The problem discusses a diagonal matrix $M$ and its commutative properties with other matrices in $M_n(\mathbb{R})$, focusing on the subspace $C(M)$. Let's analyze the tasks step by step:

1. Proof: $A \in C(M) \iff A$ is diagonal

Given the diagonal matrix $M = \text{diag}(1, 2, \dots, n)$, the objective is to show that the matrix $A \in M_n(\mathbb{R})$ commutes with $M$ if and only if $A$ is diagonal.

Proof Sketch:

• Forward Direction ($A \in C(M)$ implies $A$ is diagonal):

  • $A \in C(M)$ means $MA = AM$, or equivalently, for the entries of $M$, $m_{ik}a_{kj} = a_{ik}m_{kj}$.
  • Since $M$ is diagonal, $m_{ik}$ is zero unless $i = k$, and similarly $m_{kj}$ is zero unless $k = j$.
  • Analyzing this equation, we see that $a_{ij}$ must be zero for $i \neq j$. Thus, $A$ is diagonal.

• Reverse Direction ($A$ is diagonal implies $A \in C(M)$):

  • If $A$ is diagonal, let $A = \text{diag}(a_1, a_2, \dots, a_n)$.
  • $MA = AM$ is satisfied directly because diagonal matrices commute.

Thus, $A \in C(M)$ if and only if $A$ is diagonal.

2. Dimension of $C(M)$

Since $C(M)$ consists of all diagonal matrices, and a diagonal matrix in $M_n(\mathbb{R})$ is uniquely determined by its $n$ diagonal entries, the dimension of $C(M)$ is $n$.

Let me know if you'd like detailed step-by-step elaboration of the proofs or related examples! Here are some follow-up questions to deepen your understanding:

1. Can you construct an example of a non-diagonal matrix that does not commute with $M$?
2. How does the result change if $M$ is not diagonal?
3. Why is the set of all diagonal matrices a vector space?
4. Can you generalize the result to block-diagonal matrices?
5. How would this proof change if $M$ had repeated eigenvalues?

Tip: Diagonal matrices commute with any other diagonal matrix because their entries directly align for the commutative multiplication property. Always verify this for complex systems.