Linear algebra question?

Question

If anyone could explain how the following problem is done, I would really appreciate it!

Here is a vector candidate. The set is R, and we define scalar multiplication by ax = a * x (usual scalar multiplication) and vector addition by x (insert circle symbol with plus sign in it that means vector addition) y = max(x, y).

For each of the following three vector space axioms, either verify the axiom or show that it does not hold.

a) a(x+y) = ax + ay
b) There exists an element 0 such that for any x in the proposed vector space, x + 0 = x.
c) x+y = y+x

Answer 1

if x>y then ax >ay PROVIDED a>0
so in general a(x[+]y) is NOT ax[+]ay

the identity for [+] would have to be -oo; which I personally would not call a "member" of R

if x>y then y< x , so x[+]y = y[+]x is the only axiom that holds.

Answer 2

a: this axiom does not hold. Let x=1, y=-1, and a=-1. Then a(x⊕y) = -1, but ax⊕ay = 1

b: False. Let "0" = y, where y is any real number. Then let x
c: this one's true at least -- the maximum of two numbers does not depend on the order in which they are considered.

Answer 3

a) You do this several ways and compare the answers. So, without loss of generality, let us say that y >= x with x and y >= 0. So, let a >= 0

a(x + y) = a(max(x,y)) = ay
ax + ay = max(ax, ay) = ay OK, so far so good.

Now let a < 0
a(x + y) = a(max(x,y)) = ay
ax + ay = max(ax,ay) = ax Oops. That doesn't work! (Try x = 1, y = 2, and a = -1)

b) Let x > = 0 then
x + 0 = max(x,0) = x

Now let x < 0

x + 0 = max(x,0) = 0 so, b) fails, too.

c) Again, let y >= x

x + y = max(x,y) = y
y + x = max(y,x) = y

c holds.

HTH

Charles

Answer 4

I'm not quite sure how you would define max(x,y) among 2 vectors that also returns a vector.

a) a*(x+y)=a*max(x,y)
a*x+a*y=max(a*x,a*y)=a*max(x,y)

So this holds

b) x+0=max(x,0)

If max(x,0)<>max(-x,0) then this wouldn't hold. For example, if x and y are 2 dimensional vectors, G=max(x,y) returns the vector

G*={max(x1,y1),max(x2,y2)}

then b would not hold. Otherwise it probably would. It really depends on the definition of the max function.

c) x+y=max(x,y)=max(y,x)=y+x

So this would hold.