Show that $B$ must equal $0$ if $A$ and $C$ have no common Eigenvalues

Question

If $AB=BC$, and $A$ and $C$ have no common eigenvalues, show that $B$ must equal $0$.

How do I do this?

Answer 1

Sorry for the first comment, it is not so useful to do this by definition.

There is an answer based on $\lambda$ matrix. Let $f_A(\lambda)=|A-\lambda E|$ be the characteristic polynomial of $A$. It is easy to show, as a consquence of $AB=BC$ that $0=f_A(A)B=Bf_A(C)$. What's left is to show that $f_A(C)$ is invertable.

Let $F_A(\lambda)=\prod_{i=1}^{dim A}(\lambda-\lambda_i)$, then $F_A(C)=\prod_{i=1}^{\dim A}(C-\lambda_iE)$, then $A,C$ has non-comm. eig. implies that $C-\lambda_iE$ must be invertable. If not, then $\exists X\neq0$, such that $CX=\lambda_iX$, which means that $\lambda_i$ is an eig. value of $C$, but it is also an eig. value of $A$, contradiction!

Answer 2

$\newcommand{\VEig}{\mathrm{VEig}}$ But I also follow my first comment, just go like @AWertheim, we only need to show that if for any $v\in \VEig(C)$ (the eig. vector spance of $C$), we have $Bv=0$, then $B=0$.

Firstly, if $C$ is diagonalizable, then $\dim\VEig(C)=\dim C$, thus the results holds. But how about $C$ is not diagonalizable?

By the standard theory of LA, we can consider $C$ as its Jordan canonical form, for simplicity, let us suppose $C=\pmatrix{0&1\\0&0}$ be a Jordan block, then you can check that in this case the results holds too.

In fact, $\VEig(C)=\left\{a\pmatrix{1\\0}|a\in R\right\}$, and $B=\pmatrix{0&b_1\\0&b_2}$. Then $AB=BC=\pmatrix{0&0\\0&0}$. Suppose $A=\pmatrix{\alpha_1\\\alpha_2}$, and let $\beta=\pmatrix{b_1\\b_2}$, then $AB=0$ implies that $\alpha_1,\alpha_2\perp \beta$. If $\beta\neq0$, then $\alpha_1$ and $\alpha_2$ must be linear dependent, which implies that $A$ is of rank $1$ at most, and it must have a non-zero solution of $AX=0$, and $0$ is an eigenvalue of $A$, which is impossible since it is also an eigenvalue of $C$.

For the general case, I suppose it can work in the same way.

Answer 3

Another approach: note that $$ (A - \lambda I)B = AB - \lambda B = BC - \lambda B = B(C - \lambda I) $$ Suppose that $\lambda$ is an eigenvalue of $C$. It follows, since there are no eigenvalues in common, that $(A - \lambda I)$ is invertible. Thus, we have $$ B = (A - \lambda I)^{-1}B(C - \lambda I) $$ In particular, this is true for every eigenvalue $\lambda$ of $C$. So, it follows that for the eigenvalues $\lambda_1,\dots,\lambda_n$ of $C$, we have $$ B = (A - \lambda_1 I)^{-1} B (C - \lambda_1 I) =\\ (A - \lambda_2 I)^{-1} \overbrace{(A - \lambda_1 I)^{-1} B (C - \lambda_1 I)}^B (C - \lambda_2 I) = \\ \vdots\\ (A - \lambda_n I)^{-1} \cdots (A - \lambda_1 I)^{-1} B (C - \lambda_1 I) \cdots (C - \lambda_n I) $$ Letting $p_C(x)$ be the characteristic polynomial of $C$, this becomes $$ B = [p_C(A)]^{-1}B[p_C(C)] $$ by the Cayley-Hamilton theorem, $p_C(C) = 0$, so that the above implies $B = 0$.

Answer 4

Just for grins let's try a Kronecker approach (here is a good source). Expressing $AB-BC=0$ in vector form using Kronecker products gives us $$\mathop{\textrm{vec}}(AB-BC) = (I\otimes A-C^T\otimes I)\mathop{\textrm{vec}}(B) = 0$$ So if $I\otimes A-C^T\otimes I$ is nonsingular, then $B=0$. Note that $$I\otimes A-C^T\otimes I=A\oplus-C^T$$ where $\oplus$ denotes the Kronecker sum. If the eigenvalues of $A$ are $\lambda_1,\dots,\lambda_n$ and the eigenvalues of $C$ are $\mu_1,\dots,\mu_n$, the eigenvalues of $A\oplus -C^T$ are $$\lambda_i-\mu_j \quad \forall i,j=1,2\dots,n$$ So indeed, it is easy to see that if $A$ and $C$ have distinct eigenvalues, then $I\otimes A-C^T\otimes I$ must nonsingular, because none of the quantities $\lambda_i-\mu_j$ would be zero.

Let's prove that the eigenvalues of $A\oplus -C^T$ are what I say they are. Let $v_i$ be a right eigenvector of $A$ corresponding to $\lambda_i$, and $w_j$ be an left eigenvector of $C$ corresponding to $\mu_j$. Then

$$\begin{aligned} (I\otimes A-C^T\otimes I)(w_j\otimes v_i)&=w_j\otimes Av_i-C^Tw_j\otimes v_i\\ &=w_j\otimes\lambda_iv_i-\mu_jw_j\otimes v_i\\ &=\lambda_i(w_j\otimes v_i)-\mu_j(w_j\otimes v_i) \\ &=(\lambda_i-\mu_j)(w_j\otimes v_i)\end{aligned}$$ So $w_i\otimes v_i$ is indeed a right eigenvector of our Kronecker sum, with eigenvalue $\lambda_i-\mu_j$.