## Question:

Let u, v, and w be distinct vectors of a real vector space V . [6 marks] (a). Show that span{u, v, w}=span{u + v − w, u + w − v, v + w −

## Answer:

To show that ( \text{span}{u, v, w} = \text{span}{u + v – w, u + w – v, v + w – u} ), we need to demonstrate that any linear combination of the vectors ( u, v, w ) can be expressed as a linear combination of the vectors ( u + v – w, u + w – v, v + w – u ) and vice versa.

**Step 1: Show that ( \text{span}{u, v, w} \subseteq \text{span}{u + v – w, u + w – v, v + w – u} )**

Let ( a u + b v + c w ) be an arbitrary linear combination of ( u, v, w ). We want to express this in terms of the new set of vectors.

Consider the vectors ( x = u + v – w ), ( y = u + w – v ), and ( z = v + w – u ).

We can express ( u, v, w ) in terms of ( x, y, z ):

- From ( x + z = (u + v – w) + (v + w – u) = 2v ) implies ( v = \frac{x + z}{2} ).
- From ( y + z = (u + w – v) + (v + w – u) = 2w ) implies ( w = \frac{y + z}{2} ).
- From ( x + y = (u + v – w) + (u + w – v) = 2u ) implies ( u = \frac{x + y}{2} ).

Now, substituting ( u, v, w ) back, we have:

[ au + bv + cw = a \frac{x + y}{2} + b \frac{x + z}{2} + c \frac{y + z}{2} ] [ = \frac{a}{2}(x + y) + \frac{b}{2}(x + z) + \frac{c}{2}(y + z) ] [ = \frac{1}{2}((a + b)x + (a + c)y + (b + c)z) ]

Thus, any linear combination ( au + bv + cw ) can be expressed as a linear combination of ( x, y, z ). Therefore, we conclude that:

[ \text{span}{u, v, w} \subseteq \text{span}{u + v – w, u + w – v, v + w – u} ]

**Step 2: Show that ( \text{span}{u + v – w, u + w – v, v + w – u} \subseteq \text{span}{u, v, w} )**

Now, take ( x = u + v – w ), ( y = u + w – v ), and ( z = v + w – u ). We express these back in terms of ( u, v, w ):

- From ( x = u + v – w ), we can isolate ( w ): ( w = u + v – x ).
- From ( y = u + w – v ), we can isolate ( v ): ( v = u + w – y ).
- From ( z = v + w – u ), we can isolate ( u ): ( u = v + w – z ).

Thus, we can write each vector as a combination of ( u, v, w ). Therefore, any linear combination of ( x, y, z ) can be expressed as a linear combination of ( u, v, w ):

[ b_1 (u + v – w) + b_2 (u + w – v) + b_3 (v + w – u) = b_1 u + b_1 v – b_1 w + b_2 u + b_2 w – b_2 v + b_3 v + b_3 w – b_3 u ]

Collecting the coefficients of ( u, v, w ):

[ (b_1 + b_2 – b_3) u + (b_1 – b_2 + b_3) v + (-b_1 + b_2 + b_3) w ]

Thus, we have:

[ \text{span}{u + v – w, u + w – v, v + w – u} \subseteq \text{span}{u, v, w} ]

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