linear algebra, generated vector spaces?

Question

if x1,x2,x3 generate a vector space V then the vectors x1,x1+x2,x1+x2+x2 also generate V, how do you prove this? appreciate your help.

i need to show that every vector in V is a linear combination of the vectors xi

i already got to the point of
v=c1 x1 + c2 (x1 + x2) + c3( x1+x2 + x3)
the part i was stuck on was how do the algebra part.. to prove its generating

i get that all the coeffiecients = 0, that doesnt really get me anywhere, tells me they are linearly dependent,, what am i doing wrong?

Answer 1

Actually, you are GIVEN that every vector in V is a linear combination of the xi (that's what it means to be told that they generate V). More explicitly, you are told that given any vector v in V, there exist scalars t1, t2 , t3 such that

v= t1 x1 + t2 x2 + t3 x3

You need to show that every vector in V is a linear combination of the other three vectors. That is, you must show that for every vector v in V, there exist scalars c1, c2, c3 such that

v=c1 x1 + c2 (x1 + x2) + c3( x1+x2 + x3)

(Half the battle in doing linear algebra proofs is explicitly writing out what you are given and what you are to prove. All I have done so far is restate the question!)

Now if you just do some algebra on the second equation, you can find values of c1, c2, c3 that work, in terms of t1, t2, t3.

Answer 2

a and c are, whilst b isn't In a, basically style the 4x4 matrix by capability of making the vectors columns (or rows - the two will artwork) of the matrix. The determinant is a million. because of the fact the determinant isn't 0, the set is a foundation. a similar thought works for c even nonetheless your vectors are 2x2 matrices. those can actually be seen vectors in R^4 whether. For b the measurement of P3 is 4 and you basically have 3 vectors in S, so it could't be a foundation.