Prove for each triple of vectors $a, x, y \in \Bbb R^n$: $a\in\text{Span} \{x,y\} \Rightarrow \operatorname{Span}\{x,y\}=\operatorname{Span}\{a,x,y\}$

Question

I am curious if the proof if have is correct and enough to prove the following question:

Prove for each triple of vectors $a, x, y \in R^n\DeclareMathOperator{\Span}{Span}$: $ a \in \Span \{x,y\} \Rightarrow \Span\{x,y\} = \Span\{a,x,y\}$:

Here is my proof:

If $a \in \Span\{x,y\}$, therefore $c_1*x+c_2*y = a$, i.e a is a linear combination of all the vectors in $\Span(x,y)$. Therefore, $\Span(x,y) = \Span(a,x,y)$.

This seems too easy as a solution and I presume I a fundamental misunderstanding about what a span is.

Thanks in advance!

Answer 1

You should prove both inclusions. The inclusion $span(x,y)\subset span(a,x,y)$ is easy. Take $v=c_1x+c_2y\in span(x,y)$. Then, $$v=c_1x+c_2y=c_1x+c_2y+0a\in span(a,x,y).$$

Now for the other direction take $v=c_1x+c_2y+c_3a$ in the right. Since $a\in span(x,y)$ you find $d_1,d_2$ such that $a=d_1x+d_2y$. But then, $$v=xc_1x+c_2y+c_3a=c_1x+c_2y+c_3(d_1x+d_2y)=(c_1+c_3d_1)x+(c_2+c_3d_2)y\in span(x,y).$$

Answer 2

You proved $Span(x,y)\subseteq Span(a,x,y)$.

You need to prove vice versa: Take any $v\in Span\{a,x,y\}$, then $$v = ma+nx+ky = m(c_1x+c_2y)+nx+ky = (mc_1+n)x+(mc_2+k)y$$ so $v\in Span(x,y)$. So, since $v$ was arbitrary we have also $Span(x,y)\subseteq Span(a,x,y)$ and now we are done.