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We know that given a square matrix $A$, we can find its eigenvalues through the characteristic polynomial:

$$\text{det}(A - \lambda I) = \lambda^n + c_{n-1} \lambda^{n-1} + \cdots + c_1 \lambda + c_0$$

Suppose I wish to factorize this polynomial in term of the $n$ roots of this polynomial, which I denote as $\lambda_1, \ldots, \lambda_n$, which of the following answer is correct?

(a) $\lambda^n + c_{n-1} \lambda^{n-1} + \cdots + c_1 \lambda + c_0 = \prod_{i = 1}^n (\lambda - \lambda_i)$

(b) $\lambda^n + c_{n-1} \lambda^{n-1} + \cdots + c_1 \lambda + c_0 = \prod_{i = 1}^n (-1)^n(\lambda - \lambda_i)$

(c) $\lambda^n + c_{n-1} \lambda^{n-1} + \cdots + c_1 \lambda + c_0 = \prod_{i = 1}^n (\lambda_i - \lambda)$

(d) $\lambda^n + c_{n-1} \lambda^{n-1} + \cdots + c_1 \lambda + c_0 = \prod_{i = 1}^n (-1)^n (\lambda_i - \lambda)$

I am a bit confused because $(-1)^n$ may produce a leading coefficient of $-1$ whenever $n$ is odd.

Shamisen Expert
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  • If all you want to do is find the eigenvalues, then it doesn't matter which of those four formulas you use. If you really need the two sides of the equation to be exactly equal, regardless of value of $n$, then go with (a). – Gerry Myerson Jan 16 '24 at 08:36
  • @GerryMyerson (a) seems to be not correct. See most accepted answer on: https://math.stackexchange.com/questions/507641/show-that-the-determinant-of-a-is-equal-to-the-product-of-its-eigenvalues and the very first equation. They added a $(-1)^n$ factor – Shamisen Expert Jan 16 '24 at 08:39
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    The leading term of $\det (A-\lambda I)$ is not $\lambda^n$ but $(-1)^n\lambda^n$. Your confusion likely comes from this. – Jean-Claude Arbaut Jan 16 '24 at 08:45
  • @Jean-ClaudeArbaut Yes! Do you have a reference? – Shamisen Expert Jan 16 '24 at 08:53
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    Why would you need a reference? Expand the determinant with Leibniz formula, you have $n$ factors $-\lambda$ coming from the diagonal - or rather $\prod_{i=1}^n (a_{ii}-\lambda)$, but then expand this. – Jean-Claude Arbaut Jan 16 '24 at 08:55
  • Of course (a) is correct. What's incorrect is, as @Jean points out, your formula for the determinant. – Gerry Myerson Jan 16 '24 at 09:55

1 Answers1

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To expand a little bit on my comment:

The correct expansion of $\det(A-\lambda I)$ is $(-1)\lambda^n+c_{n-1}\lambda^{n-1}+\dots$. However, in the subsequent question, you are trying to factor the monic polynomial $\lambda^n+c_{n-1}\lambda^{n-1}+\dots$.

In general, a polynomial $P=c_n\lambda^n+c_{n-1}\lambda^{n-1}+\cdots+c_0$ has factorizarion $c_{n}\prod_{i=1}^n (\lambda-\lambda_i)$, where $\lambda_i$ are the roots of $P$.

The answer a) is thus correct, since here $c_n=1$.

But there is a trap.

b) is wrong in general because, if you get the $(-1)^n$ factor out of the product, you have then $(-1)^{n^2}\prod_{i=1}^n (\lambda-\lambda_i)$, and $(-1)^{n^2}\ne1$ for odd $n$.

The answer c) is also wrong, because by changing the sign of each factor and getting a factor $(-1)$ out, you have really $(-1)^{n}\prod_{i=1}^n (\lambda-\lambda_i)$, and again $(-1)^n\ne1$ if $n$ is odd.

And here comes the trap. For answer d), you can rewrite the product as $(-1)^{n^2+n}\prod_{i=1}^n (\lambda-\lambda_i)$, and $n^2+n=n(n+1)$ is always even, therefore it's really $\prod_{i=1}^n (\lambda-\lambda_i)$, and answer d) is correct.

However, if your goal is to factor $\det(A-\lambda I)$, and the eigenvalues are $\lambda_i$, i.e. these are the roots of the polynomial $\det(A-\lambda I)$, then you can write:

$$\det(A-\lambda I)=(-1)^n \prod_{i=1}^n (\lambda-\lambda_i)$$

Here the polynomial $\det(A-\lambda I)$ is not monic, it has leading term $(-1)^n\lambda^n$, as can be seen by expanding the determinant by Leibniz formula: the only term in the expansion with $n$ factors $\lambda$ comes from the identity permutation (and its signature is $+1$), i.e. it's $\prod_{i=1}^n (a_{ii}-\lambda)$, and when you expand this product the only term of degree $n$ is $(-1)^n\lambda^n$.

Jean-Claude Arbaut
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