# Distinct eigenvalues' eigenvector

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 $$\lambda_2$$.

• Apr 27 '20 at 8:19

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 $$\vec{x}$$.

• Nice proof, +1, endorsed! Apr 25 '20 at 5:45

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.

• And yours is just as nice! +1 :) Apr 25 '20 at 5:49