a ⃗ 1 = ( 1 , 1 , 1 ) , a ⃗ 2 = ( 0 , 1 , 1 ) , a ⃗ 3 = ( 0 , 1 , − 1 ) \vec a _1 =(1,1,1), \vec a_2=(0,1,1), \vec a_3=(0,1,-1) a 1 = ( 1 , 1 , 1 ) , a 2 = ( 0 , 1 , 1 ) , a 3 = ( 0 , 1 , − 1 )
We show that the vectors form the basis.
Consider a linear combination of vectors
α a ⃗ 1 + β a ⃗ 2 + γ a ⃗ 3 = 0 ⃗ \alpha \vec a_1+\beta\vec a_2+\gamma\vec a_3=\vec 0 α a 1 + β a 2 + γ a 3 = 0
and show that the numbers α , β , γ \alpha,\beta,\gamma α , β , γ are equal 0.
α a ⃗ 1 + β a ⃗ 2 + γ a ⃗ 3 = 0 ⃗ α ( 1 , 1 , 1 ) + β ( 0 , 1 , 1 ) + γ ( 0 , 1 , − 1 ) = ( 0 , 0 , 0 ) α = 0 α + β + γ = 0 α + β − γ = 0 α = β = γ = 0 \alpha \vec a_1+\beta\vec a_2+\gamma\vec a_3=\vec 0\\
\alpha (1,1,1)+\beta(0,1,1)+\gamma(0,1,-1)=(0,0,0)\\
\begin{matrix}
\alpha=0\\
\alpha+\beta+\gamma=0\\
\alpha+\beta-\gamma=0
\\\end{matrix}\\
\alpha=\beta=\gamma=0 α a 1 + β a 2 + γ a 3 = 0 α ( 1 , 1 , 1 ) + β ( 0 , 1 , 1 ) + γ ( 0 , 1 , − 1 ) = ( 0 , 0 , 0 ) α = 0 α + β + γ = 0 α + β − γ = 0 α = β = γ = 0
Then vectors a ⃗ 1 = ( 1 , 1 , 1 ) , a ⃗ 2 = ( 0 , 1 , 1 ) , a ⃗ 3 = ( 0 , 1 , − 1 ) \vec a _1 =(1,1,1), \vec a_2=(0,1,1), \vec a_3=(0,1,-1) a 1 = ( 1 , 1 , 1 ) , a 2 = ( 0 , 1 , 1 ) , a 3 = ( 0 , 1 , − 1 ) form the basis in R 3 R^3 R 3
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