From the National Board for Higher Mathematics (NBHM) 2022 exam:
Suppose $A \in \Bbb C^{n \times n}$ has the property that every matrix that commutes with $A$ is a polynomial in $A$. Then which of the following is/are true?
(a) Such a matrix $A$ is necessarily diagonalizable.
(b) Such a matrix $A$ cannot be nilpotent.
(c) The characteristic and minimal polynomials coincide for every such $A$.
For option a.
If a matrix A has n distinct eigenvalues, then A is diagonalizable.
If two matrices commute, then they are simultaneously diagonalizable.
First, let's prove that every matrix that commutes with A is diagonalizable. Suppose B is a matrix that commutes with A, that is, AB = BA. Then, by the second fact above, we know that A and B are simultaneously diagonalizable, meaning that there exists an invertible matrix P such that both A and B are diagonalized by P. That is, $P^{-1}AP$ and $P^{-1}BP$ are both diagonal matrices. But since AB = BA, we have
$P^{-1}ABP$ = $P^{-1}BAP$
which implies that both $P^{-1}AP$ and $P^{-1}BP$ commute with each other. Since diagonal matrices are uniquely determined by their diagonal entries, we conclude that $P^{-1}AP$ and $P^{-1}BP$ must have the same diagonal entries, which implies that A and B have the same eigenvectors. Therefore, B is also diagonalizable.
Now, let's prove that A itself is diagonalizable. Suppose A has a repeated eigenvalue λ, and let E be the eigenspace corresponding to λ. Then, we can choose a basis of E consisting of linearly independent eigenvectors, say $v_1$, $v_2$, …, $v_k$. Since every matrix that commutes with A is diagonalizable, we know that the restriction of A to E, denoted by $A|_E$, is diagonalizable. Therefore, there exists an invertible matrix Q such that both $A|_E$ and $Q^{-1}A|_EQ$ are diagonal matrices. But since A commutes with $A|_E$, we have
AQ = $QA|_E$ = $Q(Q^{-1}A|_EQ)$ = $(QQ^{-1})A|_EQ$ = $A|_EQ$
which implies that the columns of Q are eigenvectors of A that belong to E. Thus, we can extend the basis ${v_1, v_2, …, v_k}$ of E to a basis ${v_1, v_2, …, v_k, w_1, w_2, …, w_l}$ of $C^n$, where ${w_1, w_2, …, w_l}$ is a basis of the eigenspace corresponding to some other eigenvalue of A (which may or may not be distinct from λ). Let P be the matrix whose columns are the basis vectors ${v_1, v_2, …, v_k, w_1, w_2, …, w_l}$. Then, we have $P^{-1}AP = D$,
where D is a block diagonal matrix with the diagonal blocks being the diagonal matrices corresponding to the restriction of A to E and the eigenspaces corresponding to the other eigenvalues of A, respectively. But since A is assumed to commute with every matrix that commutes with it, we know that every matrix that commutes with A also commutes with $P^{-1}AP$ = D. Therefore, every such matrix must be a diagonal matrix with the same diagonal entries as D. But since the diagonal entries of the block diagonal matrix D are the eigenvalues of A, counted with multiplicity, we conclude that A has n distinct eigenvalues, and hence is diagonalizable.
Is it correct for option a?
And what about for option b?
Here is for option c Characteristic polynomial of a A agrees with its minimal polynomial if and only if all matrices that commutes with A is a polynomial of A
Answer is only option c is true and option a,b are false.
Please have a look…
Best Answer
Hint $:$ To disprove $(a)$ and $(b)$ just consider the non-diagonalizable nilpotent matrix (of index $2$) $$A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}.$$
It's easy to check that all the matrices that commute with $A$ are polynomials in $A$ which are of the form $$aI + b A = \begin{pmatrix} a & b \\ 0 & a \end{pmatrix}.$$
For $(c)$ you already have an answer.