linear transformation matrix relative to a basis Let $B={b_1,b_2,b_3}$ be a basis of the real vector space $V$ . Take the linear transformation
$\phi : V \to V$ defined by $\phi(b_1) = \phi(b_2) = \phi(b_3) = b_1 + b_2 + b_3$.
a) Write the matrix $A = [\phi]_B$ of the transformation $\phi$ relative to the basis $B$.
b) Find the characteristic polynomial, eigenvalues and eigenvectors of $\phi$.

Looking for a help to find $A$, I know that it's $(3$x$3)$-  matrix and I have to figure out 9 entries.
Now is it true to have three similar columns with entries all are variables? If yes how can I deal with it to solve the second problem, I know it might be an easy question but I am really confused.
 A: If you are familiar with change of basis you should know that the entries of a linear transformation $f : (V,B) \longmapsto (V,B')$ are the coordinates in the basis $B'$ of  of the image of the starting basis $B$.
Let's $B:=\{v_{1},v_{2},v_{3}\}$ and $\phi:=f$ . 
Let's think of $V$ as $\mathbb{R}^{3}$ thanks to the isomorphism of coordinates (given by the specific that $B$ is a basis of $V$, real vectorial space).
In our case since $f(v_{1}) = f(v_{2}) = f(v_{3}) = 1 \cdot v_{1} + 1 \cdot v_{2} + 1 \cdot v_{3}$ we have that : 
$$M_{B \to B}(f):=A= \begin{pmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix}$$
Why is that? Simply because the coordinates of $f(v_{1})$ in the basis $B$ are $1 \cdot v_{1} + 1 \cdot v_{2} + 1 \cdot v_{3}$
Of course, to find the characteristical polynomial you could compute $det(A-tI_{d})$
But in this case the job is much easier, why ? Simply note that $v_{1}+v_{2}+v_{3}$ is an eigenvector of eigenvalue $3$,
(You can notice that by seeing that $f(v_{1}+v_{2}+v_{3}) =3f(v_{1}) = 3(v_{1}+v_{2}+v_{3})$
And secondly observing the matrix we notice that $Ker(f)$ at leat dimension two, since there are 2 linear dependent vectors,each of which give us an eigenvector of eigenvalue $0$.
We've just computed the characteristic polynomial without computational effort since $dim(V) = dim(\mathbb{R}^{3}) = 3$;
But we have just found 3 independant eigenvectors, 
So the characteristic polynomial must be $p_{A}(t) = t^{2}(t-3)$ 
A: Writing the coefficients of $b_1,b_2,b_3$ columnwise from each of the three conditions, the linear map $A$ will be a $9\times 9$ matrix with all entries unity. 
The characteristic polynomial $f_A(t)=t^3-3t^2$ gives eigenvalues $0,0,3$. The corresponding eigenvectors will be $(1,-1,0)^T,(1,0,-1)^T$ and $(1,1,1)^T$ respectively.
A: Relative to the basis $B$, the coordinate vectors of the elements of $B$ are various columns of the identity matrix, and right-multiplying a matrix by the $j$th column of the identity picks out its $j$th column, so the columns of $[\phi]_B$ are just the coordinates of the images of the basis vectors. It looks like you might already have know this.  
Now recall that the coordinates of a vector relative to an ordered basis are simply the coefficients of the unique linear combination of the basis vectors that equals that vector. You’re given that the image of every basis vector $b_i$ is $b_1+b_2+b_3$. What are the coordinates of this vector relative to $B$? What does this mean for the columns of $[\phi]_B$?  
For the second part, since the columns of $[\phi]_B$ are identical and nonzero, you know that the matrix has rank one, so its null space is two-dimensional. This means that $0$ is an eigenvalue with algebraic multiplicity of at least two. You can find the third eigenvalue by examining the trace of $[\phi]_B$, which is equal to the sum of the eigenvalues. Once you have all three eigenvalues, I expect that you know how to construct the characteristic polynomial. I hope that you know how to find a basis for the null space of a matrix. This will give you two independent eigenvectors of zero. For the eigenvector corresponding to the remaining eigenvalue, once again use the fact that the matrix has rank one, and the only possibility for the eigenspace of a nonzero eigenvalue is its column space.
