Answer to Question #283848 in Linear Algebra for DElici

Question

Let W⊆R

5

W⊆R5 be the set of solutions of the linear homogeneous system given by

x1−x2+4x3−x4−x5=0

−x1+x2−x3+2x4+x5=0

x3−3x4+x5=0

Accordingly,

(i) show that W⊆R5 is a subspace,

(ii) find a basis for W

(iii) dim(W)=?

Answer 1

Consider the given homogenous system of linear equations.

$\begin{aligned} x_{1}-x_{2}+4 x_{3}-x_{4}-x_{5} &=0 \\ -x_{1}+x_{2}-x_{3}+2 x_{4}+x_{5} &=0 \\ x_{3}-3 x_{4}+x_{5} &=0 \end{aligned}$

The above system of equations can be expressed as,

$\left[\begin{array}{ccccc} 1 & -1 & 4 & -1 & -1 \\ -1 & 1 & -1 & 2 & 1 \\ 0 & 0 & 1 & -3 & 1 \end{array}\right]\left[\begin{array}{l} x_{2} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right] ...(1)$

Let W be the set of solutions of the above linear homogeneous system of equations.

$[i] W \subseteq \mathbb{R}^{5}$ is a subspace.

clearly $\overline{0}=\left[\begin{array}{l}0 \\ 0 \\ 0 \\ 0 \\ 0\end{array}\right] \in W.$

Now, let $x_{1}, x_{2} \in \mathbb{N}.$

$\Rightarrow A X_{2}=0 \text { and } A X_{2}=0$

Consider,

$\begin{gathered} A\left(X_{1}+X_{2}\right)=A X_{1}+A X_{2}=0+0=0 \\ \therefore A\left(X_{1}+X_{2}\right)=0 \Rightarrow X_{1}+X_{2} \in W . \end{gathered}$

 Thus. vector addition is satisfied.

Further. Let $\lambda \in \mathbb{R}$ and $x \in \mathbb{N}.$

Now.

 

$\Rightarrow \quad A x=0 \text {. }$

 

 

$A(\lambda x)=\lambda(A x)=\lambda(0)=0$

 

Thus, $A(\lambda x)=0 \Rightarrow \lambda x \in W.$

therefore scalar multiplication is satisfied.

Therefore, $\mid N \subseteq \mathbb{R}^{5}$ is a subspace.

ii) From (1) we have,

 

$\left[\begin{array}{ccccc} 1 & -1 & 4 & -1 & -1 \\ -1 & 1 & -1 & 2 & 1 \\ 0 & 0 & 1 & -3 & 1 \end{array}\right]\left[\begin{array}{l} x_{2} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right]$

 

Apply, $R_{2} \rightarrow R_{2}+R_{1}.$

 

$\left[\begin{array}{ccccc} 1 & -1 & 4 & -1 & -1 \\ 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 1 & -3 & 1 \end{array}\right]\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right]$

 

 

$\left[\begin{array}{ccccc} R_{3} & \rightarrow & 3 R_{3}-R_{2} & \\ 1 & -1 & 4 & -1 & -1 \\ 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & -8 & 3 \end{array}\right]\left[\begin{array}{l} x_{2} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right]$

 

$\begin{array}{r} \Rightarrow x_{1}-x_{2}+4 x_{3}-x_{4}-x_{5}=0 \\ 3 x_{3}+x_{4}=0 \\ -8 x_{4}+3 x_{5}=0 \end{array}$

 

No. of equations =3

No. of variables =5.

Therefore, No. of free variables =5-3=2.

Let us treat, $x_{2}$ and $x_{4}$ as free variables.

Thus. we have,

 

$x_{5}=\frac{8}{3} x_{4} \quad x_{3}=\frac{-1}{3} x_{4}$  

And,

 

$\begin{aligned} x_{1} &=x_{2}-4 x_{3}+x_{4}+x_{5} \\ &=x_{2}+\frac{4}{3} x_{4}+x_{4}+\frac{8}{3} x_{4} \\ x_{1} &=x_{2}+5 x_{4} \end{aligned}$

 

$\begin{aligned} X=\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5}\end{array}\right] &=\left[\begin{array}{c}x_{2}+5 x_{4} \\ x_{2} \\ -1 / 3 x_{4} \\ x_{4} \\ 8 / 3 x_{4}\end{array}\right]=\left[\begin{array}{l}x_{2} \\ x_{2} \\ 0 \\ 0 \\ 0\end{array}\right]+\left[\begin{array}{c}5 x_{4} \\ 0 \\ -1 / 3 x_{4} \\ x_{4} \\ 8 / 3 x_{4}\end{array}\right]=x_{2}\left[\begin{array}{l}1 \\ 1 \\ 0 \\ 0 \\ 0\end{array}\right]+x_{4}\left[\begin{array}{c}5 \\ 0 \\ -1 / 3 \\ 1 \\ 8 / 3\end{array}\right] \\ & \therefore\left[\begin{array}{c}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5}\end{array}\right]=x_{2}\left[\begin{array}{c}1 \\ 1 \\ 0 \\ 0 \\ 0\end{array}\right]+x_{4}\left[\begin{array}{c}15 \\ 0 \\ -1 \\ 3 \\ 8\end{array}\right] \text { where } x_{2}, x_{4} \in \mathbb{R} \end{aligned}$

Thus the solution space W is generated by two linearly independent vectors, namely

$\left[\begin{array}{l} 1 \\ 1 \\ 0 \\ 0 \\ 0 \end{array}\right] \text { and }\left[\begin{array}{c} 15 \\ 0 \\ -1 \\ 3 \\ 8 \end{array}\right]$

Therefore,

Basis for W is,

$B=\left\{\left[\begin{array}{c}1 \\ 1 \\ 0 \\ 0 \\ 0\end{array}\right],\left[\begin{array}{c}15 \\ 0 \\ -1 \\ 3 \\ 8\end{array}\right]\right\}$

$\underline{or} \quad B=\{(1,1,0,0,0),(15,0,-1,3,8)\} \subseteq \mathbb{R}^{5}$

(iii) Thus, $\operatorname{dim}(W)=2.$