Nilpotent and Diagonalizable

Algebra Level 4

Any $A\in M_n(\mathbb C)$ yields a linear transformation $T_A$ on vector space $M_n(\mathbb C)$ defined by $T_A(X)=AX-XA.$ Which of the following statements is/are true?

Ⅰ. If $A$ is nilpotent , that is, $A^k=0$ for some positive integer $k,$ then $T_A$ is also nilpotent.

Ⅱ. If $A$ is diagonalizable, then $T_A$ is also diagonalizable.

Bonus: What about the converses of Ⅰ and Ⅱ?

Neither Ⅰ only Ⅱ only Ⅰ and Ⅱ

1 solution

Otto Bretscher
Dec 30, 2018

(I) We can see, by induction, that $T_A^m(X)$ is a linear combination of expressions of the form $A^pXA^q$ where $p+q=m$ . Thus, if $A^k=0$ , then $T_A^{2k}=0$ .

(II) If $\vec{v}_1,...,\vec{v}_n$ is an eigenbasis for $A$ , with associated eigenvalues $\lambda_1,...,\lambda_n$ , consider the matrix $M_{ij}$ whose $j$ th column is $\vec{v}_i$ , with 0's elsewhere. The matrix of $T_A$ with respect to the basis $(M_{ij})$ will be diagonalizable. More precisely, it will have $n$ diagonalizable blocks $\lambda_iI_n-A^T$ along the diagonal.

The converse of (I) fails to hold, of course: If $A=I_n$ then $T_A=0$ . The converse of (II) does hold, though; consider the Jordan normal form.

Wow! Great solution! Would you be kind enough to elaborate on (II)? I'm not familiar with your technique. I see that you have brilliantly formed a basis of $n^2$ $nxn$ matrices (for $M_{nxn}(C)$ ). I also understand why $T_A$ is a linear transformation. For me to understand the remainder of your solution I must know how you determined $T_A$ as a matrix. Because $X$ and $T_A(X)$ are $nxn$ , it comes to reason that $T_A$ must be $nxn$ . Since your diagonalization has an $n^2xn^2$ matrix in the center, this can only mean that the matrices on the sides must be $nxn^2$ and $n^2xn$ . But isn't that an impossibility? I only know of cases where diagonalizations result in square matrices. How can you take the inverse of a non-square matrix? I am definitely missing something major here. Hope you or someone else is willing to help me understand.

James Wilson - 2 years, 5 months ago

If $T: V \rightarrow V$ is a linear map with $\dim V=m$ , then the matrix of $T$ with respect to a chosen basis of $V$ will be $m \times m$ . In the case of $T_A$ we have $V=\mathbb{C}^{n\times n}$ with $\dim V = n^2$ so that the matrix of $T_A$ will me $n^2 \times n^2$ .

To understand what is going on here, you may find it helpful to work out a simple case, for example, $A=\begin{bmatrix} 1 & 1\\ 0 & 0\end{bmatrix}$ .

Otto Bretscher - 2 years, 5 months ago

Ah ok, that helps a lot! Thanks for taking the time to answer my question! I tried your example with $A$ =[1 1; 0 0] and found $T_A$ =[0 1 0 0; 0 -1 0 0; -1 0 1 1; 0 -1 0 0] (by concatenating the columns of both X and AX-XA and setting TX = AX-XA). Some time in the near future, I'll try to spend some more time to try to figure out how to diagonalize $T_A$ using your method.

James Wilson - 2 years, 5 months ago

If you use an eigenbasis for $A$ , for example, $(1,0),(1,-1)$ , then the matrix of $T_A$ comes out to be $\begin{bmatrix} 0&0&0&0\\-1&1&0&0\\0&0&-1&0\\0&0&-1&0\end{bmatrix}$ , as I mention in my solution, with the $2\times2$ blocks $\lambda_iI_2-A^T$ on the diagonal, and that one is easy to diagonalise. You can also see that the eigenvalues of $T_A$ are the differences of the eigenvalues of $A$ .

Otto Bretscher - 2 years, 4 months ago

Nilpotent and Diagonalizable

1 solution

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