Linear Subspace

DEFINITION1: Let K be a field, X be a K-vector space and M\subset X. If M is also a K-vector space, then M is called a linear subspace or vector subspace (or shortly subspace) of X.

PROPOSITION1: Let K be a field, X be a K-vector space and M\subset X. M is a subspace of X if and only if

a) \theta\in M,

b) x+y\in{M} for all x,y\in{M},

c) \lambda{x}\in{M} for all {\lambda}\in{K} and {x}\in{M}.

PROOF: (will be linked)

PROPOSITION2: Let K be a field, X be a K-vector space and \varnothing\neq M\subset X. M is a subspace of X if and only if \lambda{x}+\mu y\in{M} for all {\lambda,\mu}\in{K} and {x,y}\in{M}.

PROOF: (will be linked)

EXAMPLE1: Let K be a field, X be a K-vector space. Then, there are always two subspaces of X. They are M_{0}=\left\{ \theta \right\} and M_{1}=X. Each of them is called a trivial subspace of X. A subspace different from them is called nontrivial subspace.

EXAMPLE2: Let X=\mathbb{R}^{2} and c_{1},c_{2}\in{\mathbb{R}} be arbitrary constants. Then, the set M_{c_{1},c_{2}}=\left\{ \left( x_{1},x_{2} \right) \in{\mathbb{R}^{2}}\,|\,c_{1}x_{1}+c_{2}x_{2}=0 \right\} is a subspace of \mathbb{R}^{2}. M_{c_{1},c_{2}} is different from the zero space \left\{ \left( 0,0 \right) \right\} for every c_{1},c_{2}\in{\mathbb{R}}. M_{c_{1},c_{2}} is nontrivial if and only if c_{1}^{2}+c_{2}^{2}>0.

SOLUTION: Let \lambda,\mu\in{\mathbb{R}} and x= \left( x_1,x_2 \right), y= \left( y_{1},y_{2} \right) \in{M}. Then, the equalities c_{1}x_{1}+c_{2}x_{2}=0 and c_{1}y_{1}+c_{2}y_{2}=0 are fulfilled.

\lambda x+\mu y=\lambda \left( x_{1},x_{2} \right) +\mu \left( y_{1},y_{2} \right) = \left( \lambda{x}_{1}+\mu y_{1},\lambda x_{2}+\mu y_{2} \right).

Since c_{1} \left( \lambda{x}_{1}+\mu y_{1} \right) +c_{2} \left( \lambda{x}_{2}+\mu y_{2} \right) =\lambda \left( c_{1}x_{1}+c_{2}x_{2} \right) +\mu \left( c_{1}y_{1}+c_{2}y_{2} \right) =\lambda 0+\mu 0=0, then \lambda x+\mu y\in M_{c_{1},c_{2}}. Moreover, the equality c_{1}0+c_{2}0=0 shows that the vector \theta=\left( 0,0 \right) is in M_{c_{1},c_{2}}. These prove that M_{c_{1},c_{2}} is a subspace of \mathbb{R}^{2}.

Now, we prove the assertion M_{c_{1},c_{2}}\neq\left\{ \left( 0,0 \right) \right\} for every c_{1},c_{2}\in{\mathbb{R}}. In the case c_{1}=0, the relation \left( 1,0 \right)\in M_{c_{1},c_{2}} holds because of the equality 1c_{1}+0c_{2}=0. On the other hand, in the case c_{1}\neq 0, the relation \left( c_{2},-c_{1} \right)\in M_{c_{1},c_{2}} holds because of the equality c_{2}c_{1}-c_{1}c_{2}=0. In either case, the subspace M_{c_{1},c_{2}} contains at least one vector different from the zero vector.

