\[T: C^5-->P_3\] is surjective:
\[ T\Bigg[ \begin{pmatrix} &a\\ &b\\ &c\\ &d\\ &e\\ \end{pmatrix} \Bigg] = a + (b +c)x +(c+d)x^2 + (d + e)x^3 \]
A function is surjective or “onto” for every y E Y there exist at least one x E X such that f(x) =y
\[ p(x) = \alpha + \beta x + \gamma x^2 + \sigma x^3 \]
\[ vector u = \Bigg[ \begin{pmatrix} &a\\ &b\\ &c\\ &d\\ &e\\ \end{pmatrix} \Bigg] in C^5 so that T(U) = P(x) \]
\[ a = \alpha \\ b + c = \beta\\ c + d = \gamma\\ d + e = \sigma\\ \]
\[ One of the solution is \\ a = \alpha , c = 0, b = \beta, d = \gamma, e = \sigma - \gamma\\ \]
\[ T\Bigg[ \begin{pmatrix} &\alpha\\ &\beta\\ &0\\ &\gamma \\ &\sigma - \gamma\\ \end{pmatrix} \Bigg] = \alpha + (\beta +0)x +(0+\gamma)x^2 + (\gamma + (\sigma - \gamma))x^3 \] \[ = \alpha + \beta x + \gamma x^2 + \sigma x^3 \\ = p(x) \] Therefore, T is surjective. T(u) = v.