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In Show that the determinant of $A$ is equal to the product of its eigenvalues, the accepted answer states that

$$ \det(A_{n \times n} - \lambda I_{n \times n}) = (-1)^n \prod_{i=1}^n (\lambda - \lambda_i) $$

and states that "$(-1)^n$ can be found by expanding the determinant along the diagonal."

What does "expanding the determinant along the diagonal" mean? Is there another way to see where the $(-1)^n $ came from?

David
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    In other words: the $(-1)^n$ is the coefficient of $\lambda^n$ in the expanded polynomial. Among the products in the Leibniz expansion of the determinant, the product along the main diagonal is the only one that results in a $\lambda^n$ when expanded. – Ben Grossmann Sep 10 '20 at 05:36
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    The $(-1)^n$ term, in my opinion, shouldn't be there; the characteristic polynomial should be defined as $\det(\lambda I - A)$ to fix this, so that it's always monic. – Qiaochu Yuan Sep 10 '20 at 05:37
  • @QiaochuYuan Isn't it more traditional to use $\det(A - \lambda I)$ when finding the characteristic polynomial and eigenvalues? I've never seen $\det(\lambda I - A)$ in my studies, although it seems you get the same eigenvalues, but just $\det(-A) = -\det(A)$ when you set $\lambda = 0$. – David Sep 10 '20 at 14:31
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    @David: yes, but it’s a bad tradition. The correct statement is that $\det(-A) = (-1)^n \det(A)$ which incidentally is one way to answer your question. – Qiaochu Yuan Sep 10 '20 at 15:57

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"What does 'expanding the determinant along the diagonal' mean?"

$$A_{n \times n} - \lambda I_{n \times n} = \begin{pmatrix} a_{11} - \lambda & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} - \lambda & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} - \lambda \end{pmatrix}$$

The "leading term" from Laplace expansion of $\det(A_{n \times n} - \lambda I_{n \times n})$ is the product of the main diagonal entries, $$\det(A_{n \times n} - \lambda I_{n \times n}) = (a_{11} - \lambda)\cdots (a_{nn} - \lambda) + \cdots.$$ For instance, if the matrix were diagonal, then $\cdots$ would be $0$ and this would be the entirety of the determinant. In general, this is the only term that will contribute $\lambda^n$; all the rest will have at most $n-1$ copies of $\lambda$. Hence $$\det(A_{n \times n} - \lambda I_{n \times n}) = (-\lambda)^n + \cdots = (-1)^n \lambda^n + \cdots.$$

You can also just do the $n=1$ case, which makes it very clear.

Joshua P. Swanson
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