2

I've been trying to dig and search for an answer for my bugging question: Why does the (Leibniz) formula for a determinant look like this:

$$\det (T) = \sum_{\sigma\in S_n} \text{sign}\,(\sigma) \prod_{i=1}^n a_{\sigma(i), i}.$$

The definition of the determinant by using eigenvalues is easy for me to understand: "Determinant of $T$ is the product of the eigenvalues of $T$ which means the amount by which the linear transformation $T$ scales the (vector) space.

I've been reading through Down with Determinants (eigenvalue based definition for determinant), History of Determinants, cyclic groups and permutations in abstract algebra and also previous questions on the subject but still I don't get it 100%. I couldn't for example start to derive the Leibniz formula on the whiteboard if somebody asked me to prove that the Leibniz formula for determinant equals the product of eigenvalues.

So my question about the subject is this:

  • How does one prove that the Leibniz formula for determinant is equal to the product of eigenvalues of a square matrix $T$?
Community
  • 1
jjepsuomi
  • 8,979
  • 1
    if $\lambda_s$ are defined as roots of $\det(\mathbf A-\lambda\mathbf I)$ and the determinant is computed by the Leibnitz formula, then the free term of the polynomial is $\det(\mathbf A)$. On the other hand, by Vieta's theorem, the free term is the product of all $\lambda$-s. – Ignat Domanov Aug 29 '16 at 13:24
  • Thank you for your help! =) Could you perhaps consider posting your comment as an answer? Maybe some more elaboration if you want also :) – jjepsuomi Aug 30 '16 at 14:26
  • I added an extended answer. – Ignat Domanov Aug 30 '16 at 15:37

2 Answers2

1

We show that both identities $trace(T)=\lambda_1+\dots+\lambda_n$ and $\det T=\lambda_1\cdots\lambda_n$ follow from the Leibnitz formula.

The eigenvalues of $T$ are roots of $\det(T-\lambda\mathbf I)$. Aplying the Leibnitz formula to $T-\lambda\mathbf I$ we obtain that $$ \det(T-\lambda\mathbf I) = (-\lambda)^n+(-\lambda)^{n-1}trace(T)+\dots+det(T)=(\lambda-\lambda_1)\cdots(\lambda-\lambda_n). $$ On the other hand, by Vieta's theorem, $$ \det(T-\lambda\mathbf I) = (-\lambda)^n+(-\lambda)^{n-1}(\lambda_1+\dots+\lambda_n)+\dots+\lambda_1\cdots\lambda_n. $$ Thus, $trace(T)=\lambda_1+\dots+\lambda_n$ and $\det T=\lambda_1\cdots\lambda_n$.

More generally: the sum of all $k\times k$ principal minors of $T$ is equal to $\sum \lambda_{i_1}\cdots\lambda_{i_k}$, where the sum is taken over all $k$-combinations of the set $\{1,\dots,n\}$.

Ignat Domanov
  • 1,123
0

You can also show that using this thinking:

  1. For an Upper Triangular Matrix form, which every operator over a complex vector space has for some basis, the leibniz formula indeed gives the product of the eigenvalues (each eigenvalue λ repeated M times in the multiplication, where M is λ's multiplicity).
  2. Now you can show that if you change a base, the determinant stays the same under the leibniz formula (base change formula for matrices + showin algebricly that det(AB)=det(BA)), so it's the same result for every basis.

Bottom line- it gives you the product of eigenvalues because it's like getting the Upper Triangular Matrix form and multiply the entries on the main diagonal.

Harel
  • 1
  • For real vector spaces, one can use the complexification of an operator and the rest of the proof stays as above. – Harel Nov 30 '23 at 10:41