# 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 $X=\mathbb{R}^{3}$,

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

and

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

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...