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Prove that

$A$ and $B$ have the same set of $n$ linearly independent eigenvectors $\Leftrightarrow$ $AB=BA$.


I succeed to prove the right way $(\Rightarrow)$ but failed to do the left way $(\Leftarrow)$.

Let $\Lambda^{(a)}$ and $\Lambda^{(b)}$ be the eigenvalue matrix of $A$ and $B$, respectively, i.e., $$\Lambda^{(a)} = \begin{bmatrix}\lambda_1 &0&\cdots &0 \\ 0 &\lambda_2 & \cdots &0\\ \vdots&\vdots &\ddots &\vdots\\0&0&\cdots&\lambda_n \end{bmatrix} \quad \text{and} \quad \Lambda^{(b)} = \begin{bmatrix}\lambda'_1 &0&\cdots &0 \\ 0 &\lambda'_2 & \cdots &0\\ \vdots&\vdots &\ddots &\vdots\\0&0&\cdots&\lambda'_n \end{bmatrix}.$$

Let $S$ be the eigenvector matrix of both $A$ and $B$, i.e. $$S = \begin{bmatrix}x_1&x_2&\cdots&x_n \end{bmatrix}$$ such that $AS=S\Lambda^{(i)}$ for $i\in\{a,b\}$.

Then, from the fact that the diagonal matrix, e.g. $\Lambda^{(a)}$a and $\Lambda^{(b)}$, is commutative, i.e., $\Lambda^{(a)}\Lambda^{(b)}=\Lambda^{(b)}\Lambda^{(a)}$, I can prove that \begin{align}AB&=(S\Lambda^{(a)}S^{-1})(S\Lambda^{(b)}S^{-1})\\&=S\Lambda^{(a)}\Lambda^{(b)}S^{-1}\\&=S\Lambda^{(b)}\Lambda^{(a)}S^{-1}\\&=(S\Lambda^{(b)}S^{-1})(S\Lambda^{(a)}S^{-1})\\&=BA.\end{align}


Can someone prove the left way?

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    $\begingroup$ The statement is true only if both $A$ and $B$ are diagonalisable. (If an $n\times n$ matrix is not diagonalisable, there aren't $n$ linearly independent eigenvectors in the first place.) $\endgroup$ – user1551 Dec 13 '17 at 9:01
  • $\begingroup$ @user1551 Is it the same statement between that the matrix has $n$ linearly independent eigenvectors and that the matrix is diagonalizable? I thik the both statements are the same, so the proposition that I want to prove holds true, doesn't it? $\endgroup$ – Danny_Kim Dec 13 '17 at 9:07
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    $\begingroup$ They are equivalent among themselves, but they are not equivalent to the condition that $AB=BA$. Let $A=I$ and $B=\pmatrix{0&1\\ 0&0}$. Then $AB=BA$ but $B$ is not diagonalisable. $\endgroup$ – user1551 Dec 13 '17 at 9:12
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If $AB=BA$ and $A$ is diagonalizable with distinct eigenvalues, then you can prove they share eigenvectors.

Let $x$ be an eigenvector of $A$ with eigenvalue $\lambda$. Then $A(Bx)=BAx=B\lambda x=\lambda (Bx)$. This shows that $Bx$ is also in the eigenspace corresponding to $\lambda$. But this eigenspace has dimension $1$ (by the assumption), so $Bx$ must be a scalar multiple of $x$, proving $x$ is a eigenvector of $B$.

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