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
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asked 2021-01-17
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.

Expert Answers (1)

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.

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