Question

Prove that every linear transformation T:R^2->R is differentiable

Answer 1

is this true?

Answer 2

Is is given to prove..

Answer 3

@ganeshie8

Answer 4

@SolomonZelman

Answer 5

hi

Answer 6

Well you have something of the form \(T(a, b) = c_1a + c_2b\) which is of course differentiable the derivative is just \((c_1\ \ c_2)\)

Answer 7

Prove to yourself that every linear transformation \(T : \mathbb{R}^2 \to \mathbb{R}\) can be expressed as above and you are done.

Answer 8

Every vector in \(\mathbb{R}^2\) can be expressed as a linear combination of the standard basis.

Then we have the following by the above and linearity.
\(T(a\cdot(1, 0) + b\cdot(0, 1)) = T(1, 0)\cdot a + T(0, 1) \cdot b = c_1a + c_2b \)

Then the partials are \(\frac{\partial T}{\partial x} =c_1\) and \(\frac{\partial T}{\partial y} =c_2\)

Do you understand?

Answer 9

The question is a little silly to begin with. The derivative of a function at a point is the best linear approximation at that point. However, what is the best linear approximation of a linear function, it is just the linear function itself.

Answer 10

@Alchemista I proved till T(x,y)=ax+by, where a=T(1,0) and b=T(0,1). But how to show that the partial derivatives are a and b?

Answer 11

T(x+h,y+k)-T(x,y)=ah+bk

Answer 12

Why are you going from first principles.

Answer 13

I meant, how to show that T(1,0)= del T/delx and T(0,1)=del T/del y

Answer 14

When you take the \(x\)'th partial of \(ax + by\) simply treat \(y\) as a constant and differentiate with respect to \(x\). Then do the same thing but with \(x\) as a constant for the \(y\)'th partial.

Answer 15

Oops , oh yes now I got it. We have the reduce the form of T(x,y) form to ax+by.. Yes thanks.:)