I've been trying to prove the following statement:

Let $A,B\in \mathbb R^{n\times n}$ be square matrices such that they share $n$ common linearly-independent eigenvectors. Then $AB=BA$.

Everything I thought of seemed to be unhelpful, so I have no clue what to do next.

Thank you and have a nice day!.

  • 1
    $\begingroup$ Hint: write the matrices in terms of basis given by those eigenvectors. $\endgroup$ – Wojowu Jan 13 at 18:18
  • $\begingroup$ @Wojowu I think I got it. Thanks! $\endgroup$ – Amit Zach Jan 13 at 18:26

Note that though $A, B \in \Bbb R^{n \times n}$, the eigenvalues may be complex and thus the eigenvectors may be in $\Bbb C^n$. Bearing this in mind, we proceed:


$\vec e_1, \vec e_2, \ldots, \vec e_n \tag 1$

be the $n$ common, linearly independent eigenvectors of $A$ and $B$; then we have

$A\vec e_i = \alpha_i \vec e_i, \; 1 \le i \le n, \tag 2$


$B\vec e_i = \beta_i \vec e_i, \; 1 \le i \le n, \tag 3$

for some scalars $\alpha_i, \beta_i \in \Bbb C$, $1 \le i \le n$. Thus

$AB \vec e_i = A(\beta_i \vec e_i) = \beta_i A \vec e_i = \beta_i \alpha_i \vec e_i = \alpha_i \beta_i \vec e_i = \alpha_i B\vec e_i = B \alpha_i \vec e_i = BA\vec e_i. \tag 4$

Since the $\vec e_i$ are $n$ in number and linearly independent, the form a basis of $\Bbb C^n$; thus for any

$\vec v \in \Bbb C^n \tag 5$

we may write

$\vec v = \displaystyle \sum_1^n v_i \vec e_i, \; v_i \in \Bbb C, \; 1 \le i \le n,; \tag 6$


$AB \vec v = \displaystyle \sum_1^n v_i AB \vec e_i = \sum_1^n v_i BA \vec e_i = BA\vec v; \tag 7$

therefore we conclude

$AB = BA. \tag 8$

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    $\begingroup$ Thank you very much, this is exactly what I did after the hint I was just given. I'm happy to see that my proof was similar to your professional one :) $\endgroup$ – Amit Zach Jan 13 at 19:29
  • $\begingroup$ Thanks for the kind words. If you really think my proof is "professional", you might consider "accepting it". Cheers! ;) $\endgroup$ – Robert Lewis Jan 13 at 19:31

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