I'm aware of

and likely more. However, I'd like to check the following proof; I'll gladly delete and/or post a different answer to the non-duplicate one of those if desired.

Theorem: with $\vec a,\vec b$ eigenvectors of $A$ corresponding to distinct eigenvalues $x,y$, we have $\vec a$ and $\vec b$ are linearly independent.

Proof: we have $A\vec a=\vec ax$ and $A\vec b=\vec by$ where $x-y\neq0$. Hence

$$\vec a·\vec b=\frac{x-y}{x-y}\;(\vec a·\vec b)=\frac{x(\vec a\cdot\vec b)-y(\vec b\cdot\vec a)}{x-y}=\frac{(A\vec a)\cdot\vec b-(A\vec b)\cdot\vec a}{x-y}=\frac{A(\vec a\cdot\vec b-\vec b\cdot\vec a)}{x-y}=0$$ which implies linear independence.

As an aside, I'm curious if a step such as $\vec a\cdot(A\vec b)=(\vec a A)\cdot\vec b$ is valid; I was originally experimenting with such expressions. I was interpreting matrix-vector multiplication as commutative, so we would derive $\vec a·(A\vec b)=(A\vec a)·\vec b$. Is this a valid identity?

  • 4
    $\begingroup$ The last step is totally wrong. Note that $A\vec a\cdot\vec b$ is a scalar and $A(\vec a\cdot\vec b)$ is a scalar multiple of a matrix. Your experimentation is correct if and only if $A$ is synmetric. $\endgroup$ Oct 2, 2023 at 0:24
  • 2
    $\begingroup$ Two comments: 1) your next to last equality does not make sense: $(Aa)\cdot b$ is a scalar, but $A(a\cdot b)$ is, if something, a matrix. 2) There is no need to use the dot product at all. This property is true at the level of vector spaces. $\endgroup$
    – GReyes
    Oct 2, 2023 at 0:25
  • 1
    $\begingroup$ This argument cannot possibly work, because it's just not true that eigenvectors associated to distinct eigenvalues need to be orthogonal. $\endgroup$ Oct 2, 2023 at 0:26
  • $\begingroup$ @QiaochuYuan Well.. it is true if the matrix is symmetric and your space is Euclidean, as Ted Shifrin pointed out. $\endgroup$
    – GReyes
    Oct 2, 2023 at 2:32


You must log in to answer this question.

Browse other questions tagged .