Show that the linear map $T:P_3(\Bbb R)→\Bbb R^4, p↦(p(0),p(1),p′(0),p′(1))$ is an isomorphism.

Question

Show that the linear map

$$T:P_3(\Bbb R)→\Bbb R^4,\quad p↦(p(0),p(1),p′(0),p′(1))$$ is an isomorphism. what I did following shows it's not a isomorphism ,what's wrong with it?

Answer 1

Note that, if $p(x)=ax^3+bx^2+cx+d\in P_3(\mathbb R)$, then we have $$\begin{align}&(1)\qquad p(0)=d \\&(2)\qquad p(1)=a+b+c+d \\&(3)\qquad p'(0)=c \\&(4)\qquad p'(1)=3a+2b+c\end{align}$$

Therefore, your map $T$ sends $$\begin{align}&(1)\qquad \color{red}{1\mapsto (1,1,0,0)} \\&(2)\qquad x\mapsto (0,1,1,1) \\&(3)\qquad x^2\mapsto (0,1,0,2) \\&(4)\qquad x^3\mapsto (0,1,0,3)\end{align}$$ so you mapped $1$ incorrectly under $T$. You need to remember that $p'(x)$ loses the constant coefficient from $p(x)$.

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But for your argument about mapping constant polynomials to $(0,0,0,0)$, this is false. $T$ maps constant polynomials to a scalar multiple of $(1,1,0,0)$. You should see that the matrix of $T$ relative to these standard bases is invertible, and therefore $T$ is an isomorphism (it is clearly linear).