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Question 1: Let $A$, $B$ be two $n\times n$ complex matrix satisfy: $AB-BA=0$. Then $A$, $B$ have a common eigenvector.

Question 2: Let $A$, $B$ be two $n\times n$ complex matrix satisfy: $AB-BA=B$. Then $A$, $B$ have a common eigenvector.

Question 1 It is easy to prove.

Let $\lambda$ be a eigenvalue of $A$ and $V_\lambda$ be the eigensubspace. For any $x\in V_\lambda$, we have $A(Bx)=\lambda(Bx)$. So $V_\lambda$ is the invariant subspace of $B$.

Question 2 I have some ideas but fail to solve it all.

Let $Ax=\lambda x$. Then we have $A(Bx)=(\lambda+1)(Bx)$. Assume $V_{\lambda_1}$, $V_{\lambda_2}$,..,$V_{\lambda_s}$ are all eigensubspaces of $A$ and $n_0,...,n_s$ are the index: for any $i$, there exists $x\in V_{\lambda_i}$ and we have $B^{n_i}x\not=0$ but for all $y\in V_{\lambda_i}$, $B^{n_i}y=0$. How is the next step? Or is there any different idea?

Can someone help me? Thank you.

Community
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gaoxinge
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2 Answers2

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For the second, you already have what is needed, you just need to wrap it up:

If $x$ is an eigenvector to the eigenvalue $\lambda$ of $A$, then you have

$$A(Bx) = B((I+A)x) = B((\lambda+1)x) = (\lambda+1)Bx,$$

so $Bx = 0$ or $Bx$ is an eigenvector to the eigenvalue $\lambda+1$ of $A$.

Now choose an eigenvalue $\lambda_m$ of $A$ such that $\lambda_m+1$ is not an eigenvalue of $A$ - such an eigenvalue exists since $A$ has only finitely many eigenvalues. Then if $x_m$ is an eigenvector to the eigenvalue $\lambda_m$ of $A$, the above shows $Bx_m = 0$, hence $x_m$ is an eigenvector to the eigenvalue $0$ of $B$, and thus a common eigenvector.

Daniel Fischer
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For the second question note that $1)$ $B$ is not invertible. $2)$ $$A(\ker B)\subset \ker B$$ Proof $1)$ Suppose that $B$ is invertible therefore exist a matrix $B^{-1}$ such that $B^{-1}B=I$. Multiplying both sides for $B^{-1}$ we obtain that $$B^{-1}AB=A+I$$ therefore $A$ is similar to $A+I$ but two matrix similar have the same trace $$tr(A+I)\neq tr(A)$$ therefore $B$ is not invertible. If $v$ is a vector belongs $\ker B$ then $$B(Av)=0$$ therefore $Av$ is a vector of $\ker B$. Consider now the map $f$ associated to $A$ in canonical basis and considering the map: $$f: \ker B\to \ker B$$ because $A$ and $B$ are two complex matrix $f$ has at least an eigenvector. This prove the question.

Domenico Vuono
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