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I have problem with this task:

Endomorphism $ F \in L(\mathbb R[t]_3 )$ is given by:

$$ F(p) = 2\cdot p'(t) + (1+t+t^2+t^3)\cdotp(0) $$
Check if the functional system $$ f_j^*(p) = F(p)(j) $$for $j=0,1,2,3$ is a base of $ (\mathbb R[t]_3 )^*$

I have an idea to solve this in this way $$p(t) = at^3 + bt^2 + ct + d $$ $$ F(p)(0) = ... = 2c+d = [0,0,2,1]^T $$ $$ F(p)(1) = ... = 6a+4b+2c+4d = [6,4,2,4]^T $$ $$ F(p)(2) = ... = 24a +8b +2c+13d = [24,8,2,13]^T $$ $$ F(p)(3) = ... = 54a + 12b + 2c + 40d = [54,12,2,40]^T $$ but I don't know what I should do next...

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ℝ[t]3 is of dimension 4. A set of vectors is a base of a vector space if it has the same dimension as the space and if the vectors of the set are linearly independent. In this case, you already have a set which has as many elements as the dimension (4), so you just have to check linear independency to show that it is a base.

EDIT: To see if your 4 vectors are linearly independent, you can put them in a matrix and the matrix has to be of full rank.

\begin{pmatrix} 0 &0 &2 &1 \\ 6 &4 &2 &4 \\ 24 &8 &2 &13 \\ 54 &12 &2 &40 \\ \end{pmatrix}

$\vdots$ which after few RE steps gives us the following matrix

\begin{pmatrix} 6 &4 &2 &4 \\ 0 &-8 &-6 &-3 \\ 0 &0 &2 &1 \\ 0 &0 &0 &12 \\ \end{pmatrix}

Which is indeed of full rank (4). Thanks to this, your vectors are linearly independent.

You can now conclude that your set of vectors is a base.

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  • $\begingroup$ Chmmm. Are you sure that it is correctly? $\endgroup$
    – user617243
    Commented Feb 1, 2019 at 13:58
  • $\begingroup$ This is a method which definitely works. It would maybe help me if you could tell me what you think is necessary for a set of vectors to be called a base. $\endgroup$
    – Pale
    Commented Feb 2, 2019 at 15:55

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