Now, we prove the proposition "M_{c_{1},c_{2}} is nontrivial if and only if c_{1}^{2}+c_{2}^{2}>0". We have already shown that M_{c_{1},c_{2}} is never equal to \left\{ \left( 0,0 \right) \right\}. So, it is sufficient to show that M_{c_{1},c_{2}} is different from \mathbb{R}^{2} if and only if c_{1}^{2}+c_{2}^{2}>0. For this, we must prove two propositions: First: If c_{1}=c_{2}=0, then M_{c_{1},c_{2}}=\mathbb{R}^{2}. Second: If at least one of the numbers c_{1} and c_{2} is different from zero, then M_{c_{1},c_{2}}\neq\mathbb{R}^{2}. The first proposition is obvious because c_{1}x_{1}+c_{2}x_{2}=0 for all \left( x_{1},x_{2} \right) \in\mathbb{R}^{2} if c_{1}=c_{2}=0. We will prove the second proposition as c_{1}\neq 0. The case c_{2}\neq 0 is similar. Let c_{1}\neq 0. Then, \left( 1,0 \right)\notin M_{c_{1},c_{2}} because the relation c_{1}1+c_{2}0=c_{1}\neq 0 holds. Consequently, we conclude that M_{c_{1},c_{2}}\neq\mathbb{R}^{2}.

EXAMPLE3: Consider the subset M=\left\{ \left( x,x^{2} \right) \in{\mathbb{R}^{2}}\,|\,x\in{\mathbb{R}} \right\} of \mathbb{R}^{2}. M contains the zero vector \left( 0,0 \right) because the equality 0^{2}=0 is satisfied. Now, we show that M is not a subspace of \mathbb{R}^{2}. Obviously, the vector \left( 1,1 \right) is in M. However, the scalar product 2\left( 1,1 \right)=\left( 2,2 \right) is not in M because 2^{2}=4\neq 2. This example shows, in order that a subset is a subspace, it is not sufficient that the subset contains the zero vector. On the other hand, we know from Proposition1 that a subset not containing the zero vector is not a subspace.

EXAMPLE4: Let X=\mathbb{R}^{3},

M_{1}=\left\{\left(s-t,s+5t,2t-s\right) | s,t\in{\mathbb{R}}\right\}


M_{2}=\left\{\left(s-t,s+5t,2t-1\right) | s,t\in{\mathbb{R}}\right\}.

We investigate whether each of M_{1} and M_{2} is a subspace of \mathbb{R}^{3} or not.

First, we investigate M_{1}. For s=t=0, we have \left(0-0,0+5.0,2.0-0\right)=\left(0,0,0\right)\in{M_{1}}.

Given a,b\in{\mathbb{R}} and \left(s-t,s+5t,2t-s\right), \left(s'-t',s'+5t',2t'-s'\right)\in{M_{1}}.

Then, the relation

a\left(s-t,s+5t,2t-s\right)+b\left(s'-t',s'+5t',2t'-s'\right)=\left(as+bs'-\left(at+at'\right),as+bs'+5\left(at+at'\right),2\left(at+at'\right)-\left(as+bs'\right)\right)\in M_{1}

shows that M_{1} is a subspace.

Now, we investigate M_{2}: in order that M_{2} is a subspace, the zero vector \left(0,0,0\right) must be in M_{2}. For this reason, we must find two real numbers s,t satisfying the equality \left(s-t,s+5t,2t-1\right)=\left(0,0,0\right). The equalities s-t=0 and s+5t=0 give us s=t=0. However, these values don't satisfy the equality 2t-1=0. So, the zero vector \left(0,0,0\right) is not in M_{2}. This shows that M_{2} is not a subspace.

EXAMPLE4 (SEQUENCE SPACES): In vector space-Example8, we have given all real sequences space S=\left\{\left(x_{n}\right)\,|\,\forall{n}\in{\mathbb{N}}, x_{n}\in{\mathbb{R}}\right\}. We will study some significant subspace of this space:

The space

\displaystyle{l_{\infty}=\left\{\left(x_{n}\right)\in{S}\,|\,\forall{n}\in{\mathbb{N}}, x_{n}\in{\mathbb{R}}\land{\sup_{n\in{\mathbb{N}}}\left|x_{n}\right|<+\infty}\right\}}

is the space of all bounded sequences. This space is a nontrivial subspace of S.

to be continued...

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