Determine the Eigenvalues and Eigen vectors of the operator $ \ A \ $ Consider the vector space of real polynomials of degree not greater than $ \ n \ $ given by $ \large \mathbb{R}_n[x]=\{\sum_{i=0}^{n} a_i x^i \ | a_i \in \mathbb{R}, \ \forall i \} \ $ . 
Define an operator $ \ A \in End (\mathbb{R}_n[x] ) \ $ by $ \ (Ap) (x)=P(2x+1) \ $ . 
Determine the Eigenvalues and Eigen vectors of the operator $ \ A \ $. 
Here $ \ End (\mathbb{R}_n[x] ) \ $ is an Endomorphism from $ \mathbb{R}_n[x] \to  \mathbb{R}_n[x] \ $
Answer:
The basis of $ \mathbb{R}_n[x] \ $ is given by $ \{1,x,x^2, x^3,......, x^n \} \ $ 
Now, 
$ (Ap)(x)=P(2x+1) \ $ gives the following system
$ For \ p(x)=1 , \ \ A(1) =1=1 \cdot 1+ x \cdot 0+............+x^n \cdot 0 \\ For \ p(x)=x , \ \ A(x)=2x+1=1 \cdot 1+2 \cdot x+0 \cdot x^2+........+0 \cdot x^n \\ For \ p(x)=x^2 , \ \  A(x^2)=(2x+1)^2=1+4x+4x^2=1 \cdot 1+4 \cdot x+4 \cdot x^2+......+0 \cdot x^n \\ ..............\\ p(x)=x^n \Rightarrow
A (x^n)=(2x+1)^n=1 \cdot 1+2 \binom{n}{1} \cdot x+2^2 \binom {n}{2} \cdot x^2+...........+2^n \binom{n}{n} \cdot x^n  $
Thus the coefficient matrix is 
$ A=\begin{bmatrix} 1 & 1 & 1 & 1 &  ............ & 1 \\ 0 & 2 & 4 &  8 & ...... ........ & 2n  \\ 0 & 0 & 4 & ..... \\  .... & ...... & ..... & .. & . ............ & ... \\ 0 & 0 &  0 & 0 &  ............ & 2^n \end{bmatrix} \ $
Am I right So far ?
If the above approach is correct , then how to find the Eigen values and Eigen vectors of the matrix $ \ A \ $ ?
Help me out 
 A: I like what you have so far.
The eigenvalues of an upper-triangle matrix are the values of the main diagonal
The first few eigenvectors
$\pmatrix{1&1&1&2\\&1&2&3\\&&1&3\\&&&1}$
Well that looks like Pascals triangle.
or $(x+1)^n$ appears to describe the set of eigenvectors.
$A((x+1)^n) = (2x+2)^n = 2^n(x+1)^n$ and that is the characteristic of an eigenvector.
I notice an error in your work above, in the $A_{2,4}$ place, you show an $8$
$A= \pmatrix{1&1&1&1\\&2&4&6&\cdots &2{n\choose 1}\\&&4&12&\cdots& 2^2{n\choose 2}\\&&&8&\cdots &2^3{n\choose 3}\\&&&&2^m{n\choose m}&\vdots\\&&&&&2^n}$
A: Hint: Your matrix $A$ is upper triangular, so the determinant is the product of the diagonal entries.
A: Try solving $AP = \lambda P$ instead. We can assume that $P$ is monic.
Then $P(2x+1) = \lambda P(x)$. Hence $2^k P^{(k)}(2x+1) = \lambda P^{(k)}( x)$.
Letting $x=-1$ (since $2x+1=x$ iff $x=-1$) gives 
$2^k P^{(k)}(-1) = \lambda P^{(k)}(-1 )$ for $k=0,...,n$. 
In other words, $(\lambda-2^k) P^{(k)}(-1 ) = 0$ for $k=0,...,n$.
If $\partial P = j$ we see that $P^{(j)}(-1 ) = 1$ and so $\lambda = 2^j$. Since $P^{(k)}(-1 ) =0$ for $k=0,...,j-1$ we see that $(x+1)^j$ divides $P$.
In particular, $P(x) = (1+x)^j$ is an eigenvector corresponding to
the eigenvalue $2^j$ for $j=0,...,n$.
