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In $M_n(\mathbb{C})$, define $ad_A$ is the operator $ad_A(X)=AX-XA$ in $\mathcal{L}(M_n(\mathbb{C}))$.

Does $ad_A$ semisimple imply $A$ semisimple?

Here 'semisimple' means that 'diagonalizable'.

(It has been proven that $A$ semisimple implies $ad_A$ semisimple, and if $\lambda_1,\cdots,\lambda_n$ are eigenvalues of $A$, then $\lambda_i-\lambda_j$'s are the eigenvalues of $ad_A$.)

Thanks.

dailycrazy
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2 Answers2

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Yes, the implication is true. Suppose the contrary that $A$ has a non-semisimple eigenvalue $\lambda$. Then there exist two linearly independent sets of generalised eigenvectors $\{u_1,u_2\}$ and $\{v_1,v_2\}$ such that $$ (A-\lambda I)u_1=0,\ (A-\lambda I)u_2=u_1,\ v_1^T(A-\lambda I)=0\ \text{ and }\ v_2^T(A-\lambda I)=v_1^T. $$ Let $X=u_2v_2^T$. Then $\operatorname{ad}_A(X)\ne0$ but $\operatorname{ad}_A^3(X)=0$. Hence $\operatorname{ad}_A$ is not diagonalisable.

user1551
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The other answer gives a computational approach.

A conceptual approach is: For any semisimple Lie algebra $L$ contained in some matrix algebra over a field of characteristic $0$, an element $A \in L$ is semisimple (as a matrix) if and only if $ad_A$ is semisimple. See e.g. Bourbaki's book on Lie Groups and Algebras, chapter I ยง6 no. 3, or any other good book on Lie algebras, or Does the abstract Jordan decomposition agree with the usual Jordan decomposition in a semisimple Lie subalgebra of endomorphisms? and its duplicates.

Now $M_n(\mathbb C)$ as a Lie algebra (for which one commonly uses the notation $\mathfrak{gl}_n(\mathbb C)$) is not semisimple, so the above cannot be applied directly. However, we happen to have $\mathfrak{gl}_n = \mathfrak{sl}_n \oplus \mathrm{ker}(ad)$, where $\mathfrak{sl}_n$ is semisimple, and the kernel of $ad$ consists of the scalar matrices $\mathbb C \cdot Id_n$ (which, in particular, are semisimple matrices); from which the statement follows.

Torsten Schoeneberg
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