# Prove that vector is eigenvector.

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.

• 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. :-) Sep 12, 2019 at 10:31
• But it is not mentioned that eigenvalues are distinct. Sep 12, 2019 at 10:33
• 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. Sep 12, 2019 at 10:48
• I don't mind checking, you should (the asker) it is just applying $Ay=\lambda y$ Sep 12, 2019 at 10:55
• 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? Sep 12, 2019 at 10:56

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.

• @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. Sep 12, 2019 at 13:10
• Because my proof shows that $\lambda_1 = \lambda_2 = \lambda_3$ therefore your counter example must be wrong. Sep 12, 2019 at 13:38
• Deleted my wrong comments, your proof is spot on :). Sep 12, 2019 at 18:41

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.

• I don't see how this transformation is linear. It is not additive for example for vectors $(0,0,0,0,2), (1,1,1,1,-1)$. Sep 12, 2019 at 11:58
• 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. Sep 12, 2019 at 12:23
• 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 Sep 12, 2019 at 13:11
• @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. Sep 12, 2019 at 13:20
• @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. Sep 12, 2019 at 14:03

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