About Dual Basis $V$ is real vector space of all polynomials in single variable and with real coefficients, of degree at most 3. $V^*$ its dual space. $t_1=1,t_2=2,t_3=3,t_4=4.$ Then I have to verify whether this set of functional forms a basis for $V^*$ or not. 


For $1\le i \le 4$, $\forall p \in V$,$f_i(p)=p(t_i)$ 


My approach: A standard basis of $V$ is $e_i=x^{i-1}$ for $i=1,2,3,4$ I have to verify whether $f_i(e_j)=\delta_i^j$. And I see $f_1(e_2)=x(t_1)=x(1)=1$. So I guess it does not form a basis of $V^*$, I am not sure about my approach, can any one verify it? 
 A: Verifying $f_i(e_j)=\delta_{ij}$ is necessary only if you want the "dual basis," but this is not the same thing as a basis for the dual (as Daniel Fischer noted). 
First, let's find the dual basis. This is easy to intuit: for our first dual basis we want it to be 1 on 1 and 0 on $x, x^2,$ and $x^3$, the second dual basis vector is 1 on x, and 0 on $1, x^2, $ and $x^3$, the third dual basis vector is 1 on $x^2$ and $0$ on $1,x,$ and $x^3$, and the last dual basis vector is 1 on $x^3$ and $0$ on $1,x,$ and $x^2$. That is we just want to pick out the coefficient on the $k^{th}$ degree term. So we find that $D^k=\dfrac{1}{k!}\dfrac{d^k}{dx^k}∣_{x=0}$ for $k=0,1,2,3$ is the dual basis. If we can re-express the set of dual vectors $t_1, t_2, t_3,$ and $t_4$ you gave as a linear combination of these, then we can use standard matrix techniques to check if it's linearly independent and therefore a bona fide basis for the dual space. 
Well, how do we find out what happens if we plug 1 into our polynomial if we know the coefficients (that's the information from the dual basis), it's simple: $D^0+D^1+D^2+D^3$, just the sum of the coffiencts. If we plug in 2, we have $D^0+D^1\times 2+D^2\times 2^2+D^3\times 2^3=D^0+2D^1+4D^2+8D^3$. If we plug in 3 we get $D^0+3D^1+9D^2+27D^3.$ If we plug in 4 we get $D^0+4D^1+16D^2+64D^3.$. Our matrix is
\begin{bmatrix}
1      & 1& 1 &  & 1 \\
1       & 2 & 3 &  & 4 \\
1      & 4 & 9 &  & 16 \\
1      & 8 & 27 &  & 64
\end{bmatrix}
Which has nonzero determinant [See https://en.wikipedia.org/wiki/Vandermonde_matrix], and so your proposed set of dual vectors is indeed a basis for the dual space. 
What we notice is that the dual basis gives us all the information we need to recover the polynomial since we have all of the coefficients. This other basis you suggested works because knowing the value of a 3rd degree polynomial at four points is also enough information to recover the polynomial.
