Matrix with unknown coefficients, finding another basis let $(e_1,e_2,...,e_5)$ canonical basis of $R^5$,  $V=(a,b,c,d,e)\in R^5$ with $V\neq(0,0,0,0,0)$. we consider $f:R^5\to R^5$ and its matrix : 
$$Mat(f) = M= \begin{pmatrix}
a&a&a&a&a\\
b&b&b&b&b\\
c&c&c&c&c\\
d&d&d&d&d\\
e&e&e&e&e\\
\end{pmatrix}$$
It's rank is $rg(f)=1$ because $V\neq(0,0,0,0,0)$ and all columns are the same. So $Dim(Im(f))=1$ and because of the rank theorem, $Dim(Ker(f))=4$
Basis of $Im(f)=Vect\begin{pmatrix}\begin{pmatrix}a\\b\\c\\d\\e\\\end{pmatrix}\end{pmatrix}$ and by solving with $X=\begin{pmatrix}x\\y\\z\\t\\u\\ \end{pmatrix}$, $ MX=0$ I find : $Ker(M)=Vect\begin{pmatrix}\begin{pmatrix}-1\\1\\0\\0\\0\end{pmatrix},\begin{pmatrix}-1\\0\\1\\0\\0\\\end{pmatrix},\begin{pmatrix}-1\\0\\0\\1\\0\end{pmatrix},\begin{pmatrix}-1\\0\\0\\0\\1\end{pmatrix}\end{pmatrix}$
I was then asked a condition on $a,b,c,d,e$ for $V\in Ker(f)$ and I found $a+b+c+d+e=0$ by solving $MV=0$
And next is a question I need to verify: 
Question : let this condition be fullfilled, let $(\epsilon_1,\epsilon_2,\epsilon_3,\epsilon_4)$ basis of $Ker(f)$ chosen so $\epsilon_4=V$
Complete it in a basis of E (here $R^5$) and write the matrix in the new basis
Verification :  what I did : Is it okay for me to set $V = \begin{pmatrix}-1\\0\\0\\0\\1\end{pmatrix}$ so it would mean that my matrix $M$ would become $M=\begin{pmatrix}-1&-1&-1&-1&-1\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\1&1&1&1&1\end{pmatrix}$,$\epsilon_1=\begin{pmatrix}-1\\1\\0\\0\\0\end{pmatrix}$, and then respectively for $\epsilon_2,\epsilon_3, ...$
or should I solve with $V=\begin{pmatrix}a\\b\\c\\d\\e\end{pmatrix}$ ? Because I don't understand what they're looking for if it's a general answer or a simple base to create.
because what I could find was my new basis (direcltly made into a base changing matrix because it's hard to type everything) : $P=\begin{pmatrix}-1&-1&-1&-1&-1\\1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\end{pmatrix}$
then I solved and answered $M'=P^-1MP=\begin{pmatrix}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&-1\\0&0&0&0&0\end{pmatrix}$
Then, right after this question, the big problem: 
Question: now, $V\notin Ker(f)$ I have to find a basis so B matrix would be : B = \begin{pmatrix}\gamma&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{pmatrix}
and of course I need to precise $\gamma$
So here, I thought I could use $V=\begin{pmatrix}-1\\0\\0\\0\\0\end{pmatrix}$ and keep my $Ker(f)$ basis I found before and use the same formula $B=P^-1M'P$ but it didn't sound good at all ... Because my $\gamma=0$ which is nonsense ... I'm a bit confused and it makes me stuck
Thank you in advance and sorry for the big stuff.
 A: Hints:
Question 1
I would assume this question requires a general answer, so setting a value for $V$ is incorrect.  
Before you can find the matrix of $f$ in the new basis, you need to find $\epsilon_5$.  Since $\mathbb{R}^5$ has dimension 5, it is enough that $\{\epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4, \epsilon_5\}$ be linearly independent.  How can we choose $\epsilon_5$ to make this true?  Remember $\{\epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4\}$ spans the nullspace of $f$. Let's take $\epsilon_5$ to be some arbitrary vector not in $\mathtt{Ker}(f)$. Then, if
$$\lambda_1\epsilon_1 + \lambda_2\epsilon_2 + \lambda_3\epsilon_3 + \lambda_4\epsilon_4 + \lambda_5\epsilon_5 = 0,$$
then
$$f(\lambda_1\epsilon_1 + \lambda_2\epsilon_2 + \lambda_3\epsilon_3 + \lambda_4\epsilon_4 + \lambda_5\epsilon_5) = \lambda_1f(\epsilon_1) + \lambda_2f(\epsilon_2) + \lambda_3f(\epsilon_3) + \lambda_4f(\epsilon_4) + \lambda_5f(\epsilon_5) = 0$$
$$\lambda_1f(\epsilon_5) = 0.$$
Since $\epsilon_5 \not\in \mathtt{Ker}(f)$, we have that $\lambda_5 = 0$.  Thus,
$$\lambda_1\epsilon_1 + \lambda_2\epsilon_2 + \lambda_3\epsilon_3 + \lambda_4\epsilon_4 = 0.$$
But the only way for this to happen is if $\lambda_1 = \lambda_2 = \lambda_3 = \lambda_4 = 0$ (why?) so $\{\epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4, \epsilon_5\}$ is linearly independent.
Now, remember that the matrix of $f$ with respect to this basis will be $\left[f(\epsilon_1)|f(\epsilon_2)|f(\epsilon_3)|f(\epsilon_4)|f(\epsilon_5)\right]$.
Question 2
We observe that $f(V) = MV 
= \left[\begin{array}{ccccc}
a & a & a & a & a \\
b & b & b & b & b \\
c & c & c & c & c \\
d & d & d & d & d \\
e & e & e & e & e \\
\end{array}\right]
\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right] = a\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right] + b\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right] + c\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right] + d\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right] + e\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right] = (a+b+c+d+e)\left[\begin{array}{c}
a \\
b \\
c \\
d \\
e \\
\end{array}\right]$
What does this tell use about the basis we need and the value of $\gamma$?
