Linear Transformation: Example Solved Problems

LINEAR TRANSFORMATION

WORKED EXAMPLE PROBLEMS

Example 1

Show that T is linear when T: R²→R² is defined by T (a₁, a₂)=(2a₁+a₂, α₁)

Solution:

Let c ∈ F; x, y ∈ V=R²

Let x=(b₁, b₂) and y = (d₁, d₂).

 x+y=(b₁+d₁, b₂+ d₂)

 cx+y = (cb₁ + d₁, cb₂ + d₂)

L.H.S

 T(cx+y)= T (cb₁ + d₁, cb₂ + d₂)

= (2 (cb₁ + d₁) + cb₂+ d₂, cb₁ + d₁)                              ...(1)

R.H.S

cT(x) + T(y) = c (2b₁ + b₂, b₁) + (2d₁ + d₂, d₁)

= (2cb₁+ cb₂ +2d₁ + d₂, cb₁+d₁)

= (2 (cb₁+d₁) + cb₂+ d₂, cb₁ + d₁)            ...(2)

From (1) and (2)

T(cx+y) = cT(x) + T(y)

L.H.S = R.H.S

Hence T is linear.

 

Example 2

Show that T_θ is linear when T_θ: R² → R².

Solution:

T_θ (a₁, a₂) is the vector obtained by rotating (a₁, a₂) counter clockwise by θ if (a₁, a₂) ≠ (0, 0) and

 T_θ(0, 0) = (0, 0). Then T_θ : R² → R² is a linear transformation.

 

Example 3

Show that T is linear when T: R²→R² is defined by T(a₁, a₂) = (a₁ ‒ a₂)

Solution:

Let c ∈ F: x, y ∈ V = R².

Let x=(b₁, b₂) and y = (d₁, d₂)

 x+y = (b₁ + d₁, b₂ + d₂)

 cx+y = (cb₁, cb₂) + (d₁, d₂) = (cb₁+d₁, cb₂+ d₂)

Let us consider the L.H.S

T(cx+y) = T (cb₁ + d₁, cb₂+ d₂)

= (cb₁+d₁, −cb₂−d₂)                 ……….(1)

Let us consider the R.H.S

cT(x)+T (y) = c(b₁, b₂) + (d₁, d₂)

= (cb₁,−cb₂) + (d₁,− d₂)

= (cb₁+d₁,− cb₂, −d₂)                 ……….(2)

From equation (1) and (2), we have

 T(cx+y) = cT(x) + T(y)

Hence T is linear.

 

Example 4

Show that T is not linear when T: R²→R² which is defined by T(a₁, a₂) = (sin a₁, 0).

Solution:

It is given that T (a₁, a₂) = (sin a₁, 0)

Let T(2π/2, 0) = Т (π, 0) = (sin π, 0)

= (0, 0)                   ...(1)

Now let T(2π/2, 0) = 2T(2π/2, 0)

= 2(sin(π/2), 0) =

 = 2 (1, 0)

 = (2,0)                                 ...(2)

From (1) and (2) we have (0, 0) ≠ (2, 0). Hence T is not linear.

 

Example 5

Let T: R³→R² which is defined by T(1,0,3)=(1, 1) and T(−2,0,−6)=(2, 1).

Verify whether T is a linear transformation or not?

Solution:

It is given that T(1, 0, 3) = (1, 1) and T(−2, 0, −6)=(2, 1).

To prove it is linear transformation, we have to show that

 T(x+y)=T(x)+T (y) and T (cx) = cT (x).

Now let us consider T(−2, 0, − 6) = T [− 2 (1, 0, 3)]

= −2T(1, 0, 3)

 = −2(1, 1)            (T(1, 0, 3) = (1, 1))

 = (−2,−2)

 T(−2,0,− 6) ≠ (2, 1)

Therefore T is not linear transformation on R³ → R².

 

Example 6

Define T: R² → R³ by T (x, y) = (x+2y, 2x − y, x+5y) show that T is linear.

Solution:

Let c ∈ F; x, y ∈ V=R³

where x= (a₁, a₂) and y=(b₁, b₂)

 cx+y = (ca₁, ca₂) + (b₁, b₂)

 = (ca₁ + b₁, ca₂ + b₂)

T(x+y) = (x+2y, 2x − y, x+5y)

L.H.S: T(cx+y) = (ca₁ + b₁ + 2(ca₂+b₂), 2(ca₁ + b₁) − ca₂−b₂), (ca₁ + b₁ + 5(ca₂ + b₂))                  

             ….(1)

RHS: CT (x) + T (y) = (ca₁+2a₂c, 2ca₁−ca₂, ca₁ +5ca₂) + (b₁+2b₂, 2b₁−b₂, b₁+5b₂)

= (ca₁ + b₁ + 2(ca₂+ b₂), 2(ca₁ + b₁) − ca₂ − b₂, (ca₁+b₁+ 5(ca₂+ b₂))

        ….(2)

From (1) & (2)

 T(x + y) = cT(x) + T(y)

 T is linear.

