In Linear Algebra, we consider characteristic polynomials.

Is a characteristic polynomial we consider in Linear Algebra a polynomial or a polynomial function?

I think it is a polynomial function.

I am reading "Introduction to Linear Algebra" (in Japanese) by Kazuo Matsuzaka.

In this book, the characteristic polynomial of a linear map $F$ is defined by $\det(A - \lambda I)$, where $A$ is a matrix which represents $F$.

And in this book, the author defines a determinant only for a matrix whose elements belong to a some field $K$.

If $\det(A - \lambda I)$ is a polynomial, then the elements of $A - \lambda I$ are polynomials too. But the author didn't define a determinant for a matrix whose elements are polynomials.

  • 2
    $\begingroup$ What is the difference between polynomial function and polynomial? $\endgroup$
    – user614287
    Apr 19 '20 at 13:14
  • 10
    $\begingroup$ For instance, over $\mathbb F_2$ the polynomial function $x^2-x$ is the null function, but the polynomial $x^2-x$ is not the null polynomial. $\endgroup$ Apr 19 '20 at 13:16
  • $\begingroup$ As long as the underlying field $K$ is infinite, it doesn't matter. In that case, the polynomial ring and corresponding ring of polynomial functions is isomorphic. $\endgroup$ Apr 19 '20 at 13:19
  • $\begingroup$ It's a polynomial. $\endgroup$
    – Alexey
    Apr 19 '20 at 13:49
  • 4
    $\begingroup$ Why is this question closed? It is perfectly clear and natural to ask. $\endgroup$
    – Alexey
    Apr 20 '20 at 10:11

Nice question! In many cases, that distinction is irrelevant, but in some cases it matters. And, when it matters, you are not right: it is a polynomial, not a polynomial function. For instance, polynomials have degrees, whereas polynomial functions don't (for instance, over $\mathbb F_2$ the polynomial function $x\mapsto x^2+x$ is the null function, but the polynomial $x^2+x$ still has degree $2$, whereas the null polynomial still has degree $0$). And the degree of the characteristic polynomial of a $n\times n$ matrix is $n$.

  • $\begingroup$ @JoséCarlosSantos Thank you very much for your answer again. I edited my question. From your anser, I know if $\det(A - \lambda I)$ is a polynomial function, then we cannot define the degree of $\det(A - \lambda I)$ in some case when the field we consider is finite. $\endgroup$
    – tchappy ha
    Apr 20 '20 at 10:05
  • $\begingroup$ @JoséCarlosSantos Thank you very much again. $\endgroup$
    – tchappy ha
    Apr 20 '20 at 10:09

The characteristic polynomial of $T$ (either a matrix or a linear transformation, depending on your preference) is a polynomial, not a function. What we really care about are its coefficients. For instance, the leading coefficient is always $1$ (so that is boring) but the degree of the polynomial is the dimension of the ambient vector space. The next coefficient is (up to a sign) the trace of $T$. The free coefficient is the determinant. The other coefficients also have meaning directly expressed in $T$. All of this will be lost if you considered the polynomial merely as a function since over certain fields this process destroys the coefficients.

  • 1
    $\begingroup$ Thank you very much for your anser, Ittay Weiss. $\endgroup$
    – tchappy ha
    Apr 20 '20 at 11:10

The characteristic polynomial is actually ... a polynomial!

Here are more details about the definition of the determinant and of the characteristic polynomial in general case. In the case of Linear Algebra, $M$ would be an $n$-dimensional vector space over $R$ (a field).

For every free unital module $M$ of finite rank $n$ over a commutative unital ring $R$ and for every endomorphism $a$ of $M$, the determinant of $a$ is defined by the identity $$ ax_1\wedge\dotsb\wedge ax_n = (\det a)(x_1\wedge\dotsb\wedge x_n)\qquad (x_1,\dotsc,x_n\in M). $$

If $S$ is a unital $R$-algebra, then there is a natural homomorphism $$ \operatorname{End}_{R}(M)\otimes_RS\to\operatorname{End}_{S}(M\otimes_RS). $$ Since $M$ is assumed to be free of finite rank, it can be shown that this homomorphism is an isomorphism: $$ \operatorname{End}_{R}(M)\otimes_RS\cong\operatorname{End}_{S}(M\otimes_RS). $$

The characteristic polynomial of $a\in\operatorname{End}_R(M)$ is $\chi_a\in R[X]$ defined by $$ \chi_a =\det(a - X), $$ where $a - X = (a\operatorname{id}_M)\otimes 1 -\operatorname{id}_M\otimes X\in\operatorname{End}_{R}(M)[X] =\operatorname{End}_{R}(M)\otimes_RR[X]$ is viewed as an element of $\operatorname{End}_{R[X]}(M[X])$, where $M[X] = M\otimes_RR[X]$.

  • $\begingroup$ Thank you very much for your answer, Alexey. I guess your answer is very useful for many people, but unfortunately it is too difficult for me. Thank you! $\endgroup$
    – tchappy ha
    Apr 20 '20 at 11:12
  • $\begingroup$ @tchappyha, the bottom line is that the characteristic polynomial of a matrix with coefficients in $R$ is defined as the determinant of a certain other matrix with coefficients in the commutative ring $R[X]$. The determinant of a matrix with coefficient in $R[X]$ is itself an element of $R[X]$. $\endgroup$
    – Alexey
    Apr 20 '20 at 14:25

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