MHB Two sets generate the same vector space

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The discussion focuses on proving that the sets {a,b} and {a, b, a-b} generate the same vector space. It is established that any vector from the span of {a,b} is also contained in the span of {a, b, a-b}, and vice versa. The transformation of vectors using the equation c1*a + c2*b + c3*(a-b) demonstrates this relationship. By rearranging terms, it shows that any linear combination can be expressed in terms of the original set {a,b}. Thus, the two sets indeed generate the same vector space.
Suvadip
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Show that the sets $$\{a,b\}$$ and $$\{a, b, a-b\}$$ of real vectors generate the same vector space.

How to proceed with it? I guess the following expression is helpful.

$$c1*a+c2*b+c3*(a-b)=(c1+c3)*a+(c2-c3)*b=k1*a+k2*b$$
 
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suvadip said:
Show that the sets $$\{a,b\}$$ and $$\{a, b, a-b\}$$ of real vectors generate the same vector space.

How to proceed with it? I guess the following expression is helpful.

$$c1*a+c2*b+c3*(a-b)=(c1+c3)*a+(c2-c3)*b=k1*a+k2*b$$

Hi suvadip, :)

You have to show any vector taken from $<a,\,b>$ lies in $<a,\,b,\,a-b>$ and vice versa. If you take any vector $v\in <a,\,b>$ it is clear that $x\in <a,\,b,\,a-b>$ since every vector generated by $a$ and $b$ is in $<a,\,b,\,a-b>$. Conversely if you take any vector $v\in<a,\,b,\,a-b>$ then $v=k_1 a+k_2 b+k_3 (a-b)=(k_1+k_3)a+b(k_2-k_3)\in <a,\,b>$. Therefore, $<a,\,b>=<a,\,b,\,a-b>$.
 

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