 

Example 7

Let T: R² →R² by T (x, y) = (x+2y, 2x + 3y + 2) check whether T is linear or not.

Solution:

Let c ∈ F, and x, y ∈ V = R²

where

 x = (a₁, a₂), y = (b₁, b₂) and (cx + y) = (ca₁ + b₁, ca₂+ b₂)

LHS:

 T(cx+y) = (ca₁ + 2ca₂ + b₁ +2b₂, 2ca₁ +2b₁ + 3ca₂ + 3b₂+2)

              ...(1)

R.H.S:

T(x) = (a₁ +2a₂, 2a₁ + 3α₂+2)

cT(x) = (ca₁+2ca₂, 2ca₁ + 3ca₂ + 2c)

T(y) = (b₁+2b₂, 2b₁+3b₂+2)

 cT(x)+ T(y) = (ca₁ +b₁+2ca₂ + 2b₂, 2ca₁ + 3ca₂+2c+2b₁+3b₂+2)

                   ……….(2)

From (1) and (2) we have (1) (2)

  T is not linear.

 

Example 8

Define T: P_n(R) → P_n−1 (R) by T (f(x)) =f '(x) where f '(x) denotes the derivative of  f(x). Show that T is linear.

Solution:

 Let g(x), h(x) ∈ P_n(R), c ∈ F,

T (cg(x) + h(x)) = d/dx (cg (x) + h (x))

 = cg'(x) + h'(x)

 = cT(g(x)) + T(h (x))

T is linear.

 

Example 9

Let T: R² →R² be defined by T (a₁, a₂) = (a₁, a₂²). Verify whether T is linear or not.

Solution:

Given that T(a₁, a₂) = (a₁, a₂²)

Let c ∈ F; x,y ∈ V = R²

Let x= (b₁, b₂) and y = (d₁, d₂)

 x+y=(b₁+d₁, b₂+ d₂)

 c(x+y)=(cb₁ + d₁, cb₂ + d₂)

T(cx+y)=T(cb₁ + d₁, cb₂ + d₂)

= [cb₁ + d₁, (cb₂+ d₂)²]

= ( cb₁ + d₁, c²b₂² + đ₂² + 2cb₂d₂)           ...(1)

 cT(x) + T(y) = [ c (b₁, b₂²) + ( d₁, d₂²) ]

 = (cb₁,cb₂²)+(d₁,d₂²)

= (cb₁+d₁, cb₂² + d₂²)                   ...(2)

From equations (1) & (2), we get T (cx+y) ≠ T (x) + T (y)

 T is not linear.

 

Example 10

A function defined by T: V→W if T (x + y) = T (x)+T (y) for all x,y ∈ V. Prove that if V and W are vector spaces over the field of rational numbers, then any additive function from V into W is a linear transformation.

Solution:

Given that T: V→ W.

Defined by T(x+y)=T(x) + T(y) for all x, y ∈ V.

Let c = a/b ∈ Q

Then now expand the above function upto "b" times

 (ie) T(x) = T ( (1/b)x + (1/b)x + (1/b)x + ... up to b times )

= bT[ (1/b)x ]

= b.(1/b)T(x)

= T(x)

 T(x)=T(x)

Then T (cx) = T ( (1/b)x + (1/b)x +(1/b)x +… up to a times )

= aT ( (1/b)x )

= a . (1/b)T(x)

= a/b . T(x)

 T(cx) = c T(x)

T is linear transformation.

 

Example 11

Let the function T: R³→R² defined by T (a₁, a₂, a₃) = (a₁−a₂, 2a₃). Prove that T is linear.

