This problem is about vector spaces in general. The vectors in those spaces are

Question

This problem is about vector spaces in general. The vectors in those spaces are not necessarily column vectors. In the definition of a vector space, vector addition x+y =y+x and scalar multiplication cx must obey the following 8 rules:
(1) x+y=y+x, commutative
(2) x + (y + z) = (x+ y) + z
(3) x + 0 =x for all x
(4) unique -x for each x, s.t. x+(-x) =0
(5) 1 Times x = x
(6) (c1c2)x = c1(c2x)
(7) c(x+y) =cx + cy
(8) (c1 + c2)x = c1x + c2x

Question: Suppose the multiplication cx is defined to produce (cx1, 0) instead of (cx1, cx2). With the usual addition in R^2, are the eight conditions satisfied?

Explanation / Answer

Let x = (x1,x2) be an arbitrary vector in R2.

Property: multiplicative identity -- in a vector space, 1*x=x.

In R2 with the newly defined scalar multiplication, 1*(x1,x2)=(x1,0). This is not equal to our original (x1,x2). So, R2 fails the property of having a multiplicative identity.

So, you can already conclude that the 8 conditions are not satisfied. More importantly, R2 is not a vector space with this new scalar multiplication.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.