Determine whether the fuction is a linear transformation(explain):-

Question

1-T:R²→R²,T(x,y)=(x,1)

2-T:R³→R³,T(x,y,z)=(x+y,x-y,z)

Answer 1

Hi Lavi!

All linear transformations will satisfy two properties:

1. T(u + v) = T(u) + T(v)
2. T(c*u) = c*T(u)

The first property gives us the extra fact that T(0) = 0 for all linear transformations T. This helps with the first one. There T(0) = T( (0,0) ) = (0,1) != 0. So 1-T cannot be linear. But the second satisfies T(0) = 0, so we'll need more work for that one.

The two properties above actually suffice to show that a transformation is linear.

Shown below is proof they apply to 2-T.

T( u + v ) = T( (u₁,u₂,u₃) + (v₁,v₂,v₃) )

= T( (u₁ + v₁, u₂ + v₂, u₃ + v₃) )

= ( (u₁ + v₁) + (u₂ + v₂), (u₁ + v₁) - (u₂ + v₂), u₃ + v₃)

= ( (u₁ + u₂) + (v₁ + v₂), (u₁ - u₂) + (v₁ - v₂), u₃ + v₃)

= ( (u₁ + u₂), (u₁ - u₂), u₃ ) + ( (v₁ + v₂), (v₁ - v₂), v₃ )

= T( (u₁, u₂, u₃) ) + T( (v₁, v₂, v₃) )

= T(u) + T(v), and

T(c*u) = T(c*(u₁, u₂, u₃))

= T( (c*u₁,c*u₂,c*u₃)

= ( (c*u₁ + c*u₂), (c*u₁ - c*u₂), c*u₃ )

= ( c*(u₁ + u₂), c*(u₁ - u₂), c*u₃ )

= c*( (u₁ + u₂), (u₁ - u₂), u₃ )

= c*T(u).

So 2-T satisfies both properties, and is therefore a linear transformation.

Note that 1-T satisfies the first property, as shown below:

T( u + v ) = T( (u₁,u₂) + (v₁,v₂) )

= T(u₁ + v₁, u₂ + v₂)

= (u₁ + v₁, 0)

= (u₁, 0) + (v₁, 0)

= T(u₁, u₂) + T(v₁, v₂)

= T(u) + T(v), but

T(c*u) = T(c*(u₁, u₂))

= T(c*u₁, c*u₂)

= (c*u₁, 1) which only equals c*T(u) = c*T(u₁, u₂) = c*(u₁, 1) = (c*u₁, c) if c=1.

This second property is where 1-T failed to be a linear transformation.

Hope this helps. I'll be glad to update this to make it clearer, so I'd appreciate any feedback. Thank you!