Show That a 2D Linear Transform $ T \left( \cdot \right) $ Is Homogeneous

Question

By my understanding, a transform T is homogeneous if T[0] = 0.

Then to prove that a linear transformation is homogeneous we say that:

T[ax(n1, n2) + bx(n1, n2)] = aT[x(n1, n2)] + bT[x(n1, n2)]

What I want to know is the difference between the following two proofs and which one is "more correct" or is there perhaps a even better answer?

let x(n1, n2) = 0

then we know that,

ax(n1,n2)- ax(n1,n2) = 0, T[ax(n1, n2) + ax(n1, n2)] = aT[x(n1, n2)] + aT[x(n1, n2)] = 0

VS

Simply showing that both sides are equal to 0 if x(n1, n2) = 0 through algebra.

On a side note, can someone also comment on why we would even care if a transformation is homogeneous vs heterogeneous? In reference to a transformation applied onto a 2D signal (such as an image transformation).

Maybe the better question to ask would be to prove if a transformation is heterogeneous, is it linear? (if so can someone help me define what it means for a transformation to be heterogeneous? Like how a homogeneous transform means T[0] = 0.

Answer 1

The idea here is that any Linear Operator is Homogeneous though not every Homogeneous Operator is Linear.

The classic proof indeed is to build the zero term as a sum (Difference) of 2 other elements in the domain. Then use Linearity to show it must be zero:

$$ T \left( 0, 0 \right) = T \left( {m}_{1} - {m}_{1}, {n}_{1} - {n}_{1} \right) = T \left( {m}_{1}, {n}_{1} \right) - T \left( {m}_{1}, {n}_{1} \right) = 0 $$

Maybe the better question to ask would be to prove if a transformation is heterogeneous, is it linear?

As I wrote above, if an operator is heterogeneous is must not be linear.

Think of the simplest Linear Operator in 1D: $ T \left( x \right) = a x $. Clearly it is a linear function and indeed homogeneous.
Let's look on Affine Function: $ T \left( x \right) = a x + b $. Usually this is also called Linear Function while it is not unless $ b = 0 $. Why? It is not heterogeneous.

It is easy to see that $ T \left( {x}_{1} - {x}_{1} \right) \neq T \left( {x}_{1} \right) - T \left( {x}_{1} \right) $.