Let u, v and w be distinct vectors of a vector space V. Show that if?

Question

{u, v, w} is a basis for V, then { u+v+w, v+w, w} is also a basis for V.

Answer 1

If {u,v,w} Is a basis of V, then any vector in V can be spanned by au+bv+cw where a,b,c are arbitrary constants.

Now, lets see what linear combinations of {u+v+w, v+w, w} look like. For constants d,e,f we have:

d(u+v+w) + e(v+w) + fw = du+dv+dw+ev+ew+fw

= du + (d+e)v + (d+e+f)w

So we can see that if we let a=d, b=d+e, and c=d+e+f that we have a similar linear combination to the span of the basis {u,v,w}. Therefore {u+v+w, v+w, w} is also a basis.

Answer 2

{u+v+w,v+w,w} form a basis for V if any arbitrary vector A can be expressed as linear combination of u+v+w, v+w, and w. Well, we know that u,v,w form a basis so that

A = au+bv+cw

But u = (u+v+w)-(v+w) and

v = (v+w) - w so we can rewrite A as

A = a[(u+v+w)-(v+w)] +b[(v+w)-w] + cw or, rearranging

A = (a+b)(u+v+w) +(b-a)(v+w) +cw

So we have expressed an arbitrary vector in terms of u+v+w, v+w, and w, so they form a basis for V also.

Answer 3

first prepare that {5–x, 6+x, x^2+4x+a million} is lineary self retaining then take any arbitrary polynomials with diploma 2 and prepare that it fairly is written as a linear mix of given set