User:IssaRice/Linear algebra/Trace of matrix equals sum of eigenvalues

This is chapter 4 exercise 1.11 in Linear Algebra Done Wrong (p. 105).

Theorem. Let $A$ be a matrix. Then $\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{e}A={\lambda }_1+\cdots +{\lambda }_n$, where ${\lambda }_1,\dots ,{\lambda }_n$ are the eigenvalues of $A$ (counting multiplicities).

Proof. We already know that $det(A-\lambda I)=({\lambda }_1-\lambda )\cdots ({\lambda }_n-\lambda )$. Multiplying out the right hand side, the ${\lambda }^{n-1}$ term is ${\lambda }_1(-\lambda {)}^{n-1}+{\lambda }_2(-\lambda {)}^{n-1}+\cdots +{\lambda }_n(-\lambda {)}^{n-1}=({\lambda }_1+\cdots +{\lambda }_n)(-1{)}^{n-1}{\lambda }^{n-1}$.

Now consider equation 4.2 from the book (p. 89):

$detA={\sum }_{\sigma \in \mathrm{P}\mathrm{e}\mathrm{r}\mathrm{m}(n)}{a}_{\sigma (1),1}{a}_{\sigma (2),2}\cdots a_{\sigma (n),n}\mathrm{s}\mathrm{i}\mathrm{g}\mathrm{n}(\sigma )$

See also

• https://machinelearning.subwiki.org/wiki/User:IssaRice/Linear_algebra/Determinant_of_matrix_equals_product_of_eigenvalues