Matrix Representation of a Linear Transformation: Example Solved Problems Using Theorems

MATRIX REPRESENTATION OF A LINEAR TRANSFORMATION

EXAMPLE PROBLEMS USING THEOREMS

Example 13

Let T: R² → R³ and U: R² → R³ be the linear transformations respectively defined by T(a₁, a₂) = (a₁ +3a₂, 0, 2a₁−4a₂) and U(a₁, a₂) = (a₁−a₂, 2α₁, 3a₁+2a₂)

Then prove that [T+U]^γ_β = [ T ]^γ_β  + [U]^γ_β

Solution:

Let T: R² → R³ and U: R² → R³.

T(a₁, a₂)=(a₁+3a₂, 0, 2a₁−4a₂) and

U (a₁, a₂) = (a₁−a₂, 2a₁, 3a₁+2a₂)

Let β and γ be the standard ordered bases for R² and R³ respectively.

Let β = {(1, 0), (0, 1)} and

 γ = { (1, 0, 0), (0, 1, 0), (0, 0, 1)}

T (v₁) = T (1, 0) = 1 (1, 0, 0) + 0 (0, 1, 0) + 2 (0, 0, 1)

 [ T(a₁, a₂) = (a₁+3a₂, 0, 2a₁ − 4a₂)]        (ie), a₁ = 1 and a₂ =0

= (1, 0, 0) + (0, 0, 0) + (0, 0, 2)

= (1,0, 2)

T(v₂) = T(0, 1) = 3(1, 0, 0) + 0(0, 1, 0) −4(0, 0, 1)

= (3, 0, 0) + (0, 0, 0)+(0, 0, −4)

= (3, 0, −4)

                  ...(1)

Let β ={(1, 0), (0, 1))

 γ = { (1, 0, 0), (0, 1, 0), (0, 0, 1) }

U (v₁) = U (1, 0) = 1 (1, 0, 0) + 2 (0, 1, 0) + 3 (0, 0, 1)

        [  U (a₁, a₂) = (a₁ − a₂, 2a₁, 3a₁ +2a₂)]

  = (1, 0, 0) + (0, 2, 0) + (0, 0, 3)

 = (1, 2, 3)

 (ie) a₁ = 1 and a₂ = 0

U (v2)=U (0, 1) = −1 (1, 0, 0) + 0 (0, 1, 0) + 2 (0, 0, 1)

=(−1, 0, 0)+(0, 0, 0)+(0, 0, 2)

= (−1, 0, 2)

              ...(2)

Now let us compute T+U, then

(T+U) (a₁, a₂) = (a₁ +3a₂, 0, 2a₁−4a₂)+(a₁−a₂, 2a₁, 3a₁+2a₂)

= (2a₁+2a₂, 2a₁, 5a₁−2a₂)

(T+U) (v₁) = (T+U) (1, 0) = 2 (1, 0, 0)+2 (0, 1, 0) + 5 (0, 0, 1)

= (2, 0, 0)+(0, 2, 0)+(0, 0, 5)

= (2,2,5)

(T+U) (v₂) = (T+U) (0, 1) = 2 (1, 0, 0)+0 (1, 0, 0)−2 (0, 0, 1)

= (2, 0, 0) + (0, 0, 0) + (0, 0, − 2)

= (2, 0, −2)

              ...(3)

Now let us compute

  From equations (3) & (4) we have proved that

[T+U]^γ_β = [ T ]^γ_β  + [U]^γ_β

Example 14

Let T: R²→R³, be the linear transformation defined by T (a₁, a₂) = (a₁+3a₂, 0, 2α₁−4a₂). Let β be the standard ordered basis of R² and R³. Then prove that [aT]^γ_β =a[T]^γ_β for all scalars a.

Solution:

Let T: R²→R³

T (a₁, a₂) = (a₁ +3a₂, 0, 2a₁ − 4a₂)

Let

β = {(1, 0), (0, 1)} and

γ= {(1, 0, 0), (0, 1, 0), (0, 0, 1)}

aT (v₁) = aT (1, 0) = (a, 0, 2a)         (Refer the previous example)

aT (v₂) = aT (0, 1) = (3a, 0, −4a)       (Refer the previous example)

               ...(1)

We know that from the previous example

                  ...(2)

From equations (1) & (2) it is proved as

[aT]^γ_β =a[T]^γ_β for all scalars a.

Example 15

Let V be the vector space of complex number over R. Let T: V→ V defined by T (Z)=, where  is the complex conjugate. Prove that T is linear and then compute [T]_β where β={1,i}

Solution:

Let us consider the complex number Z= a + ib

Here a and b are all real numbers.

Given that T: V→V defined by T(Z)= = (a−ib)

. T (a+ib) = a – ib    ⇒       T (a, b) = (a, − b)

To prove T is linear let us consider

T[x (a + ib) + y (c+id)] = T [(xa+yc) + i (xb+yd)]

= (xa + yc) ‒ i(xb + yd)

= (xa ‒ ixb) + (yc−iyd)

= x(a−ib) + y (c−id)

= xT(a+ib) + yT(c+id)

T is linear.

It is given that the basis is β = { 1, i }

⇒ {1+0i, 0+ li}

= {(1, 0), (0, 1)}

Let v₁ = (1, 0) and v₂ = (0, 1)

T(v₁)=T(1, 0) = (1, 0) = 1

T(v₂)=T(0, 1) = (0,−1) = −i                  (T(z)=)

T(v1)=T(1, 0) = T (1, 0) = 1 (1, 0)+0 (0, 1)

= (1, 0)+(0, 0)

= (1, 0)

T(v₂) = T (0, − 1) = 0 (1, 0) ‒ 1 (0, 1)

= (0, 0) + (0, − 1)

= (0,−1)

Example 16

 Let V be n−dimensional vector space with an ordered basis β, which is defined by T : V→Fⁿ by T (x)= [x]_β. Prove that it is linear.

Solution:

It is given that V is an n−dimensional vector space over F and sional vector B = { v₁, v₂ ... v_n } is its basis.

Every vector in V can be written as a linear combination of β = { v₁, v₂, v₃ ... v_n }.

Also T: V→Fⁿ by T(x) = [x]_β.

Here [x]_β is the coefficient of linear combination of

β = { v₁, v₂... v_n}

We know that T(x+y) = T(x)+T (y)

and also we know that [cx]_β = c[x]_β

Hence T is linear.