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Assume that $\lambda_1$ and $\lambda_2$ are distinct eigenvalues of the $n\times n$ matrix $A$. Prove that an $n\times 1$ vector $x$, which is not a null vector, cannot be an eigenvector of both $\lambda_1$ and $\lambda_2$.

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Suppose for contradiction that $\vec{x}$ is an eigenvector of both $\lambda_1$ and $\lambda_2$. Then we have $A\vec{x}= \lambda_1 \vec{x}$ and $A\vec{x} = \lambda_2 \vec{x}$. So $\lambda_1 \vec{x} = \lambda_2 \vec{x}$. Multiplying both sides by $\vec{x}^*$ we obtain:

$\lambda_1 \vec{x} (\vec{x}^*) = \lambda_2 \vec{x} (\vec{x}^*)$. (i)

Note that $\vec{x}(\vec{x}^*) = ||\vec{x}||^2 \neq 0$ as $\vec{x}$ is non-zero. Multiplying both sides of (i) by $\frac{1}{||\vec{x}||^2}$ we obtain $\lambda_1=\lambda_2$. This is a contradiction as $\lambda_1$ and $\lambda_2$ are distinct. It follows that there cannot be a non-zero eigenvector (in this space) possessing multiple eigenvalues. Here $\vec{x}^*$ denotes the complex conjugate of $\vec{x}$. If the entries of $\vec{x}$ are real, then $\vec{x}^*$ is the transpose of $\vec{x}$.

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    $\begingroup$ Nice proof, +1, endorsed! $\endgroup$ Apr 25 '20 at 5:45
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If

$Ax = \lambda_1 x, \tag 1$

and

$Ax = \lambda_2 x, \tag 2$

then

$0 = Ax - Ax = \lambda_1x - \lambda_2x = (\lambda_1 -\lambda_2)x, \tag 3$

if

$\lambda_1 \ne \lambda_2, \tag 4$

so that

$\lambda_1 -\lambda_2 \ne 0, \tag 5$

then from (3),

$x = 0; \tag 6$

but this is precluded by the hypothesis

$x \ne 0; \tag 7$

we conclude that (1) and (2) cannot both bind.

Note added in Edit; Saturday 25 April 2020 8:38 PM PST: It is worth noting that the preceding demonstration does not require the entries of $A$ or $x$, or $\lambda_1$ and $\lambda_2$, to take values in $\Bbb Q$, $\Bbb R$, or $\Bbb C$, or indeed in any field. What does appear to be necessary is that $\lambda_1 - \lambda_2 \ne 0$ can be cancelled from both sides of (3), yielding (5). So for example if we are working over an integral domain $D$, with $A \in M_n(D)$ etc. then the argument succeeds; this is true even if $D$ does not support an inner product on vectors such as $x$. Such would be the case if for instance $D = \Bbb F_p$, the finite field with prime $p$ elements, and $n = p$. End of Note.

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    $\begingroup$ And yours is just as nice! +1 :) $\endgroup$
    – Jephph
    Apr 25 '20 at 5:49

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