Linear transformation diagonalization with unknown vectors in basis I've been working for some hours with this problem but I still can't get it. The problem says as follows:
Given $ B=\{V_1, V_2, V_3\} $ and $ B'= \{V_1, V_1+V_2,-V_1-2V_2-V_3\} $ , basis of a vector space $ V $, and $ f:V \mapsto V $ a linear transformation such thath $M_(BB)'= $$\begin{pmatrix}5 & -2 & 2\\\ 0 & 1 & a\\\ 0 & -1 &-4\end{pmatrix}$$ $ find, if possible, $ a \in \Re $ such that $ 2V_2 - V_3 $ is an autovector.
What I did: I know that for a linear transformation to be diagonalizable, then the standard matriz associated to said transformation must be diagonalizable too. However, in this case I am completely unable to build the standard matriz, because I don't know the components of the vectos V1, V2 and V3 which work as basis for V. I have checked my bibliography, because I feel that there must be some other way to find a diagonalization of a LT without resorting to the standard matrix, but I haven't found anything yet. Could someone lead me in the right way of approaching this exercise?
Lots of thanks in advance. 
 A: Let $B=\{v_1, v_2, v_3\}$ and $B'= \{v_1,v_1+v_2,-v_1-2v_2-v_3\}$ be bases of a vector space $V$ over $\mathbb{R}$, and suppose $f:V \to V$ is a linear transformation such that $M_{B'}^B$ is given by
$$
\begin{bmatrix}
5 & -2 & 2\\ 
0 & 1 & a\\
0 & -1 &-4\\
\end{bmatrix}
$$
for some $a\in\mathbb{R}$.

The objective is to find $a\in\mathbb{R}$, if any, such that $2v_2-v_3$ is an eigenvector of $f$.

Letting $A=M_{B'}^B$, we get
\begin{align*}
f(v_1)&=
A
\begin{bmatrix}
1\\
0\\
0\\
\end{bmatrix}_B
=
\begin{bmatrix}
5\\
0\\
0\\
\end{bmatrix}_{B'}
\\[6pt]
&
\phantom{
\;\;\;\;=
A
\begin{bmatrix}
0\\
\end{bmatrix}_B
}
=
(5)(v_1)+(0)(v_1+v_2)+(0)(-v_1-2v_2-v_3)
\\[4pt]
&
\phantom{
\;\;\;\;=
A
\begin{bmatrix}
0\\
\end{bmatrix}_B
}
=
5v_1
\\[12pt]
f(v_2)&=
A
\begin{bmatrix}
0\\
1\\
0\\
\end{bmatrix}_B
=
\begin{bmatrix}
-2\\
1\\
-1\\
\end{bmatrix}_{B'}
\\[6pt]
&
\phantom{
\;\;\;\;=
A
\begin{bmatrix}
0\\
\end{bmatrix}_B
}
=
(-2)(v_1)+(1)(v_1+v_2)+(-1)(-v_1-2v_2-v_3)
\\[4pt]
&
\phantom{
\;\;\;\;=
A
\begin{bmatrix}
0\\
\end{bmatrix}_B
}
=
3v_2+v_3
\\[12pt]
f(v_3)&=
A
\begin{bmatrix}
0\\
0\\
1\\
\end{bmatrix}_B
=
\begin{bmatrix}
2\\
a\\
-4\\
\end{bmatrix}_{B'}
\\[6pt]
&
\phantom{
\;\;\;\;=
A
\begin{bmatrix}
0\\
\end{bmatrix}_B
}
=
(2)(v_1)+(a)(v_1+v_2)+(-4)(-v_1-2v_2-v_3)
\\[4pt]
&
\phantom{
\;\;\;\;=
A
\begin{bmatrix}
0\\
\end{bmatrix}_B
}
=
(a+6)v_1+(a+8)v_2+4v_3
\\[4pt]
\end{align*}
hence $2v_2-v_3$ is an eigenvector of $f$ if and only if, for some $t\in\mathbb{R}$,
\begin{align*}
&f(2v_2-v_3)=t(2v_2-v_3)\\[4pt]
\iff\;&2f(v_2)-f(v_3)=2tv_2-tv_3\\[4pt]
\iff\;&2(3v_2+v_3)-\bigl((a+6)v_1+(a+8)v_2+4v_3\bigr)=2tv_2-tv_3\\[4pt]
\iff\;&(-a-6)v_1+(-a-2)v_2-2v_3=2tv_2-tv_3\\[4pt]
\iff\;&(-a-6)v_1+(-2-a-2t)v_2+(t-2)v_3=0\\[4pt]
\iff\;&
\begin{cases}
-a-6=0\\
-2-a-2t=0\\
t-2=0\\
\end{cases}
\\[4pt]
\iff\;&
\begin{cases}
t=2\\
a=-6\\
\end{cases}
\\[4pt]
\end{align*}
so the answer is $a=-6$.
