Let ß = ( b 1 , . . . , b n ) be a basis of the vector space V, let T : V → V be a linear transformation of V, and let B be the ß-matrix of T.(a) Prove that v ∈ ker ⁡ ( T ) if and only if [ v ] B ∈ ker ⁡ ( B ).(b) Prove that v ∈ Im(T) if and only if [ v ] B ∈ Im(T).(c) Prove that T is an isomorphism if and only if B is invertible.

Question

Let   ß  =  (      b    1    ,  .  .  .  ,      b    n    ) be a basis of the vector space   V, let   T  :  V  →  V be a linear transformation of   V, and let B be the   ß-matrix of   T.(a) Prove that   v  ∈  ker  ⁡  (  T  ) if and only if   [  v      ]    B    ∈  ker  ⁡  (  B  ).(b) Prove that   v  ∈  Im(T) if and only if   [  v      ]    B    ∈  Im(T).(c) Prove that   T is an isomorphism if and only if   B is invertible.

Keyon Fitzgerald · Accepted Answer

If Tv=w, then because B is the β-matrix of T, B[v]B=[w]B.(a) Assume first that   v  ∈  ker  ⁡  (  T  )  .. Then   T  v  =  0  ,, so   B  [  v      ]    B    =  [  0      ]    B    =  0.. Hence,   [  v      ]    B    ∈  ker  ⁡  (  B  )  .. Conversely, assume that   [  v      ]    B    ∈  ker  ⁡  (  B  )  .. Then   B  [  v      ]    B    =  0  =  [  0      ]    B    ,, so   T  v  =  0.. Hence,   v  ∈  ker  ⁡  (  T  )  .. Thus,   v  ∈  ker  ⁡  (  T  ) if and only if   [  v      ]    B    ∈  ker  ⁡  (  B  )  ..(b) Assume first that   v  ∈      Im    (  T  )  .. Then there exists a   u  ∈  V such that   T  u  =  v  ,, so   B  [  u      ]    B    =  [  v      ]    B    .. Hence,  [  v      ]    B    ∈      Im    (  B  ). Conversely, assume that   [  v      ]    B    ∈      Im    (  B  )  .. Then there exists a   u  ∈  V such that   B  [  u      ]    B    =  [  v      ]    B    ,, so   T  u  =  v  .. Hence,   v  ∈      Im    (  T  )  .. Thus,   v  ∈      Im    (  T  ) if and only if   [  v      ]    B    ∈      Im    (  B  )  ..(c) Assume first that   T is an isomorphism. Then by definition,   T is a non-singular linear transformation of   V onto   V  ,, so   T is invertible. So if   T  v  =  w  ,, then   v  =      T          −      1        w  ,, i.e., if   B  [  v      ]    B    =  [  w      ]    B    ,, then   [  v      ]    B    =      B          −      1        [  w      ]    B    .. Hence,   B is invertible. Conversely, assume that   B is invertible. Then by the same reasoning as above,   T is invertible, and by the same Theorem 9,   T is non-singular and onto. Hence,   T is an isomorphism. Thus,   T is an isomorphism if and only if   B is invertible.

Let ß = ( b 1 </msub> , . . . , b n

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