Let f : R → R3 be defined by f(x) = <4x,−3x,5x+5>. Is f a linear transformation

since each component is linear, I'd say yes.

of course, to prove it, you'd need to show that

f(cx) = c*f(x)
f(x+y) = f(x)+f(y)
...
huh. I guess not linear.

To determine if f is a linear transformation, we need to check two conditions:

1. f(u + v) = f(u) + f(v) for any vectors u and v.
2. f(ku) = k f(u) for any scalar k and vector u.

Let's check these conditions for the given function f(x) = <4x, -3x, 5x + 5>:

1. Add two vectors u = a and v = b:
f(u + v) = f(a + b)
= <4(a + b), -3(a + b), 5(a + b) + 5>
= <4a + 4b, -3a - 3b, 5a + 5b + 5>

On the other hand, calculate f(u) + f(v):
f(u) + f(v) = <4a, -3a, 5a + 5> + <4b, -3b, 5b + 5>
= <4a + 4b, -3a - 3b, 5a + 5b + 5>

Since f(u + v) = f(u) + f(v), the first condition is satisfied.

2. Multiply a vector u = a by a scalar k:
f(ku) = f(ka)
= <4(ka), -3(ka), 5(ka) + 5>
= <4ka, -3ka, 5ka + 5>

On the other hand, calculate k f(u):
k f(u) = k <4a, -3a, 5a + 5>
= <4ka, -3ka, 5ka + 5k>

Since f(ku) = k f(u), the second condition is satisfied.

Since f satisfies both conditions, f is indeed a linear transformation from R to R^3.

To determine whether the function f is a linear transformation, we need to check if it satisfies two properties:

1. Additivity: f(u + v) = f(u) + f(v) for any vectors u and v.
2. Homogeneity: f(ku) = k * f(u) for any scalar k and vector u.

Let's check these properties for the given function f(x) = <4x, -3x, 5x+5>.

1. Additivity:
Let's consider two vectors, u = a and v = b, where a and b are any real numbers.
f(u + v) = f(a + b) = <4(a + b), -3(a + b), 5(a + b) + 5>
= <4a + 4b, -3a - 3b, 5a + 5b + 5>

f(u) + f(v) = f(a) + f(b) = <4a, -3a, 5a + 5> + <4b, -3b, 5b + 5>
= <4a, -3a, 5a + 5> + <4b, -3b, 5b + 5>
= <4a + 4b, -3a - 3b, 5a + 5b + 5>

Since f(u + v) = f(u) + f(v), the function satisfies the additivity property.

2. Homogeneity:
Let's consider a scalar k and a vector u = a, where a is any real number.
f(ku) = f(ka) = <4(ka), -3(ka), 5(ka) + 5>
= <4ka, -3ka, 5ka + 5>

k * f(u) = k * f(a) = k * <4a, -3a, 5a + 5>
= <4ka, -3ka, 5ka + 5>

Since f(ku) = k * f(u), the function satisfies the homogeneity property.

Therefore, since the function f satisfies both the additivity and homogeneity properties, we can conclude that f is a linear transformation.