Solution:

Given that T (a₁, a₂, a₃) = (a₁, ‒ a₂, 2a₃)

Let c ∈ F; x, y ∈ V=R³

Let x = (a₁, a₂, α₃) and y = (b₁, b₂, b₃)

 x+y = (a₁, a₂, a₃) + (b₁, b₂, b₃)

= (a₁+b₁, a₂+ b₂, a₃ + b₃)

c(x+y)=(ca₁+b₁, ca₂+ b₂, ca₃ +b₃)

Now we have to prove that T(cx + y) = cT(x) + T (y)

Now T (cx+y)= T (ca₁ + b₁, ca₂ + b₂, ca₃ + b₃)

= ( (ca₁ + b₁) − (ca₂ + b₂), 2 (ca₃ + b₃) )

= ((ca₁ − ca₂) + b₁ − b₂, 2(ca₃ + b₃))

T(cx+y)=c(a₁−a₂) + b₁ − b₂, 2 (ca₃ + b₃)                  ………..(1)

Then cT(x)+T (y) = cT (a₁, a₂, a₃) + T(b₁, b₂, b₃)

= c(a₁ − a₂, 2a₃) + (b₁ − b₂, 2b₃)

= (ca₁−ca₂, 2ca₃) + (b₁−b₂, 2b₃)

= ca₁−ca₂+b₁−b₂, 2ca₃−2b₃    

= c(a₁ − a₂) + b₁ − b₂, 2(ca₃ + b₃)              ...(2)

From equations (1) & (2) we have

 T(cx+y) = c T(x) + T(y)

Hence T is linear.

 

Example 12

Let T=R² → R³. Define T (a₁, a₂) = (a₁+a₂, 0, 2α₁ − a₂) Verify whether T is linear or not.

Solution:

Given that T (a₁, a₂) = (a₁ + a₂, 0, 2a₁−a₂)

Let c ∈ F; x, y ∈ R³.

Let x = (a₁, a₂) & y = (b₁, b₂)

⇒ x+y = (a₁, a₂) + (b₁, b₂)

= (a₁+b₁, a₂+b₂)

c(x+y) = (ca₁+b₁, ca₂+ b₂)

T(cx+y)=T(ca₁ + b₁, ca₂+b₂)

= (ca₁ + b₁ + ca₂ + b₂, 0, 2 (c a₁ + b₁) − (c a₂+b₂)

= ( c (a₁ + a₂) + b₁ + b₂, 0, c (2a₁−a₂) + (2b₁−b₂))

                            ........(1)

Now cT(x)+T(y) = cT(a₁, a₂, a₃) + T(b₁, b₂, b₃)

= c(a₁+a₂, 0, 2a₁−a₂) + (b₁+b₂, 0, 2b₁−b₂)

= ( c (a₁ + a₂) + b₁ + b₂, 0, c (2a₁ – a₂) + (2b₁ − b₂))

                      ………(2)

From equations (1) & (2)

We get T(cx+y)= cT(x)+T(y)

  T is linear.

 

Example 13

Let T=M_2×3 (F) → M_2×2 (F).

Prove that T is linear.

Solution:

From equations (1) & (2) we get

T(cx+y) = cT(x) + T(y)

Hence T is linear.

 

Example 14

Let T: M_n×n(F) → F. Define T(A) = Trace(A).

Where Trace (4)= ⁿΣ_i=1 A_ii. Prove that T is linear.

Solution:

Given that T: M_n×n(F) → F and T(A)= Trace (A).

Let R, S ∈ M_n×n (F) such that T (R) = Trace (R) & T (S) = Trace (S).

Now we know that Tc (cx+y) = cT(x) + T(y)

T (R + S) = Trace (R + S) =

= (R_ii + S_ii)

= R_ii + S_ii

= T (R) + T(S)

  T (R+ S) = T (R) + T (S)              ………….(1)

Let c ∈ F and R ∈ M_n×n(F)

Now T (cR) Trace (cR)

= (cR_ii)

= c R_ii

= cT(R)

From (1) & (2): T(cR) = cT(R)

Hence T is linear.

 

Example 15

Let T: P₂(R) → P₃(R). Let us define T[f(x)]=xf(x)+f '(x). Verify that whether T is linear or not.

Solution:

Given that T: P₂ (R) → P₃ (R) & T[f(x)]=xf(x) +ƒ′(x).

To prove it is linear, we have to prove that

 T(cx+y) = cT(x) + T(y)

T[(f+g) x] = x(f + g) (x) + (f + g)′ (x)

= xf(x)+x g(x) +f '(x)+g'(x)

= xf(x)+f '(x)+ x g(x) + g'(x)

= T [f(x)] + T [g (x)]                 ………..(1)

Now T[cf(x)] = cxf(x)+cf '(x)

= c[xf(x)+f '(x)]

= c T [f(x)]                     ………..(2)

  From (1) & (2) it is clear that T is linear.

