if $V=\mathbb{R}^3=\lbrace (a,b,c)\ | \ a,b,c \in \mathbb{R} \rbrace$, we know that $V$ is a vector space over $\mathbb{R}$ under the componentwise addition and scalar multiplication. Show that the vectors $(1,-1,2), \ (1,3,1), \ (1,2,1)$ span $V$.
My attempt:
Let $u=(1,-1,2), \ v=(1,3,1), \ w=(1,2,1)$
we say that $u, v, w \in \mathbb{R}^3$ span $\mathbb{R}^3$ if every vector in $\mathbb{R}^3$ can be written as a linear combination of $u, v, w$.
let $(a,b,c) \in \mathbb{R}^3$, if we find $\alpha_1,\ \alpha_2, \ \alpha_3 \in \mathbb{R}$, satisfying that;
$(a,b,c)=\alpha_1 u+ \alpha_2 v+ \alpha_3 w$
Then $u,v$ and $w$ span $V$.
Now,
$(a,b,c)=\alpha_1 (1,-1,2)+ \alpha_2 (1,3,1)+ \alpha_3 (1,2,1)$
This implies that,
$\alpha_1+\alpha_2+\alpha_3=a$
$-\alpha_1+3\alpha_2+2\alpha_3=b$
$2\alpha_1+\alpha_2+\alpha_3=c$
And by solving this system, we find that;
$\alpha_1=c-a$
$\alpha_2=-5a+b+3c$
$\alpha_3=7a-b-4c$
This show that any vector in $\mathbb{R}^3$ is a linear combination of $(1,-1,2), \ (1,3,1), \ (1,2,1)$, hence the vectors $(1,-1,2), \ (1,3,1), \ (1,2,1)$ span $\mathbb{R}^3$
Is that true 100% ?
Best Answer
Yes it is.
You have proved the statement clearly and correctly.
You could have checked the determinant made by your three vectors and show that the determinant is non zero.