Below is a proof that any linear operator must have an eigenvalue. The proof obviously contains a mistake because the statement is wrong. But I do not see the mistake. Please point it out if you see it.

Assumption: $V$ is a finitely dimensional $K$-linear space ($K$ is a field with infinitely many elements, e.g. $\mathbb R$), $V \ne \{0\}$, $L:V \to V$ is a linear operator.

Claim: $L$ must have an eigenvalue $\lambda \in K$, i.e. there must be $\lambda \in K$ and $0 \ne v \in V$ such that $L(v)=\lambda v$.

Proof: Assume that $L$ does not have an eigenvalue. Then for all $\lambda \in K$ it should hold: $L-\lambda I$ is bijective ($I$ is the identity operator).

So we have an infinite family of bijective linear operators $S:=(L-\lambda I)_{\lambda \in K}$ in the linear space $H$ of linear operators $V\to V$ that is finitely dimensional ($\dim H = (\dim V)^2$).

$S$ is a linear independent family. Indeed, let $a_1,\ldots,a_n \in K$ such that $0=\sum_{i=1}^n a_i (L-\lambda_i I)=(\sum_{i=1}^n a_i)L - (\sum_{i=1}^n a_i\lambda_i) I$. If not all $a_i$ are $0$, the $S \ni L - \frac{ \sum_{i=1}^n a_i\lambda_i}{\sum_{i=1}^n a_i} I=0$. Therefore $S$ contains a zero operator. This contradicts the statement above that all elements in $S$ are bijective and $V \ne \{0\}$.

So we have found an infinite linearly independent family $S$ in a finitely dimensional linear space $H$. That is a contradiction. This $L$ has an eigenvalue.

Addition: $H$ is finitely dimensional

Proof: fix a basis of $V$ say $e_1,\ldots,e_m$. Consider linear operator $L_{i,j}$ for which hold $L(e_i)=e_j$ and $L(e_k)=0$ for $k \ne i$. Than $H$ is panned by the finite family $(L_{i,j})_{j,i = 1,\ldots m}$

  • $\begingroup$ Why is the statement false? It is true in the complex plane $\endgroup$
    – FShrike
    Apr 27, 2022 at 19:16
  • $\begingroup$ I think $H$ is not finite $\endgroup$
    – FShrike
    Apr 27, 2022 at 19:18
  • $\begingroup$ @FShrike I have not assumed anything about $K$, it can be just $\mathbb R$. $\endgroup$
    – zesy
    Apr 27, 2022 at 19:19
  • 9
    $\begingroup$ $a_i\ne0$ for some $i$ does not imply $\sum_{i=1}^na_i\ne0$. $\endgroup$ Apr 27, 2022 at 19:24
  • 1
    $\begingroup$ By the way: you are implicitly assuming your field is infinite. Finite fields exist, so it may also be the case that you have only finitely many operators, and that there are fewer than $n^2$ of them. $\endgroup$ Apr 27, 2022 at 20:55

1 Answer 1


The statement that $S$ is linearly independent is false. $\sum_{i=1}^na_i$ can be zero even if not all the $a_i$ are zero. As a specific example of how this fails, (over $\mathbb R$) set $\lambda_1=2,\lambda_2=3,\lambda_3=1$ with $a_1=1,a_2=-\frac 12,a_3=-\frac 12.$ Then $$a_1(L-\lambda_1 I)+a_2(L-\lambda_2 I)+a_3(L-\lambda_3 I)=0$$


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