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Let $A$ be a matrix of size $5$ (real values) and real $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$, such that:

vector $(1,1,1,1,1)$ is eigenvector of $A$ corresponding to eigenvalue $\lambda_{1}$

vector $(1,2,3,4,5)$ is eigenvector of $A$ corresponding to eigenvalue $\lambda_{2}$

vector $(1,3,5,7,9)$ is eigenvector of $A$ corresponding to eigenvalue $\lambda_{3}$

Prove that vector $(43,53,63,73,83)$ is eigenvector of $A$ corresponding to eigenvalue $3\lambda_1 + 5\lambda_2 - 7\lambda_3$.

I know that this set of eigenvectors is not lineary independent and I think that this might be important but I don't know how to use this information.

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  • $\begingroup$ A selection of eigenvectors from distinct eigenvalues are always linearly independent, which is why I'm confident that this exercise is just bogus. Coming up with a counterexample should be straight forward enough, but I'm lazy. :-) $\endgroup$
    – Theo Bendit
    Sep 12 '19 at 10:31
  • $\begingroup$ But it is not mentioned that eigenvalues are distinct. $\endgroup$
    – Berto
    Sep 12 '19 at 10:33
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    $\begingroup$ If the exercise isn't bogus, we at least know that the eigenvalues are not distinct. But I am having a hard time believing that the exercise isn't bogus. Seems too good to be true that we could find a 4th eigenvector and know its eigenvalue just by knowing the other $3$ eigenvectors and eigenvalues. $\endgroup$
    – Charles Hudgins
    Sep 12 '19 at 10:48
  • $\begingroup$ I don't mind checking, you should (the asker) it is just applying $Ay=\lambda y$ $\endgroup$
    – Toni Mhax
    Sep 12 '19 at 10:55
  • $\begingroup$ So let me change my question, in this exercise is it possible that, $\lambda_1 = \lambda_2 \neq \lambda_3$ or all eigenvalues must be equal because eigenvectors form lineary dependent set? $\endgroup$
    – Berto
    Sep 12 '19 at 10:56
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Let $v_1, v_2, v_3, v_4$ be vectors mentioned in exercies(in order).

We have $v_3$ = $2v_2 - v_1$

$A(2v_2 - v_1) = \lambda_3(2v_2 - v_1)$

$2\lambda_2v_2 - \lambda_1v_1 = 2\lambda_3v_2 - \lambda_3v_1$

$v_2(2\lambda_2 - 2\lambda_3) + v_1(\lambda_3 - \lambda_1) = 0$

$\lambda_1 = \lambda_2 = \lambda_3$

$3\lambda_1 + 5\lambda_2 -7\lambda_3 = \lambda_1$

$Av_4 = A(av_1 + bv_2 + cv_3)$

$Av_4 = \lambda_1(av_1 + bv_2 + cv_3)$

This is my solution to this problem, because I am almost sure that this exercies is not bogus. If this solution is wrong i would appreciate feedback, if it is right it will be amazing if somebody could tell why counterexample in previous answer is wrong.

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  • $\begingroup$ @RichardJensen Can you tell exactly what is wrong with my answer? If your example is indeed correct there have to be something wrong with my solution. "It is wrong, because I found a counter example" is of course correct(if it's in fact counter example) but doesn' tell me why my solution is wrong. $\endgroup$
    – Berto
    Sep 12 '19 at 13:10
  • $\begingroup$ Because my proof shows that $\lambda_1 = \lambda_2 = \lambda_3$ therefore your counter example must be wrong. $\endgroup$
    – Berto
    Sep 12 '19 at 13:38
  • $\begingroup$ Deleted my wrong comments, your proof is spot on :). $\endgroup$
    – Richard Jensen
    Sep 12 '19 at 18:41
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Edit: My counterexample is not a counterexample, and the exercise is correct.

