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I have attempted to prove the following Theorem is my argument correct?

NOTE: The argument below makes use of the following notation and theorems.

  • $T\in\mathcal{L}(V)$ denotes that $T$ belongs to the set of all linear operators on the vector space $V$.

  • $5.10$ Given that $T\in\mathcal{L}(V)$ such that $\lambda_1,\lambda_2,...,\lambda_m$ are distinct eigenvalues of $T$ then the corresponding list of eigenvectors $v_1,v_2,...,v_m$ is linearly independent.

Theorem. Given that $T$ is a linear operator over the vector space $V$ where $u$ and $v$ are eigenvectors of $T$ such that $u+v$ is also an eigenvector of $T$, then $u$ and $v$ are eigenvectors corresponding to the same eigenvalue.

Proof. Assume on the contrary that $u$ and $v$ are eigenvectors of $T\in\mathcal{L}(V)$ such that they do not correspond to the same eigenvalue consequently $$Tu = \lambda_1u\ \operatorname{and}\ Tv = \lambda_2v\ \operatorname{where}\ \lambda_1\neq\lambda_2$$

more over since $u+v$ is an eigenvector of $T$ it follows that $T(u+v) = \lambda_3(u+v)$. The Linearity of $T$ implies that $T(u+v) = Tu+Tv = \lambda_1u+\lambda_2v = \lambda_3(u+v)$ which implies that $$(\lambda_3-\lambda_1)u+(\lambda_3-\lambda_2)v = 0$$

but $\lambda_1\neq\lambda_2$ which by $5.10$ implies that $u$ and $v$ are linearly independent consequently $$\lambda_3-\lambda_1 = 0$$ $$\lambda_3-\lambda_2 = 0$$ which implies that $\lambda_1 = \lambda_2$ a contradiction.

$\blacksquare$

Source: Linear Algebra Done Right by Sheldon Axler 5-A Exercise 25.

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  • $\begingroup$ yes, you are correct. $\endgroup$ – Tony S.F. Jan 30 '18 at 14:10
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It is perfect, except that if you look closely there is no need to work by contradiction. You never actually use the fact that $\lambda_1\neq\lambda_2$, therefore you can remove this assumption and simply come to the conclusion that $\lambda_1= \lambda_2$ without changing a word.

Edit since it was not clear for everyone: "if the vectors are linearly dependent then obviously $\lambda_1=\lambda_2$, and if they are not (insert your own words here) again $\lambda_1=\lambda_2$".

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  • $\begingroup$ But did i not use the assumption $\lambda_1\neq\lambda_2$ to justify the linear independency of $u$ and $v$ ? $\endgroup$ – Atif Farooq Jan 30 '18 at 14:14
  • $\begingroup$ There is no guaranteed linear independence of eigenvectors without distinctness of eigenvalues. $\endgroup$ – Tony S.F. Jan 30 '18 at 14:14
  • $\begingroup$ @TonyS.F. I am sorry i dont seem to follow $\endgroup$ – Atif Farooq Jan 30 '18 at 14:15
  • $\begingroup$ @AtifFarooq "if the vectors are linearly dependent then obviously $\lambda_1=\lambda_2$, and if they are not (insert your own words here) again $\lambda_1=\lambda_2$". $\endgroup$ – Arnaud Mortier Jan 30 '18 at 14:16
  • $\begingroup$ @AtifFarooq you are agreeing with me, it is necessary to assume that $\lambda_1\neq \lambda_2$ in the proof to justify that $u$ and $v$ are linearly independent. Your proof is correct as written. $\endgroup$ – Tony S.F. Jan 30 '18 at 14:17

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