Determinants and Eigenvalues

Theorem

If $A$ has Eigenvalues, ${\lambda }_1,\dots ,{\lambda }_n$, then $\det (A)={\Pi }_{i=1}^n{\lambda }_i$ Refer to Pi Notation

Proof

Consider characteristic polynomial $X(\lambda )=\det (A-\lambda I)$ If we input $\lambda =0$, we get $X(0)=\det (A-0I)=\det (A)$ Given polynomail is also factor $X(\lambda )=({\lambda }_1-\lambda )({\lambda }_2-\lambda )\dots ({\lambda }_n-\lambda )$ These are eigenvalues of $X(\lambda )$ $X(0)=({\lambda }_1-0)({\lambda }_2-0)\dots ({\lambda }_n-0)$