Let $\{(1,1,1,1,1), v_2, v_3, v_4, v_5 \}$ be a basis of $\mathbb{R}^5$. A linear transformation is uniquely determined by its values on a basis, so let $L$ be the operator which sends $(1,1,1,1,1)$ to itself, and $v_2, v_3, v_4, v_5$ to $0$. Let A be the matrix representation of $L$ with respect to the above mentioned basis. This $A$ satisfies the conditions of the exercise with $\lambda_1 = 1$ and $\lambda_2 = \lambda_3 = 0$.

But then the vector $(43,53,63,73,83)$ has eigenvalue $3$, which is impossible since it is linearly independent of $(1,1,1,1,1)$. So the statement is false.

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  • $\begingroup$ I don't see how this transformation is linear. It is not additive for example for vectors $(0,0,0,0,2), (1,1,1,1,-1)$. $\endgroup$
    – Berto
    Sep 12 '19 at 11:58
  • $\begingroup$ Right, I was messed up a bit. Here is a corrected version of the proof (I'll update OP): Let $\{ (1,1,1,1,1), v_2, v_3, v_4, v_5 \}$ be a basis of $\mathbb{R}^5$. A linear transformation is uniquely determined by its values on a basis, so let $L$ be operator which sends $(1,1,1,1,1)$ to itself, and $v_2, v_3, v_4, v_5$ to $0$. Let A be the matrix representation of $L$ with respect to the above mentioned basis, and follow the proof. $\endgroup$
    – Richard Jensen
    Sep 12 '19 at 12:23
  • $\begingroup$ That's messy you have not given your $A$, as i said the three eigenvalues are indeed equal. Maybe one can prove the existence etc $\endgroup$
    – Toni Mhax
    Sep 12 '19 at 13:11
  • $\begingroup$ @ToniMhax The problem is that I have shown here that there exists a matrix $A$ satisfying the assumptions, and shown that this does not satisfy the conclusion, hence exercise is wrong. The problem, as I have also written to Bertos solution is that when we conclude that $\lambda_1 = \lambda_2 = \lambda_3$, we reach a contradiction, since this is not always the case, as shown by my example. $\endgroup$
    – Richard Jensen
    Sep 12 '19 at 13:20
  • $\begingroup$ @RichardJensen You say it is linear therefore $A(1,2,3,4,5) =A( \alpha_1(1,1,1,1,1) + \alpha_2v_2 + \dots + \alpha_5v_5) = \alpha_1(1,1,1,1,1)$ For $\alpha_1 \neq 0$ $(1,2,3,4,5)$ is not even eigenvector so your counter example doesn't fit exercise. $\endgroup$
    – Berto
    Sep 12 '19 at 14:03
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I'll post a short hint it is easy to show that $\lambda_3=\lambda_1$, and so $\lambda_2=\dfrac{\lambda_1+\lambda_3}{2}$. Taking entrywise the first two rows of $Ax_i=\lambda_ix_i$, for $i=1,2,3$. Proof: $(a_1,\cdots,a_5)$ resp. $(b_1,\cdots,b_5)$ are the first resp. second row of $A$. If $\lambda_1=\lambda_3$ we got that $v_2\in\text{span}\{v_1,v_3\}$ or $\lambda_2=\lambda_1=\lambda_3=\lambda$.

So for the given eigenvectors $Av_1=\lambda_1v_1$ $\implies $ $\sum_{i=1}^5a_i=\sum_{i=1}^5b_i=\lambda_1$. $Av_2=\lambda_2v_2$ $\implies$ $\sum_{i=1}^5ia_i=\lambda_2$, $\sum_{i=1}^5ib_i=2\lambda_2$. $Av_3=\lambda_3v_3$ $\implies$ $\sum_{i=1}^5(2i-1)a_i=\lambda_3$, $\sum_{i=1}^5(2i-1)b_i=3\lambda_3$ so $\lambda_3+\lambda_1=2\lambda_2$ and $3\lambda_3+\lambda_1=4\lambda_2$

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