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

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.

• What is the difference between polynomial function and polynomial? Apr 19 '20 at 13:14
• 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. Apr 19 '20 at 13:16
• 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. Apr 19 '20 at 13:19
• It's a polynomial. Apr 19 '20 at 13:49
• Why is this question closed? It is perfectly clear and natural to ask. 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 $$n$$.

• @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. Apr 20 '20 at 10:05
• @JoséCarlosSantos Thank you very much again. 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.

• Thank you very much for your anser, Ittay Weiss. 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 $$M[X] = M\otimes_RR[X]$$.

• 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! Apr 20 '20 at 11:12
• @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 $R[X]$. Apr 20 '20 at 14:25