Final answer:
We proved that if T is a linear transformation, then S(v) = T(T(v)) is also a linear transformation by verifying that it preserves vector addition and scalar multiplication. An example of a transformation T such that T(T(v)) = 0 for all v in R2 is a projection onto the x-axis followed by projection to the zero vector.
Step-by-step explanation:
To prove that S(v) = T(T(v)) is a linear transformation, we need to verify two properties:
- Linearity over vector addition: S(u + v) = S(u) + S(v) for all vectors u, v in V.
- Linearity over scalar multiplication: S(\(\alpha\)v) = \(\alpha\)S(v) for all vectors v in V and all scalars \(\alpha\).
Since T is already a linear transformation, it preserves vector addition and scalar multiplication:
T(u + v) = T(u) + T(v) and T(\(\alpha\)v) = \(\alpha\)T(v).
Now, we examine S:
S(u + v) = T(T(u + v))
= T(T(u) + T(v))
= T(T(u)) + T(T(v))
= S(u) + S(v).
For scalar multiplication:
S(\(\alpha\)v) = T(T(\(\alpha\)v))
= T(\(\alpha\)T(v))
= (\alpha\)T(T(v))
= (\alpha\)S(v).
Therefore, S preserves both vector addition and scalar multiplication, thus S is a linear transformation.
Now, consider a non-zero transformation T : R2 -> R2 such that T(T(v)) = 0. An example is the transformation that projects every vector onto a line and then projects to the zero vector from that line. Mathematically, consider T defined by:
T([x, y]) = [x, 0]
Applying T twice, we get:
T(T([x, y])) = T([x, 0]) = [0, 0],
which satisfies the condition that T(T(v)) = 0 for every vector v in R2.