Problem statement:

Let $n \ge 2 $ be an integer. Suppose that A is an $n \times n$ matrix and that $\lambda_1$, $\lambda_2$ are eigenvalues of A with corresponding eigenvectors $v_1$, $v_2$ respectively. Prove that if $v_1$, $v_2$ are linearly dependent then $\lambda_1 = \lambda_2$.

I have an intuition as to why this is true, but am having difficulty formalizing a proof. What I have doesn't seem tight enough.

If $v_1$ and $v_2$ are linearly dependent then $v_1$ lies in the span of $v_2$. If two eigenvectors lie in the span of one another then only one of them is required in order to form a basis of the eigenspace.

All eigenvalues correspond to a single $n\times 1$ eigenvector or a set of $n\times 1$ linearly independent vectors.

Since $v_1$ and $v_2$ are linearly dependent, we know that there can only be one eigenvalue that corresponds to the single eigenvector.

Thus $\lambda_1$ must equal $\lambda_2$.

Any thoughts or criticism are welcome.



1 Answer 1


You know that $$Av_1=\lambda_1v_1\\Av_2=\lambda_2v_2$$

If $v_1,v_2$ are linearly dependent, then $v_1=\mu v_2$ for some scalar $\mu$. Putting this in the first equation,

$$A(\mu v_2) = \lambda_1(\mu v_2) \implies Av_2 = \lambda_1 v_2$$

This gives $\lambda_1=\lambda_2$ as desired.

I think your idea is on the right track, but putting it in the above way gives more clarity.


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