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Let $A$ be a matrix that satisfies $A^3 = 25 A$. Find the form of its minimal polynomial. Is the matrix diagonalizable?

I said that its characteristic polynomial is $$P_A(x) = x^3 - 25x = x(x-5)(x+5)$$ so its minimal polynomial polynomial is $m_A(x) = x(x-5)(x+5)$, but i'm not sure if that's correct.

Rodrigo de Azevedo
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Alex Matt
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  • If the characteristic polynomial has a factor $x-r$, then the equation $(A-rI)x=0$ has non-trivial solutions. Therefore, $x-r$ has to be a factor of the minimal polynomial because $0=m(A)x=m(r)x$. – orole Jan 12 '18 at 16:24
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    How about the $1\times1$ matrix $(5)$? And $(0)$? Do you want examples? Or do you want a characterization of the set of square matrices that hold the equation? – ajotatxe Jan 12 '18 at 16:26
  • I guess that the matrix will have some sort of formula for its minimal polynomial, based on its dimension(?). – Alex Matt Jan 12 '18 at 16:43
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    How do you know that $x^3-5x$ is the characteristic polynomial? – David C. Ullrich Jan 12 '18 at 16:56
  • I don't, that why i said i wasn't sure if it's correct. – Alex Matt Jan 12 '18 at 17:09

2 Answers2

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You cannot conclude that $t^3-25t$ is the characteristic polynomial; in fact that's impossible unless $A$ is $3\times 3$, which is not given.

It's easy to give examples (diagonal matrices) showing that the minimal polynomial can be any non-trivial polynomial that divides $t^3-25t$; since you have a factorization of $t^3-25t$ you can enumerate the seven possibilities if you want.

Something I figured out a few days ago:

(Over any field) Suppose $p$ is a polynomial that factors $p(t)=\prod_{j=1}^n(t-\lambda_j)$, where $\lambda_j\ne\lambda_k$ for $j\ne k$. If $A$ is a square matrix and $p(A)=0$ then $A$ is diagonalizable.

(Of course this is very familiar if $p$ is the characteristic polynomial.)

Proof: The familiar partial-fractions expansion shows that there exist scalars $\alpha_j$ such that$$\frac1{p(t)}=\sum_{j=1}^n\frac{\alpha_j}{t-\lambda_j}.$$(At least that looks familiar from calculus. It was just a few days ago I finally saw a purely algebraic proof; see below)

That says $$\sum\alpha_jp_j(t)=1,$$where $p_j$ is the polynomial such that $$p(t)=(t-\lambda_j)p_j(t).$$So for any vector $x$ we have $$x=\sum x_j,$$where $x_j=\alpha_jp_j(x)$.

It follows that $$(A-\lambda_jI)x=0,$$or $Ax_j=\lambda_jx_j$. So each $x_j$ is an eigenvector of $A$, unless of course $x_j=0$. In any case, each $x_j$ lies in an eigenspace. So the eigenspaces span, hence $A$ is diagonalizable.

Proof of that partial-fractions thing: Since $p_1,\dots,p_n$ have no common factor and the polynomials form a PID there exist polynomials $q_j$ with $$1=\sum q_jp_j.$$There exist polynomials $r_j$ and scalars $\alpha_j$ such that $$q_j(t)=(t-\lambda_j)r_j(t)+\alpha_j.$$ Since $(t-\lambda_j)p_j(t)=p(t)$ this shows that $$1=p(t)\sum r_j(t)+\sum\alpha_jp_j(t),$$or$$1-p(t)\sum r_j(t)=\sum\alpha_jp_j(t).$$

If $\sum r_j\ne0$ then the left side above has degree strictly larger than the right side. So $\sum r_j=0$, hence $\sum\alpha_jp_j=1$.

Exercise Suppose just that $p(t)=\prod(t-\lambda_j)^{n_j}$ and $\deg(d)<\deg(p)$, and show that $$\frac{d(t)}{p(t)}=\sum\frac{d_j(t)}{(t-\lambda_j)^{n_j}}$$with $\deg(d_j)<n_j$.

David C. Ullrich
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  • In the first sentence, do you mean characteristic polynomial? Or do you mean "at least $3\times3$" in the next sentence. – Erick Wong Jan 12 '18 at 17:24
  • @ErickWong Yes, I meant characteristic polynomial, thanks. – David C. Ullrich Jan 12 '18 at 17:48
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    More generally, it's a well-known fact that if $p_i$ are pairwise relatively prime, and $p = p_1 \cdots p_n$, then $\ker p(T) = \ker p_1(T) \oplus \cdots \oplus \ker p_n(T)$. And also, if the minimal polynomial $p$ of $T$ is square-free (i.e. $p$ and the formal derivative $p'$ are relatively prime, which is true for a product of distinct linear factors) then $T$ is diagonalizable. – Daniel Schepler Jan 12 '18 at 18:25
  • @DanielSchepler Right. Of course I assumed all this was welll known - been using MSE questions to teach myself a little algebra. Seems to me what you say all follows by arguing more or less as above, yes? (Of course when you say diagonalizable you mean diagonalizable over the algebraic completion; I explicitly assumed that factorization so I wouldn't have to get into that.) – David C. Ullrich Jan 12 '18 at 18:55
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Because $p(A)=0$ where $p(x)=x(x-5)(x+5)$, it follows that the minimal polynomial $m$ divides $p$. So $m$ could be $x$, $x-5$, $x+5$, $x(x-5)$, $x(x+5)$, $(x-5)(x+5)$, or $p$. Without further information, that's about all you can say. The matrix is diagonalizable because a matrix is diagonalizable iff the minimal polynomial contains no repeated factors, which must be the case here.

Disintegrating By Parts
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