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Consider vector space $V$ consist all $n \times n$ matrix (real or complex).

What is the rank of the linear transformation $f(X)=AX-XA$ ($A\in V$)? ($A$ is a given matrix, which means we can have information about it)

I have tried to consider the basis of $V$ but it doesn't work.

EDITED:

$\operatorname{rank} f = n^2 - \operatorname{dim}N(f)$, which means we just need to find the demension of $N(f)$, which I think is much easier.

It turns out that it's not that easier.

EDIT 2: In the first answer it has been shown that the matrix version of the transformation has at least $n$ zero roots in the characteristic polynomial, however it's not true that these roots are accompanied with eigenvectors (Consider a nilpotent matrix of degree $n$ always have $n$ zero roots but may have only $1$ eigenvector). Therefore unless $A$ is diagonalisable, I think the problem is not solved yet.

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    $\begingroup$ Considering $X$ is a $n$ x $1$ vector, and $A$ a $n$ x $n$ matrix, how can $XA$ be defined? $\endgroup$ – MathMan Jul 7 '14 at 10:57
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    $\begingroup$ @VHP No, $X \in \mathbb K^{n\times n}$, so $f \colon V \to V$. $\endgroup$ – martini Jul 7 '14 at 10:58
  • $\begingroup$ One thing which I can think of at this moment is : let rank $A =r_1$ and rank $X = r_2$ $r_1 +r_2 -n< rank ~AX < min (r_1,r_2)$ ; ; ; $r_1 +r_2 -n< rank ~XA < min (r_1,r_2)$ $max~(0, r_1+r_2-n-min~(r_1,r_2))< rank ~AX - rank ~XA$$<min~(r_1,r_2)-(r_1 +r_2 -n)$ Also $rank~(A+B) \leq rank~A + rank~B$ We should be able to use these inequalities $\endgroup$ – MathMan Jul 7 '14 at 11:14
  • $\begingroup$ It depends on $A$ of course. If $A$ is a multiple of $I$, then the rank is zero. The rank must be less than $n^2$ because $f(A)=0$. But $n^2-1$ is possible, e.g. for $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$. $\endgroup$ – Henning Makholm Jul 7 '14 at 11:19
  • $\begingroup$ Calculating rank(AX-XA) seems to be useless. It almost has nothing to do with the rank of the linear transformation $f$ @HenningMakholm Of course, let's say what characters of A would determine the rank of f $\endgroup$ – Leaning Jul 7 '14 at 11:20
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Whether the underlying field is $\mathbb R$ or $\mathbb C$ makes no difference. The rank of $f$ is equal to the rank of its matrix representation $M=I\otimes A-A^T\otimes I$, but the rank of a real matrix over $\mathbb R$ is equal to its rank over $\mathbb C$. So, we may simply assume that the ground field is complex.

By a change of basis, we may further assume that $A$ is already in its Jordan form (in the sequel, if the Jordan form contains a submatrix of the form $\lambda_k I_{m_k}$, this submatrix will be viewed as a direct sum of $m_k$ trivial Jordan blocks). By the rank-nullity theorem, the problem boils down to finding all centralisers of a Jordan form $A$. Denote a Jordan block with eigenvalue $\lambda$ and size $p$ by $J_p(\lambda)$. If we impose a block structure on $X$ that conforms to the block structure of $A$, we can inspect the equation $AX=XA$ blockwise. Now, it is a technically straightforward but very dull exercise to show that:

  1. If $Y$ is a $p\times q$ rectangular matrix, then $J_p(\lambda_1)Y=YJ_q(\lambda_2)$ for some $\lambda_1\ne\lambda_2$ if and only if $Y=0$.
  2. If $Y$ is a $p\times q$ rectangular matrix, then $J_p(\lambda)Y=YJ_q(\lambda)$ if and only if $Y$ has the following form, where $T$ denotes any $\min(p,q)\times\min(p,q)$ upper triangular Toeplitz matrix: $$ \begin{cases} Y=T & \text{if } p=q>1,\\ Y=\pmatrix{T\\ 0} & \text{if } p>q>1,\\ Y=\pmatrix{0, T} & \text{if } 1<p<q,\\ \text{the first entry of } Y \text{ is zero} & \text{if } p>q=1,\\ \text{the last entry of } Y \text{ is zero} & \text{if } p=1<q,\\ Y \text{ is any scalar } & \text{if } p=q=1. \end{cases} $$

It follows that if $A$ is decomposed into a direct sum of Jordan blocks $A = \bigoplus_{i=1}^m J_{p_i}(\lambda_i)$, the nullity of $f$ is given by $\sum_{i=1}^m\sum_{j=1}^m r(i,j)$, where $$ r(i,j)= \begin{cases} 0 & \text{if } \lambda_i\ne \lambda_j,\\ \min(p_i,p_j) & \text{if } \lambda_i=\lambda_j \text{ and } \min(p_i,p_j)>1,\\ \max(p_i,p_j)-1 & \text{if } \lambda_i=\lambda_j \text{ and } \max(p_i,p_j)>\min(p_i,p_j)=1,\\ 1 & \text{if } \lambda_i=\lambda_j \text{ and } p=q=1 \end{cases} $$ and $\operatorname{rank}(f)=n^2-\sum_{i=1}^m\sum_{j=1}^m r(i,j)$. In the special case where $A$ is diagonalisable (so that $m=n$ and $p_i=1$ for all $i$), this reduces to $\operatorname{rank}(f)=n^2-\sum_{i=1}^n\sum_{j=1}^n 1_{\lambda_i=\lambda_j}=\sum_{i=1}^n\sum_{j=1}^n 1_{\lambda_i\ne\lambda_j}$.

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  • $\begingroup$ so the reasoning about $n$ zero eigenvalues is no useful in the general case? $\endgroup$ – Leaning Sep 12 '14 at 14:04
  • $\begingroup$ I only need to understand the bound of daw's answer. What I don't understand is: Why the nullity of $f$ is at least $n$? When I write down the form of the vectorization of operator, I can see that it has $n$ $0$'s value in the diagonal, and the characteristic polynomial has $x^n$. But how exactly does it lead to the fact that $dim ker f$ is at least n? $\endgroup$ – Leaning Sep 12 '14 at 14:14
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    $\begingroup$ For each Jordan block $J$ of size $p$, there exists $p$ linearly independent Toeplitz matrices $T=T_1,T_2, \ldots, T_p$ (namely, $T_p=I$ and $T_j$ with $j<p$ is the upper triangular matrix with ones on its $j$-th super diagonal and zeros elsewhere) such that $JT=TJ$. So, for every $n\times n$ Jordan form $A$, there are at least $n$ blockwise Toeplitz matrices $X$ (in the form of $0\oplus\cdots\oplus0\oplus T\oplus0\oplus\cdots\oplus0$ for some Toeplitz matrix $T$ that is zero except on a certain super diagonal) such that $AX=XA$, i.e. $M\operatorname{vex}(X)=0$. $\endgroup$ – user1551 Sep 12 '14 at 14:45
  • $\begingroup$ That's the point I'm waiting for. How can we prove that $JT=TJ$? $\endgroup$ – Leaning Sep 12 '14 at 14:49
  • $\begingroup$ seem like I just need to write it down explicitly. Thanks for the answer!!!!!! $\endgroup$ – Leaning Sep 12 '14 at 14:57
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The equation is a special form of a Sylvester equation. Using vectorization operation, the equation $$ AX-XA=0 $$ is equivalent to $$ (I\otimes A - A^T\otimes I)vec(X)=0. $$ Denote the eigenvalues of $A\in \mathbb C^{n,n}$ by $\lambda_1\dots \lambda_n$. Then the matrix $I\otimes A - A^T\otimes I$ has eigenvalues $\lambda_{ij} = \lambda_i - \lambda_j$, $i,j=1\dots n$.

Thus, the dimension of the null space of $f$ depends on the spectrum of $A$. The dimension of the null space is at least $n$, which appears if the eigenvalues of $A$ are distinct. On the other extreme, the dimension of the null space is $n^2$ if $A$ is a multiple of the identity.

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  • $\begingroup$ Can you explain the vectorization operation in basic language? $\endgroup$ – Leaning Jul 7 '14 at 12:18
  • $\begingroup$ See en.wikipedia.org/wiki/Vectorization_%28mathematics%29 , the idea is to rewrite $AX-XA=0$ as $vec(AX-XA)=0$, then use $vec(AX)=(I\otimes A)vec(X)$ and $vec(XA)=(A^T\otimes I)vec(X)$. $\endgroup$ – daw Jul 7 '14 at 12:27
  • $\begingroup$ Vectorization: Take a matrix, then stack all columns into one large column vector of size $n^2\times 1$. $\endgroup$ – daw Jul 7 '14 at 12:28
  • $\begingroup$ Thanks for an awesome solution. $\endgroup$ – Leaning Jul 7 '14 at 12:48
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I just read this file; here the Kronecker product is useless. Indeed $rank(f)=n^2-dim(C(A))$ where $C(A)$ is the commutant of $A$. According to the Jordan decomposition theory, it suffices to obtain $dim(C(A))$ when $A$ is nilpotent. Thus, let $n_0=0,n_k= dim(ker(A^k))$. It is well-known that $dim(C(A))=\sum_{k\geq 0}(n_{k+1}-n_k)^2$. In particular, $dim(C(A))$ is a sum of squares ${m_k}^2$ s.t. $\sum_km_k=n$. Conversely, if we give such a sequence, then there is an associated matrix $A$.

$rank(f)$ has a geometric sense: let $V$ be the algebraic variety of matrices that are similar to $A$; then the elements of $V$ depend on $rank(f)$ free parameters.

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