Verified answer

Another way to do this (since P3 is finite dimensional) is to find the matrix for L with respect to the standard basis {1, x, x^2, x^3} for P3:

L(1) = 1 - 0 = 1 + 0x + 0x^2 + 0x^3

L(x) = x - 1 = -1 + x + 0x^2 + 0x^3

L(x^2) = x^2 - 2x = 0 - 2x + x^2 + 0x^3

L(x^3) = x^3 - 3x^2 = 0 + 0x - 3x^2 + x^3.

So, the matrix for L is

[1 -1 0 0]

[0 1 -2 0]

[0 0 1 -3]

[0 0 0 1].

Row reducing this matrix to reduced row echelon form yields

[1 0 0 0]

[0 1 0 0]

[0 0 1 0]

[0 0 0 1], which has no free variables.

Hence, ker L = {0}.

Remark:

Indeed, there is no polynomial for which L(p(x)) = p(x) - p'(x) = 0 <==> p'(x) = p(x).

(The only solutions to y' = y are of the form y = Ce^x for some constant C, and e^x is not a polynomial.)

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Since ker L = {0}, we conclude that L is 1-1.

Next, since L maps from P3 to itself, and P3 has finite dimension, we conclude that P is also onto; hence Range L = P3.

I hope this helps!