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.