Recently I was bored and thought about how to find eigenvalues.
The casual equation is: $|A-\lambda I|=0$.
But I thought to myself, what would happen if we do: $|A-\lambda I|= \lambda$ .?
I tried to do it on all sorts of matrices, and my conclusion was that (Pay attention we do not compare to $0$ but:
$$|A-\lambda I|= \lambda$$
And it gives: $$\lambda_1 , \lambda_2, ..., \lambda_n$$
And so this equation pops up to mind: $$\prod_{k=1}^n \lambda_k = \text{det}(A)$$
Why does it happen? I don't know how to even start..
Thank you! (It's something I found just by messing around with this material)
Consider that $|A - 0 I| = \det A$, of course. Recall also that the product of the roots of a polynomial (up to multiplicity) is the constant coefficient, divided by the leading coefficient. In the case of the polynomial $|A - \lambda I|$, it must be $\pm \det A$ (the $\pm$ depending on the dimension of $A$, i.e. the degree of the polynomial).
Now, provided $A$ is larger than $1 \times 1$, the polynomial $|A - \lambda I| - \lambda$ has exactly the same constant term, and the same leading term, which means the product of the roots, up to multiplicity, is still the same. It stands to reason that this product must still be $\det A$.