Prove that vector is eigenvector.

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $3\lambda_3+\lambda_1=4\lambda_2$