Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ (over the field $\mathbb{Q}$) be the $6 \times 6$ block matrix below: $$A=\left(\begin{array}{rrrr|rr} -3 &3 &2 &2 & 0 & 0\\ -1 &0 &1 &1 & 0 & 0\\ -1&0 &0 &1 & 0 & 0\\ -4&6 &4 &3 & 0 & 0\\ \hline 0 & 0 & 0 & 0 & 0 &1 \\ 0 & 0 & 0 & 0 & -9 &6 \end{array}\right).$$

I found out that the minimal polynomial of $A$ is $(x-3)^3(x+1)^2$, and now let $f(x)=2x^9+x^8+5x^3+x+a$ a polynomial , $a\in N$. I need to find out for which $a$ the matrix $f(A)$ is invertible.

It has some similarity to to my last question, but I still can't understand and solve it..

Thanks again.

share|improve this question
    
Hi Nir! I’d say that $f(A)$ is invertible iff $f(3)$ and $f(-1)$ are nonzero. –  Pierre-Yves Gaillard Aug 13 '11 at 12:21
    
That's what I thought also.. but I'm no completely sure why.. and It drives me mad! –  user6163 Aug 13 '11 at 12:29
1  
@Nir: "it's" is a contraction of "it is", not a possessive. The possessive is "its". –  Arturo Magidin Aug 13 '11 at 18:22
    
@Arturo: Ok, thanks for correcting me. –  user6163 Aug 13 '11 at 18:33

4 Answers 4

up vote 1 down vote accepted

Hint

Let $A$ be an square matrix with coefficients in a field $K$, and let $g$ be its minimal polynomial.

Then the epimorphism $K[X]\to K[A]$, $f\mapsto f(A)$ induces an isomorphism $K[X]/(g)\to K[A]$.

Assume that $g$ splits over $K$.

Then the Chinese Remainder Theorem says that this algebra is isomorphic to the product of the $K[X]/(X-\lambda)^{m(\lambda)}$, where $\lambda$ is an root of $g$ and $m(\lambda)$ its multiplitity.

Moreover the natural morphism from $K[X]$ to $K[X]/(X-\lambda)^{m(\lambda)}$ attaches to $f\in K[X]$ its degree $ < m(\lambda)$ Taylor polynomial at $\lambda$.

EDIT 1. The interest of the above observation (which is of course entirely classical) is that it gives you a formula for $f(A)$. [If $K=\mathbb C$ the formula holds also for the functions which are holomorphic on the spectrum --- like the exponential.]

[Technical point: In positive characteristic, $$\frac{f^{(n)}}{n!}$$ is not defined as $f^{(n)}$ divided by $n!$ (Here $f$ is in $K[X]$.)]

EDIT 2. This is to explain how Andrea's very nice answer can be obtained in this setting. Once you've noticed the isomorphism $K[A]=K[X]/(g)$, it's clear that $f(A)$ is invertible iff $f$ is invertible mod $g$, iff $f$ is prime to $g$.

share|improve this answer

Expanding on the comment:

If $A$ has eigenvalue $\lambda$, then $f(A)$ has eigenvalue $f(\lambda)$. So $f(A)$ is not invertible if $f(\lambda)=0$.

share|improve this answer
    
"If A has eigenvalue $λ$, then $f(A)$ has eigenvalue $f(λ)$"- How do you know that? –  user6163 Aug 13 '11 at 13:07
    
+1. Very concise! --- What about the other direction? –  Pierre-Yves Gaillard Aug 13 '11 at 13:10
2  
@Nir: Take an eigenvector. Check that it works when $f$ is a monomial. Check that it works for $f+g$ if it works for $f$ and $g$. –  Pierre-Yves Gaillard Aug 13 '11 at 13:12
1  
@Nir, what is $f(A)$ when $A$ is diagonal? Also, let $A = S^{-1}DS$, where $D$ is diagonal, then what is $A^n$? You can use those results to show that $f(\lambda)$ is an eigenvalue of $f(A)$. –  rcollyer Aug 13 '11 at 14:13
    
@rcollyer: This technique does not apply when $A$ is not diagonal, nor of the form $A=S^{-1}DS$ with $D$ diagonal. –  Did Aug 13 '11 at 15:18

Theorem. Let $V$ be a finite $\mathbb{K}$-vector space and let $f \in \mathrm{End}(V)$ an endomorphism with minimal polynomial $m_f(t) \in \mathbb{K}[t]$. If $a(t) \in \mathbb{K}[t]$, then $a(f) \in \mathrm{GL}(V)$ if and only if $\gcd(a,m_f)=1$.

Proof. $\Leftarrow$) Since Bezout's identity, $1 = \lambda m_f + \mu a$ for some polynomials $\lambda, \mu$. So, evaluating in $f$, one has $\mathrm{id}_V = \mu(f) \circ a(f)$, that proves that $a(f)$ is invertible.

$\Rightarrow$) Let $d$ the greatest common divisor between $a$ and $m_f$. One has $a = \tilde{a} d$ for a polynomial $\tilde{a}$, then $a(f) = \tilde{a}(f) \circ d(f)$, so $\ker d(f) \subseteq \ker a(f)$. But $a(f)$ is invertible, so also $d(f)$ is invertible. One has $m_f = \tilde{m} d$, so $0 = \tilde{m}(f) \circ d(f)$; but $d(f)$ is invertible, so $\tilde{m}(f) = 0$. But $m_f$ is the minimal polynomial, so $m_f = \tilde{m}$ and then $d = 1$. $\square$

share|improve this answer

Put $A$ in Jordan form. The diagonal is made of $3$s and $-1$s. In this vector basis, $f(A)$ is also upper triangular and its diagonal is made of $f(3)$s and $f(-1)$s. Hence $f(A)$ is invertible if and only if there is no zero on the diagonal of its Jordan form if and only if $f(3)$ and $f(-1)$ are nonzero (and this condition is equivalent to the fact that the gcd of $f$ and the minimal polynomial of $A$ is $1$).

share|improve this answer
1  
"triangular superior" - "upper triangular" is the more customary English term methinks. :) –  J. M. Aug 13 '11 at 14:35
    
@J. M. Thanks! $ $ –  Did Aug 13 '11 at 15:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.