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.
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 $a=-6$.