How Can Scalars Represent Any Linear Transformation in R^3?

[loli12]

Let T:R^3 -> R be linear. Show that there exist scalars a, b, and c such that T(x, y , z) = ax + by + cz for all (x, y, z) in R^3. State and prove an analogous result for T: F^n -> F^m.

I know that we just have to multiply by a matrix then we can get the desired transformation. But how would I go around to show that such scalars a, b and c exists?

 

[Galileo]

Use the fact that T is linear. This means that T(v+w)=T(v)+T(w).
So if you know a basis for R^3 (there's an obvious one) and you know how T acts on this basis, you know how T acts on every vector in R^3.

 

[quicknote]

As a scientist, it is important to provide a clear and rigorous response. To show that there exist scalars a, b, and c such that T(x, y, z) = ax + by + cz for all (x, y, z) in R^3, we need to prove that T is a linear transformation.

Firstly, let's define T as a linear transformation from R^3 to R. This means that T must satisfy two properties:

1. T(u + v) = T(u) + T(v) for all u, v in R^3
2. T(ku) = kT(u) for all u in R^3 and k in R

To prove the first property, let u = (x1, y1, z1) and v = (x2, y2, z2) be two arbitrary vectors in R^3. Then,

T(u + v) = T(x1 + x2, y1 + y2, z1 + z2)
= a(x1 + x2) + b(y1 + y2) + c(z1 + z2) (since T is a linear transformation)
= (ax1 + by1 + cz1) + (ax2 + by2 + cz2)
= T(u) + T(v)

Therefore, T satisfies the first property.

To prove the second property, let u = (x1, y1, z1) be a vector in R^3 and k be a scalar in R. Then,

T(ku) = T(kx1, ky1, kz1)
= a(kx1) + b(ky1) + c(kz1) (since T is a linear transformation)
= k(ax1 + by1 + cz1)
= kT(u)

Therefore, T satisfies the second property.

Since T satisfies both properties, it is a linear transformation. Now, let's define the scalars a, b, and c as a = T(1,0,0), b = T(0,1,0), and c = T(0,0,1).

Then, for any vector (x, y, z) in R^3, we have

T(x, y, z) = T(x, 0, 0) + T(0, y