Linearity and linear differential operators

In terms of linear transformations, linearity means one very specific thing:

for any vectors $𝐮,𝐯$ and scalars $𝑎,𝑏$.

That single equation contains the whole idea. It says:

Transforming a linear combination gives the same result as transforming each piece separately and then taking the same linear combination.

There are two properties hidden inside it:

and

So a linear transformation preserves addition and scalar multiplication.

For example, suppose

Take

Then

Separately,

Same answer. That’s linearity.

This also explains the matrix interpretation you’ve been using. If

then

So every matrix transformation is linear.

And geometrically, this means something quite powerful. Suppose your basis vectors are

Every vector can be written as

Once you know where the transformation sends the basis vectors,

linearity forces

So the transformation has no freedom left. Knowing where the basis vectors go tells you where every vector goes.

That’s why the columns of a matrix are the transformed basis vectors:

There’s another extremely useful consequence:

Because

So a genuinely linear transformation must leave the origin fixed.

For example,

is linear, but

is not, because

That second transformation contains a translation. It’s affine, not linear.

This connects directly back to your ODE question. When we write

we can think of $𝐿$ as a transformation whose “vectors” are functions:

It’s linear because

just like

So

in linear algebra and

in differential equations are conceptually the same kind of problem: you’re looking for the kernel/null space of a linear transformation.

That’s the deeper reason homogeneous linear ODEs behave so much like linear algebra.

For a linear ODE operator

linearity gives

Now suppose we have the nonhomogeneous equation

Pick one particular solution ${𝑦}_{𝑝}$ such that

Then any other solution $𝑦$ satisfies

Subtract:

So

Therefore,

where

So the full solution set is

And that is literally an affine subspace: a vector subspace, $ker𝐿$, shifted by ${𝑦}_{𝑝}$.

It is the exact same theorem as for a linear system

If ${𝐱}_{𝑝}$ is one particular solution, then every solution is

with

So you can line them up:

The important subtlety is that the affine-subspace result works because $𝐿$ itself is a linear transformation. The equation does not need to be homogeneous. In fact, the interesting affine-space case is specifically when

When $𝑓=0$, the shift disappears:

so the solution set is an actual vector subspace rather than merely an affine one.

This is also why whether $𝑓$ is a constant or depends on $𝑥$ makes no difference. You could have

and as long as a particular solution exists, the solution set is still

So the real distinction is not “constant forcing vs variable forcing.” It’s: