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I initially encountered a problem in this question, which I couldn't resolve using fundamental concepts. However, thanks to the results obtained from this question, I can now demonstrate that the dimension of the commutant space, denoted as $C(A)$, equals $n$ when $A$ is a diagonal matrix with pairwise distinct entries on its main diagonal. To establish this, I introduced the following basis:

$$B = \{E_{11}, E_{22}, \dots, E_{nn}\}$$

Each $E_{ij}$ represents an $n\times n$ matrix with a value of one at position $(i, j)$ and zeros elsewhere. It is evident that we have $n^2$ such matrices of the form $E_{ij}$. Notably, these matrices are linearly independent and collectively generate all $n\times n$ matrices over real numbers. Consequently, I was able to demonstrate that $B$ generates $C(A)$ while maintaining linear independence. Thus, the dimension of $C(A)$ is $\text{dim }C(A)=|B|=n$.

In the more general case (as described in the initial question), I employed the following approach:

Matrix $A$ can fall into one of three categories:

  1. $A$ is a diagonal matrix with pairwise distinct entries on its main diagonal: In this case, I have shown that $\text{dim }C(A)=n$.

  2. $A$ is not a diagonal matrix: As indicated in the second question linked to this one, $A$ does not commute with diagonal matrices having different entries on the main diagonal. As a result, the basis should not generate such matrices. Therefore, we cannot retain those $n$ matrices from $B$ introduced at the beginning, i.e., $E_{11}, E_{22}, \dots, E_{nn}$. However, we need to introduce additional elements into the basis to account for the matrices we have removed, those $n$ elements that commute with $A$. This leads to the conclusion that $\text{dim }C(A)\geq n^2-n=n(n-1)$. Since $n>1$ and is a natural number, this shows that $\text{dim }C(A)\geq n$.

  3. $A$ is a diagonal matrix that may have equal entries on the main diagonal: These matrices commute with all other diagonal matrices and some other matrices. Consequently, the basis must include $E_{11}, E_{22}, \dots, E_{nn}$ along with some other matrices. Thus, $\text{dim }C(A)\geq n$ in this case as well.

I acknowledge that this reasoning may not be entirely rigorous mathematically. If you spot any errors, I would greatly appreciate your feedback. Additionally, verifying the correctness of this reasoning would be highly valuable to me.

EDIT

I encourage you to read the comments below, particularly my second comment.

Mason Rashford
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  • Your conclusion from part 2 that $\dim C(A) \geq n^2 - n$ first of all makes sense and second is demonstrably incorrect. However, your work for 1 and 3 is fine: it is indeed true that if $A$ is is diagonal, then its commutant will include every element of $B$ and therefore have dimension at least $n$. – Ben Grossmann Nov 06 '23 at 19:10
  • Also, it is not necessarily true that if $A$ is not diagonal, then $C(A)$ does not contain the elements $E_{ii}$. For instance, the matrix $$ A = \pmatrix{0&0&0\0&1&1\0&1&1} $$ is not diagonal, but $E_{11} \in C(A)$ – Ben Grossmann Nov 06 '23 at 19:13
  • A typical way to generalize your result is to use it to conclude that if $A = SDS^{-1}$ for an invertible matrix $S$ and diagonal $D$, then we can similarly conclude that $\dim C(A) \geq n$ using the fact that $DB = BD$ if and only if $(SDS^{-1})(SBS^{-1}) = (SBS^{-1})(SDS^{-1})$. That is, we can conclude that $\dim C(A) \geq n$ for all diagonalizable $A$. – Ben Grossmann Nov 06 '23 at 19:18
  • @BenGrossmann Is $A=SDS^{-1}$ a well-known decomposition? Or is there a method to find it? In this way we proved for case 1, 2 and diagonalizable matrices. What about other matrices? – Mason Rashford Nov 06 '23 at 20:24
  • Yes, it's well known; see section 6.2 of Hoffmann and Kunze. Taking an analogous approach for other matrices would require some kind of canonical form like JCF or the cyclic decomposition, both of which you're apparently not allowed to use here. You could also use the fact that "most" matrices are diagonalizable and make an appeal to continuity, but that to is beyond what you have studied so far. – Ben Grossmann Nov 06 '23 at 21:16
  • @BenGrossmann about the counter-example you said for second case, can we say the basis can contain some of $E_{ii}$ but surely not all of them. Hence the statement becomes true. – Mason Rashford Nov 07 '23 at 07:05
  • Ok, so $C(A)$ doesn't include every matrix $E_{ii}$. How do you start from there and land on the conclusion that $\dim C(A) \geq n^2 - n$? – Ben Grossmann Nov 07 '23 at 15:43
  • It is notably not necessarily the case that $\dim C(A) \geq n^2 - n$, even for non-diagonalizable matrices. Some counterexamples with $\dim C(A) = n \neq n^2 - n$ for $n = 3$: $$ A = \pmatrix{0&1&0\0&0&1\0&0&0}, A = \pmatrix{1&0&0\0&1&1\0&1&1}. $$ – Ben Grossmann Nov 07 '23 at 16:02
  • @BenGrossmann I got a hint! Although they didn't teach diagonalization, we can use it. So I can say if $A$ is diagonalizable, $A=SDS^{-1}$ then continue like you said. But I didn't understand how to prove also for not diagonalizables. The teaching assistant of the course told me "you can show all the elements of $C(A)$ can be diagonalized using an invertible matrix". But I can't find out why all of them are diagonalizable. – Mason Rashford Nov 08 '23 at 21:08
  • The teaching assistant's hint doesn't help with the case that $A$ is not diagonalizable. If $A$ is not diagonalizable, then $C(A)$ will necessarily contain elements that are not diagonalizable. The typical approaches to extending this argument to that case are using something more general than diagonalization (e.g. JCF or cyclic decomposition) or making an argument via limits or continuity, both of which were suggested on your original post. – Ben Grossmann Nov 09 '23 at 17:45
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    By the way, even if $A$ is diagonalizable, then $C(A)$ might contain non-diagonalizable elements. For example, denote $$ A = \pmatrix{1&0&0\0&2&0\0&0&2}, \quad B = \pmatrix{1&0&0\0&2&1\0&0&2}. $$ $A$ is diagonalizable, $B \in C(A)$, but $B$ is not diagonalizable. – Ben Grossmann Nov 09 '23 at 17:46

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