 

Example 16

Let T: C→ C, defined by T (z) =. Show that T is not linear with respect to the additive operation.

Solution:

It is given that T: C→C and T (z) = 

where  is the complex conjugate of Z.

We know that T (x, y) = T (x) + T (y) for all x, y ∈ C.

Let us consider the following complex numbers as

 x= a + ib and y=c+id for a, b, c, d ∈ R.

Then T(x+y)=T(a+ib)+T (c+id)

= T[(a+c) + i(b + d)]

Since it is given that T (z) = , then

T(x+y)=T[(a+c)+i (b+d)]

= (a+c)−i (b+d)

= (a−ib)+(c−id)

= 

T(x+y)=T(x)+T (y)

T is additive.

Now we have to consider T (cx) = cT (x)

Let x = a + ib, α = i ∈ C

T (αx) = T (ix)= T [i (a + ib)]

= T[ia+i²b]

= T (ia − b)

= T (−b + ia)

                   ('.'T(z) =)

T (αx) = −b−ia         ……………(1)

 α T(x) = iT(a + ib) = i (a − ib) = b + ia             (2)

From equations (1) and (2), we have

T(αx) ≠ αT(x)

Hence T is not linear.

 

Example 17

Let V and W be two vector spaces. Let T: V→W be linear and { w₁, w₂, w₃ ... w_n} be a linearly independent subset of R (T). Then prove that if S={v₁, v₂, v₃... v_n} is chosen so that T (v_i)=w_i for i = 1,2... n, then S is linearly independent.

Solution:

Given that T: V→ W is linear and V and W are two vector spaces.

Let {w₁, w₂, w₃ ... w_n} be a linearly independent subset of R (T).

Let S = { v₁, v₂, v₃... v_n} be a subset of V such that

T (v_i) = w_i for i = 1, 2, 3 ...n.

To show that S is linearly independent, we know that

 a₁v₁ + a₂ν₂ + a₃v₃ ... + a_nv_n = 0

T (a₁v₁ + a₂ν₂ + a₃v₃ ... + a_nv_n) = T (0)

T (a₁ v₁) +T (a₂ v₂) +T (a₃v₃) + ... + T(a_nv_n) = 0

 a₁T (v₁) + a₂T(v₂) + ... + a_nT(v_n)=0

 a₁w₁ + a₂w₂ + a₃w₃ ... + a_nw_n = 0

. The set {w₁, w₂, w₃... w_n} is linearly independent.

         ⇒      a₁=α₂=a₃= …. = a_n = 0

  S = { v₁, v₂, v₃ ..... v_n } is linearly independent.

 

Example 18

Let V be the vector space of sequences {a_n}+{ b_n } = { a_n + b_n } and t{a_n}={ta_n} where {a_n} and {b_n} ∈ V and t ∈ F. The function T, U: V→V is defined by T (a₁, a₂, a₃...) = (a₂, a₃, a₄ ...) and U (a₁, a₂, a₃) = (0, a₁, a₂...). Prove that T and U are linear.

Solution:

It is given that T, U: V→ V and V is a vector space of sequences {a_n}.

Given that T (a₁, a₂, a₃ ...) = (α₂, a₃, a₄ ...)

  U (a₁, a₂, a₃...) = (0, a₁, a₂...)

To prove T is linear, let (a₁, a₂, a₃ .....), (b₁, b₂, b₃ ...) be any set of V and c is a scalar.

We know that T (cx) = cT (x)

.. T (c (a₁, a₂ ...) + (b₁, b₂, ...))

= T (ca₁ + b₁, ca₂ + b₂, ca₃ + b₃ ...)

= (ca₂ + b₂, ca₃ + b₃ ...)

= (ca₂, ca₃, ca₄...) + (b₂, b₃, b₄ ...)

= c(a₂, a₃, a₄...) + (b₂ b₃, b₄ ...)

= cT (a₁, a₂...) + T (b₁, b₂, b₃ ...)

Hence T is linear.

Now U (c (a₁, a₂, ...) + (b₁, b₂, b₃ ...))

= U (ca₁+b₁, ca₂+ b₂, ca₃+b₃, ...)

= (0, ca₁ + b₁, ca₂ + b₂, ca₃ + b₃ ...)

= (0 ca₁, ca₂, ca₃...) + (0, b₁, b₂, b₃ ...)

= c(0, a₁, a₂...) + (0, b₁, b₂, b₃ ...)

= cU (a₁, a₂, a₃ ...) + U (b₁, b₂, b₃ ...)

Hence U is also linear.