A [[Basis]] of a [[Linear Subspace|Subspace]] $U$ is a [[Linear Independence|Linearly Independent]] [[Set]] of [[Basis Vectors|Vectors]] that [[Span]] $U$. The [[Basis]] of a [[Linear Subspace|Subspace]] with only one element ([[Zero Linear Subspace]]), is the [[Empty Set]]. >[!example] Example: Finding a [[Basis]] of a [[Linear Subspace|Subspace]] >$ >\let U = \left\{ >\vec x \in \R^5 >\left| \begin{split} >x_{1}-2x_{3}+x_{4}-x_{5}&=0\\ >x_{1}+2x_{2}-x_{4}-3x_{5}&=0 >\end{split} >\right.\right\} >$ >Write a [[Basis]] of $U$. > >>[!check]- Solution >>$\begin{align} >>T: \R^5 &\to \R^{2} \\ >>T: \mat{x_{1}\\x_{2}\\x_{3}\\x_{4}\\x_{5}} &\mapsto \mat{ >>x_{1}-2x_{3}+x_{4}-x_{5}\\ >>x_{1}+2x_{2}-x_{4}-3x_{5} >>}\\ >> >>\therefore >>\ker(T) &= U \\ >>\op{Nul}(A) &= U\\ >>\end{align}$ >>$\begin{align} >>A &= \mat{ >>1&0&-2&1&1\\ >>1&2&0&-1&-3 >>} \\ >>\augmented{c|c}{A|\vec 0} &\sim >>\augmented{ccccc|c}{1&0&-2&1&1&0\\1&2&0&-1&-3&0}\\ >>&\sim \augmented{ccccc|c}{1&0&-2&1&1&0\\0&1&-1&0&-1&0}\\ >>&\sim \begin{cases} >> >>x_{1}-2x_{3}+x_{4}+x_{5}&=0\\ >>x_{2}-x_{3}-x_{5} &=0 >>\end{cases}\\ >>&\sim \begin{cases} >>x_{1}&=2t-s-r \\ >>x_{2}&=t+r\\ >>x_{3}&=t\\ >>x_{4}&=s\\ >>x_{5}&=r >>\end{cases} >>\end{align}$