Showing that a Transformation Matrix is injective and surjective.

Question

I have the following linear transformation:

$F: \mathbb P_{3} \to \mathbb R^{3}$ where $\mathbb P_{3}$ is the set of all polynomials with degree at most 3.

$F(p) = (p(0),p(1),p(2))$.

I arrived at the following Transformation Matrix:

$F = \begin{bmatrix}1 & 0 & 0 & 0\\ 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \end{bmatrix}$.

Is this F correct for the Transformation? Furthermore is it injective and /or surjective? My assumption is that F is both injective and surjective since the coloumns are lineary independent which makes it injective and F also has rank 3 which means that it produces the entire Imagespace.

$F = \begin{bmatrix}1 & 0 & 0 & 0\\ 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \end{bmatrix} = \begin{bmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 1 & 1 \\ 0 & 2 & 4 & 8 \end{bmatrix} = \begin{bmatrix}1 & 0 & 0 & 0\\ 0 & 1 & 1 & 1 \\ 0 & 0 & 2 & 6 \end{bmatrix} $. Is the Gauss-elimination and I assume that F has 3 pivots.

Answer 1

your transformation matrix is correct. The columns of $F$ cannot be linearly independent: you have a set of $4$ vectors in a linear space of dimension $3$ !

By the way, $F$ is not a matrix, you should give a different name to the representation matrix, even if I'm still going to use $F$ here.

The fact that the rank of $F$ is $3$ is correct, so $F$ is surjective. Be careful: written like this, your proof is false. YOu don't have a series of equalities. You should rewrite your proof in a correct way.

To see if $F$ is injective or not, you have many possibilities:

• you may compute the nullspace of $F$ by solving a linear system

• if you believe it is not, you may try to find a polynomail with degree less than 3 whose image is (0,0,0)

• you can also provide a more conceptual proof: if a linear transformation $F:E\to E'$ between to vector spaces is injective, can you compare $\dim(E)$ and $\dim(E')$ ?

Can your $F$ be injective here ?

Answer 2

Let A the matrix of the homomorphism. If you solve the linear system $$Ax=0$$

you will obtain $\ker(f)$. Then a basis for $\ker(f)$ is $$\ker(f) = \langle ( 0, \ 2, \ -3, \ 1)^{t}\rangle$$

With this, is $f$ inyective? Remember the caracterisation for the $\ker(f)$ when $f$ is inyective.

And it is surjective, take any vector in $\mathbb{R}^{3}$ $$v = ( a_{1}, \ a_{2}, \ a_{3})^{t}$$

then a polynomial whose image is $v$ can be obtained by interpolation.

As you saw, $F$ is surjective.