Examples from Linear independence All examples from Linear independence.
[ EXAMPLE ] Let
u 1 = ( 1 1 ) , u 2 = ( 2 α ) . u_1=\begin{pmatrix}1\\1\end{pmatrix},
\qquad
u_2=\begin{pmatrix}2\\\alpha\end{pmatrix}. u 1 = ( 1 1 ) , u 2 = ( 2 α ) . The homogeneous relation
λ 1 u 1 + λ 2 u 2 = 0 \lambda_1u_1+\lambda_2u_2=0 λ 1 u 1 + λ 2 u 2 = 0 gives
λ 1 + 2 λ 2 = 0 , λ 1 + α λ 2 = 0. \lambda_1+2\lambda_2=0,
\qquad
\lambda_1+\alpha\lambda_2=0. λ 1 + 2 λ 2 = 0 , λ 1 + α λ 2 = 0. If α = 2 \alpha=2 α = 2 α ≠ 2 \alpha\neq2 α = 2 λ 2 = 0 \lambda_2=0 λ 2 = 0 λ 1 = 0 \lambda_1=0 λ 1 = 0
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Consider
u 1 = ( 1 − 2 3 ) , u 2 = ( − 1 2 − 4 ) . u_1=\begin{pmatrix}1\\-2\\3\end{pmatrix},
\qquad
u_2=\begin{pmatrix}-1\\2\\-4\end{pmatrix}. u 1 = 1 − 2 3 , u 2 = − 1 2 − 4 . The equations in
λ 1 u 1 + λ 2 u 2 = 0 \lambda_1u_1+\lambda_2u_2=0 λ 1 u 1 + λ 2 u 2 = 0 include λ 1 − λ 2 = 0 \lambda_1-\lambda_2=0 λ 1 − λ 2 = 0 3 λ 1 − 4 λ 2 = 0 3\lambda_1-4\lambda_2=0 3 λ 1 − 4 λ 2 = 0 λ 1 = λ 2 \lambda_1=\lambda_2 λ 1 = λ 2 λ 1 = 0 \lambda_1=0 λ 1 = 0 λ 2 = 0 \lambda_2=0 λ 2 = 0
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] Let
S = { ( 1 2 ) , ( 1 3 ) , ( 1 4 ) } . S=\left\{
\begin{pmatrix}1\\2\end{pmatrix},
\begin{pmatrix}1\\3\end{pmatrix},
\begin{pmatrix}1\\4\end{pmatrix}
\right\}. S = { ( 1 2 ) , ( 1 3 ) , ( 1 4 ) } . The third vector is twice the second minus the first, so it adds no new
direction. The first two vectors are independent because neither is a
scalar multiple of the other. Therefore
{ ( 1 2 ) , ( 1 3 ) } \left\{
\begin{pmatrix}1\\2\end{pmatrix},
\begin{pmatrix}1\\3\end{pmatrix}
\right\} { ( 1 2 ) , ( 1 3 ) } is independent and has the same span as S S S
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] The set
S = { ( 1 2 ) , ( 1 3 ) , ( 1 4 ) } S=\left\{
\begin{pmatrix}1\\2\end{pmatrix},
\begin{pmatrix}1\\3\end{pmatrix},
\begin{pmatrix}1\\4\end{pmatrix}
\right\} S = { ( 1 2 ) , ( 1 3 ) , ( 1 4 ) } is dependent. For example,
( 1 4 ) = − ( 1 2 ) + 2 ( 1 3 ) = − 2 ( 1 2 ) + 4 ( 1 3 ) − ( 1 4 ) . \begin{pmatrix}1\\4\end{pmatrix}
=-\begin{pmatrix}1\\2\end{pmatrix}
+2\begin{pmatrix}1\\3\end{pmatrix}
=-2\begin{pmatrix}1\\2\end{pmatrix}
+4\begin{pmatrix}1\\3\end{pmatrix}
-\begin{pmatrix}1\\4\end{pmatrix}. ( 1 4 ) = − ( 1 2 ) + 2 ( 1 3 ) = − 2 ( 1 2 ) + 4 ( 1 3 ) − ( 1 4 ) . The relation ( 1 , 2 ) T − 2 ( 1 , 3 ) T + ( 1 , 4 ) T = 0 (1,2)^T-2(1,3)^T+(1,4)^T=0 ( 1 , 2 ) T − 2 ( 1 , 3 ) T + ( 1 , 4 ) T = 0
[RIGHT CLICK] [ASK AI] [ EXAMPLE ] The set
{ x , 1 + x , 1 − x } \{x,\ 1+x,\ 1-x\} { x , 1 + x , 1 − x } is dependent: equating coefficients in
λ 1 x + λ 2 ( 1 + x ) + λ 3 ( 1 − x ) = 0 \lambda_1x+\lambda_2(1+x)+\lambda_3(1-x)=0 λ 1 x + λ 2 ( 1 + x ) + λ 3 ( 1 − x ) = 0 gives the non-trivial choice λ 1 = − 2 \lambda_1=-2 λ 1 = − 2 λ 2 = 1 \lambda_2=1 λ 2 = 1 λ 3 = − 1 \lambda_3=-1 λ 3 = − 1 { x , 1 + x } \{x,1+x\} { x , 1 + x } 1 + x 1+x 1 + x x x x
[RIGHT CLICK] [ASK AI] Linear independence Builds on Spans .