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I am interested in a linear mapping of the form $$ T(A) = A M - M A, \ \ A \in V $$ where $V = M^{n,n}$.

First, we suppose $n = 2$ and consider a simple numerical problem.

I was solving the problem to find a basis for the kernel of $T$, where $$ M = \left[ \begin{array}{cc} 1 & 2 \\ 0 & 3 \\ \end{array} \right] $$

So, we take $$ A = \left[ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right] $$

We solve the equation $$ A M - M A = 0 $$

After simplification, we get $$ \left[ \begin{array}{cc} -c & a + b - d \\ -c & c \end{array} \right] = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array} \right] $$ which reduces to two linear equations $$ a + b = d, \ \ c = 0 $$

Hence, a basis for $\mbox{Ker}(T)$ is given by $$ \mathcal{B} = \left\{ \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right], \left[ \begin{array}{cc} 1 & -1 \\ 0 & 0 \\ \end{array} \right] \right\} $$

Thus, for this example, nullity of $T$ = $2$.

In general, is there any relation between the nullity of $T$ and the rank of $M$ for the mapping (this looks like the Lie bracket) $$ T(A) = A M - M A, \ \ A \in V = \mathcal{M}^{n,n} $$

Dr. Sundar
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    Relevant: https://math.stackexchange.com/questions/92480/given-a-matrix-is-there-always-another-matrix-which-commutes-with-it – Matthew H. Apr 22 '22 at 00:57
  • The question in that exchange was this: "Given a matrix $A$ over a field $F$, does there always exist a matrix $B$ such that $A B=B A$?" In my question, the matrix $M$ is fixed, and $A$ ranges over the vector space of $n \times n$ matrices. If $M = D$ is a diagonal (commuting) matrix, then $T(A) = 0$ for all $A$ and naturally, nullity of $T$ is $n$. Interestingly, rank of $M$ is irrelevant in this calculation... Thanks for this example!! (This could be an answer to my question?) – Dr. Sundar Apr 22 '22 at 02:42
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    "is there any relation between the nullity of and the rank of " No. The spectrum of $M$ does play a decisive role though. Variations on this have been asked many times, e.g. https://math.stackexchange.com/questions/969714/the-rank-and-eigenvalues-of-the-operator-tm-am-ma-on-the-space-of-matric?rq=1 . An alternative approach is to view $T$ as isomorphic to $T'\big(\text{vec}(A)\big) = \big(-M \oplus M^T\big)\text{vec}(A)$ and use properties of the Kronecker sum. – user8675309 Apr 22 '22 at 03:19
  • Thank you, I shall read the discussion there! – Dr. Sundar Apr 22 '22 at 03:27
  • Somewhat related – PinkyWay Apr 24 '22 at 05:46

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