able premises and various cases. Might be tedious, I haven't tried, but we can't analyze the properties of the problem without having well-defined subspaces :-)

Answer 18

Thank you. @Spacelimbus

Answer 19

let dim (M) =m , dim (H) =h , dim (K) =k
y \(\in\)H \[y = m + \alpha x\]
y \(\notin\) M and y \(\in\) H , therefore \(\alpha \neq 0\)
\(x = \dfrac{1}{c} (y-m)\) and y -m is k, so x \(\in\) K.
Am I right? @Spacelimbus