Linear Transformation: Example Solved Problems

Example Problems for Linear Transformation

Example 29

Verify the dimension theorem for T: R³→R² defined by T (a₁, a₂, α₃) = (a₁−a₂, 2a₃).

Solution:

We know that the dimension theorem, is stating that, let V and W be vector spaces, and let T: V→ W be linear. If V is finite dimensional then,

 nullity (T) + rank (T) = dim (V)

(ie) N(T)+R(T) = dim (V).

Here R² and R³ are two vector spaces and T: R³ → R² is linear.

The dimension of R³ is 3⇒ dim (V) = 3.

From Example 10, we have N(CT) = 1 and

From Example 14, we have R (CT) = 2

  By dimension theorem we have

N(T)+R(T) = dim (V)

 1+2=3

Hence the dimension theorem is verified.

Example 30

Verify the dimension theorem defined for T: R²→ R³ where T (a₁, a₂)= (a₁+a₂, 0, 2a₁−a₂).

Solution:

We know that the dimension theorem is

  N(T) + R (T) = dim (V)

From Example 12, we know that N (T) = 0

From Example 15, we have R (T) = 2.

Here R², R³ are two vector spaces and T: R² → R³ is linear, The dimension of R² is 2. ⇒ dim (V) = 2.

  N (T) + R (T) = dim (V)

 0+2=2

Hence the dimension theorem is verified.

Example 31

Let V be the vector space of sequences. Let T, U: V→ V be defined by T (a₁, a₂, a₃...) = (a₂, a₃, a₄...) and U(a₁, a₂, a₃ ...) = (0, a₁, a₂, а₃ ...), prove that T is not one−to−one and T is onto.

Solution:

(i) To prove T is not 1−1 and on to

Let us consider (a₂, a₃, α₄ ... ) = (a₂, a₃, a₄ .......)

T (a₁, a₂, a₃ ...) = T (b₁, a₂, a₃ ...)

Here a₁ ≠ b₁.

  (a₁, a₂, a₃ ... ) ≠ (b₁, a₂, a₃ ... )

Hence T is not one−to−one.

Let (a₂, a₃, a₄ ... ) ∈ V. Then we can choose a scalar a₁ ∈ F and T(a₁, a₂, a₃ ...) = (a₂, a₃, α₄ ...)

Here T is onto.

(ii) To prove U is 1−1 not on to

Let (0, a₁, a₂, a₃ ......) = (0, b₁ b₂, B₃ .....)

U (0, a₁, a₂, a₃...) U (0, b₁, b₂, b₃ ...)

On other hand (0, a₁, a₂, a₃ ...) = (0, b₁, b₂, b₃ ...)

⇒  (a₁, a₂, a₃ ...) = (b₁, b₂, b₃ ...)

  U is one to one.

We know that if {a_n} is a sequence in V, then 0 is not the first number of this sequence.

Hence (0, a₁, a₂, a₃ ...) ∉ V such that

 U(a₁, a₂, a₃ ...) ∉ (0, a₁, a₂, a₃ ... )

Hence U is not onto.

Example 32

Let W be a subspace of V. Define T: V→V is linear. Prove that the subspace {0}, V, R (T) and N (T) are T−invariant.

Solution:

If W is T−invariant, define

 T_w: W→W by T_w(x) = T(x) for all x ∈ W.

Let W={0}. Since T is linear T (0) = 0, T(x) ∈ W.

W= {0} is T−invariant.

Let the subspace W= V.

Since T: V→ V, T (V)  V.

 W=V is T−invariant.

Let the subspace W = R(T)

Let x ∈ R(T)  ⇒  y ∈ V such that T (y) = x.

T [T (y)] = T (x)

Since T(y) ⇒ V, T(x) ∈ R(T) and so T (R (T))  R (T)

 W = R(T) is T−invariant.

Let the subspace W = N(T).

Let x ∈ N (T)  ⇒  T(x) = 0.

Since T (0)=0      ⇒    0 ∈ N (T), T(x)=0 ∈ N(T)

 TIN (T)]  N (T)

W= N(T) is T−invariant.