existence proof of minimal polynomial for an invertible linear transformation

Question

I'm having problems showing the existence of the polynomial $q$ in exercise $36.9$ from Halmos' Finite-dimensional vector spaces. I note that matrices, determinants, and characteristic polynomials have not (yet) been introduced.

If $A$ is an invertible linear transformation on a finite-dimensional vector space $\mathcal{V}$, then there exists a polynomial $p$ such that $A^{-1}=p(A)$. (Hint: find a non-zero polynomial $q$ of least degree such that $q(A)=0$ and prove that its constant term cannot be 0.)

If there exists a polynomial $q'$ such that $q'(A)=0$, I know that we can find a polynomial $q$ with non-zero constant term simply by multiplying $q'(A)=0$ by $A$'s inverse as many times as is necessary to obtain a non-zero leading term. If the leading coefficient is not 1, then we can divide by the leading coefficient to make it so. Multiplying again by $A$'s inverse, then rearranging, we have an expression for $A$'s inverse. So far so good, we didn't even need $q$ to be of least degree.

But the exercise requires we show the existence of the polynomial $p$. We just moved the goalpost, and now need to show the existence of a $q$ or $q'$. This is where I'm stuck. Is there a simple way to show this without invoking determinants or characteristic polynomials?

Answer 1

Hint: $A \in L(\mathcal{V})$, which has finite dimension, because so has $V$. Therefore, the powers of $A$ cannot be linearly independent. This gives you a polynomial $f$ such that $f(A)=0$. Thus, the set of polynomials killing $A$ is non-empty and so has an element of minimum degree.

Answer 2

If

$\dim \mathcal V = n, \tag 1$

then $\mathcal L(\mathcal V)$, the space of linear maps $L: V \to V$ is of dimension $n^2$; this may be seen using the matrix representation for $\mathcal L( \mathcal V)$; thus the $n^2 + 1$ transformations

$I, A, A^2, A^3, \ldots, A^{n^2} \in \mathcal L(\mathcal V) \tag 2$

must be linearly dependent; hence, if $\Bbb F$ is the field of scalars, we have

$a_i \in \Bbb F, \; 0 \le i \le n^2, \tag 3$

with not all the $a_i = 0$, and

$\displaystyle \sum_0^{n^2} a_i A^i = 0; \tag 4$

now take $p(x) \in \Bbb F[x]$ to be

$p(x) = \displaystyle \sum_0^{n^2} a_i x^i; \tag 5$

then clearly (4) asserts that

$p(A) = 0. \tag 6$

Having shown the existence of some $p(x)$ satisfying (6), we can then affrim the existence of $q(x) \in \Bbb F[x]$ of minimal degree $m$ such that

$q(A) = 0, \tag 7$

and if

$q(x) = \displaystyle \sum_0^m q_i x^i, \tag 8$

then

$q_0 \ne 0; \tag 9$

otherwise, if $d > 1$ were the least degree in $x$ of the non-zero terms occurring in (8), then

$q(x) = x^d \displaystyle \sum_d^m q_i x^{i - d}, \tag{10}$

so

$A^d \displaystyle \sum_d^m q_i A^{i - d} = q(A) = 0, \tag{11}$

and since $A$ is invertible we may write

$\displaystyle \sum_d^m q_i A^{i - d} = 0; \tag{12}$

the degree of

$\displaystyle \sum_d^m q_i x^{i - d} = 0 \tag{13}$

is $m - d < m$, contradicting the minimaliy of $\deg q(x)$; thus $d = 0$ and $q_d = q_0 \ne 0$; thus from (7) - (8) we have

$\displaystyle \sum_1^m q_i A^i = - q_0 I, \tag{14}$

or

$\displaystyle \sum_1^m -\dfrac{q_i}{q_0} A^i = I, \tag{15}$

or

$\displaystyle A \sum_1^m -\dfrac{q_i}{q_0} A^{i -1} = I, \tag{16}$

which shows that

$A^{-1} = \displaystyle \sum_1^m -\dfrac{q_i}{q_0} A^{i -1}. \tag{17}$