Recipe

Verify a claimed identity between maps in L(V,W)

Use when: you're given a candidate map built from other linear maps, such as a sum, a scalar multiple, or a claimed identity element, and need to confirm a stated equality actually holds

Goal: a vector-by-vector confirmation that the claimed equality holds for every vector in the domain, not just a few checked numbers

Taught in L(V,W) is itself a vector space

1. 1

   State how the new map acts on a general vector, using the pointwise definitions of addition and scalar multiplication given earlier in this section.

   Example

   For T_1(x,y)=(x+y,x) and T_2(x,y)=(x-y,y), the definition (T_1+T_2)(\mathbf v)=T_1(\mathbf v)+T_2(\mathbf v) gives (T_1+T_2)(x,y) = (x+y,x)+(x-y,y) = (2x,\ x+y)

   Why: pointwise addition of maps is only a definition until it is applied to an actual vector, so writing the formula out is a real step, not a formality.

2. 2

   Apply the candidate map to (x_1,y_1)+(x_2,y_2), then separately apply it to each piece and add the two results together.

   Example

   (T_1+T_2)\big((x_1,y_1)+(x_2,y_2)\big) = (2(x_1+x_2),\ (x_1+x_2)+(y_1+y_2)), which splits into (2x_1,x_1+y_1)+(2x_2,x_2+y_2), matching (T_1+T_2)(x_1,y_1)+(T_1+T_2)(x_2,y_2) exactly.

   Why: additivity is one of the two defining conditions of linearity, and it must hold for every pair of vectors, not just one.

3. 3

   Apply the candidate map to c(x,y) for an arbitrary scalar c, then compare the result to c times the map applied to (x,y) directly.

   Example

   (T_1+T_2)(cx,cy) = (2cx,\ cx+cy) = c(2x,\ x+y) = c\,(T_1+T_2)(x,y), so the scalar factors out cleanly.

   Why: homogeneity is the second defining condition of linearity, and additivity alone can never guarantee it on its own.

4. 4

   Look back at the vector used in Steps 1 through 3. If it was written with general letters rather than specific numbers, the confirmed equality already covers every vector in the domain at once.

   • the vector used throughout was fully general (letters, not numbers) — the identity is proved for all of V, and no further checking is needed.
   • specific numbers were substituted somewhere along the way — redo that computation with general letters before treating anything as proved.

   Why: an identity confirmed on one numeric example says nothing about a different vector, but one confirmed on a genuinely arbitrary vector covers the whole space at once.