---
tag: 7NSB724DV3
kind: proof
name: Proof that the eigenvalues of a self-adjoint operator are real
msc: [47B15, 15A18]
proves: 0C4NT4GT3X
uses: [0VGSYBWHV2, 0MB8HQW495, 0ZND7NQ970, 0BJV24C4SY]
sources:
  - url: https://linear.axler.net/LADR4e.pdf
    locator: Theorem 7.12
---

## Proof

We compute $\ip{T(v)}{v}$ in two ways for an eigenvector $v$.

1. Let $V$ be a finite-dimensional inner product space over $\mathbb{K} = \R$ or $\C$, with [[inner product|0VGSYBWHV2]] $\ip{\cdot}{\cdot}$. Let $T \colon V \to V$ be [[self-adjoint|0MB8HQW495]], so $\ip{T(v)}{w} = \ip{v}{T(w)}$ for all $v, w \in V$.
2. If $\mathbb{K} = \R$, every scalar is real, so there is nothing to prove. Let $\mathbb{K} = \C$.
3. Let $\lambda$ be an [[eigenvalue|0ZND7NQ970]] of $T$. So there is $v \in V$ with $v \ne 0$ and $T(v) = \lambda v$.
4. Linearity in the first argument gives $\ip{T(v)}{v} = \ip{\lambda v}{v} = \lambda \ip{v}{v}$.
5. Conjugate symmetry and linearity in the first argument give $\ip{v}{\lambda v} = \overline{\lambda \ip{v}{v}} = \overline{\lambda}\, \ip{v}{v}$, because $\ip{v}{v}$ is real.
6. By step 1, $\ip{T(v)}{v} = \ip{v}{T(v)} = \ip{v}{\lambda v}$. So steps 4 and 5 give $\lambda \ip{v}{v} = \overline{\lambda} \ip{v}{v}$.
7. Since $v \ne 0$, positivity gives $\ip{v}{v} > 0$. Dividing by it gives $\lambda = \overline{\lambda}$.
8. A complex number equal to its [[complex conjugate|0BJV24C4SY]] has imaginary part $0$. So $\lambda$ is real.

## Idea

Self-adjointness lets $T$ move from one side of the inner product to the other. On an eigenvector, moving $T$ to the left gives the factor $\lambda$, and moving it to the right gives $\overline{\lambda}$. So $\lambda = \overline{\lambda}$.