Question # Guided Proof Let {v_{1}, v_{2}, .... V_{n}} be a basis for a vector space V. Prove that if a linear transformation T : V rightarrow V satisfies T (v_{

Transformation properties
ANSWERED Guided Proof Let $${v_{1}, v_{2}, .... V_{n}}$$ be a basis for a vector space V.
Prove that if a linear transformation $$T : V \rightarrow V$$ satisfies
$$T (v_{i}) = 0\ for\ i = 1, 2,..., n,$$ then T is the zero transformation.
To prove that T is the zero transformation, you need to show that $$T(v) = 0$$ for every vector v in V.
(i) Let v be the arbitrary vector in V such that $$v = c_{1} v_{1} + c_{2} v_{2} +\cdots + c_{n} V_{n}$$
(ii) Use the definition and properties of linear transformations to rewrite T(v) as a linear combination of $$T(v_{j})$$ .
(iii) Use the fact that $$T (v_{j}) = 0$$
to conclude that $$T (v) = 0,$$ making T the zero transformation. 2021-01-18

a)Given:
The linear transformation $$T : V \rightarrow V$$
represented as $$T (v_{i}) = 0\ for\ i = 1, 2, ..., n.$$
Approach:
Consider an arbitrary $$v = {v_{1}, v_{2},..., v_{n}}$$ is basis for V.
The function T is said to be linear transformation if it satisfies the vector addition and scalar multiplication properties.
The linear transformation is given by,
$$T(v_{i}) = 0, i = 1, 2, \cdots,n...(1)$$
Calculation:
As the vector set v is the subspace of V, the vector v can be written linear combination.
Write the subspace vas linear combination.
$$v = c_{1} v_{1} + c_{2} v_{2} + \cdots + c_{n} v_{n} \cdots, (2)$$
Here, $$c_{1}, c_{2}, \cdots c_{n}$$ are arbitrary scalars.
Conclusion:
Hence, it is proved above that the set $$(v_{1}, v_{2}, \cdots v_{n})$$ is represented as
$$v = c_{1} n_{1} + c_{2} v_{2}, + \cdots + c_{n} v_{n}.$$
b)Given:
The linear transformation $$T : V \rightarrow V$$
represented as $$T (v_{i}) = 0\ for\ i = 1, 2, \cdots, n.$$
Approach:
Consider an arbitrary $$v = {v_{1}, v_{2},\cdots, v_{n}}$$ is basis for V.
The function T is said to be linear transformation if it satisfies the vector addition and scalar multiplication properties.
The linear transformation is given by,
$$T(v_{i}) = 0, i = 1, 2, \cdots,n \cdots(1)$$
The vector additinon is given by,
$$T (u + v) = T (u) + T (v)$$
The scalar multiplication is given by,
$$T (cu) = cT (u)$$
Calculation:
As the vector set v is the subspace of V, the vector v can be written linear combination.
Write the subspace vas linear combination.
$$T(v) = (c_{1} v_{1} + c_{2} v_{2} + \cdots + c_{n} v_{n})$$
$$= T(c_{1} v_{1}) + T(c_{2} v_{2}) + \cdots + T (c_{n} v_{n})$$
$$= c_{1} T(v_{1}) + c_{2} T( v_{2}) + \cdots + c_{n} T(v_{n})....(3)$$
Conclusion:
The transformation form of linear combination $$v = c_{1} v_{1} + c_{2} v_{2} + \cdots + c_{n} v_{n}$$ is
$$T(v) = c_{1} T(v_{1}) + c_{2} T( v_{2}) + \cdots + c_{n} T(v_{n})$$
c) Given:
The linear transformation $$T : V \rightarrow V$$
represented as $$T (v_{i}) = 0\ for\ i = 1, 2, \cdots, n.$$
Approach:
Consider an arbitrary $$v = {v_{1}, v_{2},\cdots, v_{n}}$$ is basis for V.
The function T is said to be linear transformation if it satisfies the vector addition and scalar multiplication properties.
The linear transformation is given by,
$$T(v_{i}) = 0, i = 1, 2, \cdots,n \cdots(1)$$
The vector additinon is given by,
$$T(u + v) = T(u) + T(v)$$
The scalar multiplication is given by,
$$T(cu) = cT(u)$$
Calculation:
Solve formula (3) with use of formula(1)
$$T(v) = c_{1} T(v_{1}) + c_{2} T( v_{2}) + \cdots + c_{n} T(v_{n})$$
$$= c_{1} (0) + c_{2} (0) + \cdots + c_{n} (0)$$
= 0
From above calculation is is clear linear transformation $$T : V \rightarrow V$$
satisfies $$T (v_{i}) = 0\ for\ i = 1, 2, \cdots, n,$$ than T is the zero transformation.
Conclusion:
Hence, ithe solution of $$T(v) = c_{1} T(v_{1}) + c_{2} T( v_{2}) + \cdots + c_{n} T(v_{n})$$ is 0 which shows that T is zero transformation.