1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $a=-6$.

$\endgroup$
  • $\begingroup$ Lots of thanks I was really messed up there! $\endgroup$ – Artem Aug 19 at 16:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.