Linear Algebra Question

Question

Let G be a function from R2 to R2, G: R2->R2 so, if xER2 then G(x) is uniquely defined. and G(x)ER2 (for example the velocity of a boat as a function of its position on the ocean)

We say that G is linear to mean that

1. For all x1,x2, ER2, G(x1+x2) = G(x1)+G(x2), and
2. For all xER2, For all cER, G(cx1) = cG(x1)

a) Show that the function G1 (a,b) = ((a)(b),b) is not linear (try using simple numbers)

b) Show that the function G2 (a,b) = (a+1,b) is not linear (try using simple numbers)

c) Show that the function G3 (a,b) = (3a+2b,4b) is linear (use the definition)

Answer 1

The first two can be done simply by plugging in.

a) If we plug in a₁ = 1 and b₁ = 2

Let a₂ = 2 and b₂ = 3 and plug those in:

So G(a₁,b₁) = (1(2), 2) = (2,2)

G(a₂, b₂) = (2*3 , 3) = (6,3)

G(a₁ + a₂, b₁+b₂) = (G(3, 5)) = (15,5)

G(a₁, b₁) + G(a₂, b₂) = (2,2)+(6,3) = (8,5) ≠(15,5)

So, G is not linear.

b) Take a =1 , b = 1, c = 2

G(1,1) = (1+1,1 ) = (2,1)

G(c(1,1)) = G(2(1,1)) = (G(2,2)) = (3,2)

2G(1,1) = 2(2,1) = (4,2) ≠ (3,2)

Therefore G(2(1,1)) ≠ 2G(1,1) so this function is not linear

c) This is a bit more rigorous:

Take any two sets of numbers (a₁, b₁) and (a₂, b₂) ∈ R²

G(a₁+a₂ , b₁+b₂) = (3(a₁+a₂)+ 2(b₁+b₂), 4(b₁+b₂))

G(a₁, b₁) + G(a₂, b₂) = (3a₁ + 2 b₁, 4b₁) +(3a₂ + 2 b₂, 4b₂) = (3a₁+3a₂ + 2 b₁+ 2 b₂, 4b₁+ 4 b₂)

= (3(a₁+a₂)+ 2(b₁+b₂), 4(b₁+b₂)) = G(a₁+a₂ , b₁+b₂)

Therefore our first property of linearity is proven.

Now onto the second. Take any (a,b) ∈ R² and c ∈ R.

G(c(a,b)) = G(ca, cb) = (3ca + 2cb, 4cb) = c(3a + 2b, 4b) = c G(a,b)

Therefore G(c(a,b)) = c G(a,b), so our second property of linearity is proven.

Since both properties of linearity are satisfied, G is a linear function.