1
$\begingroup$

How can I show that the minimal polynomial of a diagonal matrix is the product of the distinct linear factors $(A-\lambda_{j}I)$? In particular, if we have a repeated eigenvalue, why is it that we only count the factor associated with that eigenvalue once?

I know by the Cayley Hamilton theorem that the characteristic polynomial $p(t)$, i.e. the product of all the linear factors, not necessarily distinct, yields $p(A) = 0$. But I'm uncertain how this can be simplified for diagonal matrices when there is a repeated eigenvalue.

$\endgroup$
  • $\begingroup$ Maybe the example of a scalar matrix $A = \lambda{}I$ will enlighten you : what would $A-\lambda{}I$ be ? $\endgroup$ – Traklon Jan 30 '14 at 8:50
  • $\begingroup$ Hint: when a polynomial has multiple roots, after removing the repeated factors, the roots remain the same. E.g. any root of $(x-1)^3(x+3)$ is a root of $(x-1)(x+3)$. $\endgroup$ – Yves Daoust Jun 24 '15 at 7:01
1
$\begingroup$

To start with the minimal polynomial is the "Polynomial of the smallest degree" which satisfies the relation P(M)=0, where M is your matrix. In case of the diagonal matrix the polynomial consisting of the factors (x-a) where a are diagonal entries (eigenvalues) these need to be written only once and the relation P(M)=0 is satisfied (proof is trivial) .

$\endgroup$
0
$\begingroup$

HINT:

$$\begin{pmatrix}D_m&0&0\\0&\color{red}0&0\\0&0&D_{n-m-1}\end{pmatrix}\cdot D'_n=\begin{pmatrix}D''_m&0&0\\0&\color{red}0&0\\0&0&D''_{n-m-1}\end{pmatrix}$$ Where $D_i$ are any diagonal matrices of dimension $i$

$\endgroup$
  • $\begingroup$ Since the eigenvalues of a diagonal matrix are the elements of the diagonal, you have that for any eigenvalue $\lambda$, $A-\lambda I$ will set to zero exactly all diagonal terms $a_{ii}$with are equal to $\lambda$. Now you multiply $A-\lambda I$ by any diagonal matrix $B$ and you will still have that the $c_{ii}$ term of $(A-\lambda I)B$ will be equal to zero, hence doing this once for all eigenvalues leads to the null-matrix $\endgroup$ – b00n heT Jan 30 '14 at 9:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.