show that each function is a linear tranformation T(x,y)=(2x,x+y,x-2y) (T:R²=>r³)
(2) T:R³=>M^(2,2) given by T(x,y,z)=[y 2]
[2 0]
i tried to make a matrice bracket but i couldn't
(1) since each of the images is a linear combination of x,y T is a linear transformation.
(2) is the same
Remember, all you have to show is that
T(a+b) = T(a)+T(b)
T(ka) = kT(a)
i think i get it now perfectly
To show that a function is a linear transformation, we need to demonstrate two properties:
1. Preservation of addition: T(u + v) = T(u) + T(v), for any vectors u and v.
2. Preservation of scalar multiplication: T(cu) = cT(u), for any scalar c and vector u.
Let's go step by step to verify these properties for each function:
1. T(x, y) = (2x, x+y, x-2y)
a) Preservation of addition:
To show that T(u + v) = T(u) + T(v), we need to compute both sides of the equation.
Let's consider u = (x₁, y₁) and v = (x₂, y₂):
T(u + v) = T(x₁ + x₂, y₁ + y₂)
= (2(x₁ + x₂), (x₁ + x₂) + (y₁ + y₂), (x₁ + x₂) - 2(y₁ + y₂))
= (2x₁ + 2x₂, x₁ + x₂ + y₁ + y₂, x₁ + x₂ - 2y₁ - 2y₂)
T(u) + T(v) = T(x₁, y₁) + T(x₂, y₂)
= (2x₁, x₁ + y₁, x₁ - 2y₁) + (2x₂, x₂ + y₂, x₂ - 2y₂)
= (2x₁ + 2x₂, x₁ + y₁ + x₂ + y₂, x₁ - 2y₁ + x₂ - 2y₂)
= (2x₁ + 2x₂, x₁ + x₂ + y₁ + y₂, x₁ + x₂ - 2y₁ - 2y₂)
We can see that T(u + v) = T(u) + T(v) holds true, so the preservation of addition property is satisfied.
b) Preservation of scalar multiplication:
To show that T(cu) = cT(u), we need to compute both sides of the equation.
Let's consider u = (x, y) and c be a scalar:
T(cu) = T(cx, cy)
= (2(cx), cx + cy, cx - 2cy)
= (2cx, cx + cy, cx - 2cy)
cT(u) = cT(x, y)
= c(2x, x + y, x - 2y)
= (2cx, cx + cy, cx - 2cy)
Again, we can see that T(cu) = cT(u) holds true, so the preservation of scalar multiplication property is satisfied.
Therefore, T(x, y) = (2x, x+y, x-2y) is a linear transformation from R² to R³.
2. T(x, y, z) = [y 2]
[2 0]
a) Preservation of addition:
To show that T(u + v) = T(u) + T(v), we need to compute both sides of the equation.
Let's consider u = (x₁, y₁, z₁) and v = (x₂, y₂, z₂):
T(u + v) = T(x₁ + x₂, y₁ + y₂, z₁ + z₂)
= [y₁ + y₂ 2]
[2 0]
T(u) + T(v) = T(x₁, y₁, z₁) + T(x₂, y₂, z₂)
= [y₁ 2]
[2 0] + [y₂ 2]
[2 0]
= [y₁ + y₂ 2]
[2 0]
Once again, T(u + v) = T(u) + T(v) holds true, so the preservation of addition property is satisfied.
b) Preservation of scalar multiplication:
To show that T(cu) = cT(u), we need to compute both sides of the equation.
Let's consider u = (x, y, z) and c be a scalar:
T(cu) = T(cx, cy, cz)
= [cy 2]
[2 0]
cT(u) = cT(x, y, z)
= c[y 2]
[2 0]
= [cy 2]
[2 0]
Once again, T(cu) = cT(u) holds true, so the preservation of scalar multiplication property is satisfied.
Therefore, T(x, y, z) = [y 2]
[2 0] is a linear transformation from R³ to M^(2, 2).