Degree of the determinant of a matrix

Question

Given a matrix $A\in C^{n\times n}$, the characteristic polynomial of $A$, $P_A$, is a polynomial of degree $n$.

By the definition of determinants, $P_A$ must be a polynomial. However, I don't completely understand how its degree cannot be larger than $n$.

It is not only no larger than $n$, it is always equal to $n$. Now have you seen a proof? What part exactly do you not understand? — Arnaud Mortier, Mar 22 '18 at 23:13
$r^n\det(A-\frac{1}{r}I)|{r=0}=\det(rA-I)|{r=0}=\det(I)=1$. That is the coefficient of degree $n$. And if you do the same but multiplying by a larger power of $r$ you would get $0$, which means there are not terms of larger degree. — blueInk, Mar 22 '18 at 23:16
@blueInk The second last equality is not correct. It should be $(-1)^n \det(I)$. — , Mar 22 '18 at 23:25
@blueInk, unless you know that the $r^n\det(A-1/rI)$ is a polynomial you cannot evaluate it at $0$. Some argument is necessary for that. — Mariano Suárez-Álvarez, Mar 23 '18 at 00:02
@MarianoSuárez-Álvarez Please, spare me that nonsense. Just look at the first equality for your "argument". — blueInk, Mar 23 '18 at 00:03
Your first equality sign has on the left hand side something which does not mean anything unless you prove it means something. It should be obvious that to write an equality of two things, one first has to know that the two things one is trying to equate make sense at all... — Mariano Suárez-Álvarez, Mar 23 '18 at 00:04
@MarianoSuárez-Álvarez HA! Really? Please tell me more about it. I am learning so much! Should I start from a system of axioms? Ridiculous! — blueInk, Mar 23 '18 at 00:07
@MarianoSuárez-Álvarez Yes, wow. You get what you deserve. — blueInk, Mar 23 '18 at 00:10
I doubt there is a way to show that $r^n\det(A-\frac1rI)$ is a polynomial which does not go through the motions needed to show that the degree of $\det(r A-I)$ is a polynomial of degree at most $n$, which is precisely, let us recall, what the question is asking. — Mariano Suárez-Álvarez, Mar 23 '18 at 00:16
@MarianoSuárez-Álvarez Open the most mediocre book in complex analysis and review pretty much the beginning. Heck, what complex analysis, just a Calculus as bad a Stewart's probably has it in the section of power series. Then use the first equality. And you are very wrong. One can know that something is a polynomial and the degree be completely unknown, even unprovable of what it is. — blueInk, Mar 23 '18 at 00:49
I know that one can prove that it is a polynomial. What I am saying —I really thought this was obvious— is that you didn't, and in the context of this particular question (as opposed to a book on complex analysis or the last of Voevodky's papers) that is a gap in the reasoning. That there exist contexts in which something is obvious is irrelevant: notice that the actual question asked by the OP is essentially the same thing you are dismissing as self evident, and you'll agree that doing that is of no use to anyone. — Mariano Suárez-Álvarez, Mar 23 '18 at 01:17
In any case, after this cozy interchange with you I have to say that I don't care if you see my point or not. Indeed, my original comment was more for the benefit of others than you. — Mariano Suárez-Álvarez, Mar 23 '18 at 01:22
Yes, that is precisely why I think there is no point in talking to you. Attitude does not fill gaps in arguments, though. — Mariano Suárez-Álvarez, Mar 23 '18 at 16:25

score 3 · Accepted Answer · answered Mar 22 '18 at 23:22

3

To compute the determinant of a matrix, one formula consists of considering an alternate sum of products of entries, where you take the entries $n$ at a time, one in each row and one in each column.

See this question.

Because of the condition "no more than $n$ entries at a time", you can already see that the degree is no more than $n$, since each entry is a polynomial of degree at most $1$.

Moreover, there is exactly one choice that will achieve a degree equal to $n$, that is by taking all the diagonal elements. This choice gives you $(-X)^n$, and because all other terms will be of lower degree, this $X^n$ will never be cancelled.

answered Mar 22 '18 at 23:22

Arnaud Mortier

27,276

Why can't an entry be a polynomial of degree greater than $1$? Is it because $A\in C^{n\times n}$? But this $n$ only points out the number of rows and columns of $A$. – Mar 22 '18 at 23:45
1

Because each entry is either an entry of A (hence of degree $0$) or an entry of $A$ minus $X$, hence of degree $1$. Remember that to get the characteristic polynomial you consider the determinant of $A$ minus $X$ times the identity. – Arnaud Mortier Mar 22 '18 at 23:47

Degree of the determinant of a matrix

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