Relationship between Trace, Determinant and Eigenvaues

THeorem

• With $A$ as a $n\times n$ matrix in $M_{n,n}(\mathbb{C})$
• With ${\lambda }_1,\dots ,{\lambda }_n$ as Eigenvalues for $A$ repeating the terms if there is algebraic multiplicity Then,

1. $trace(A)={\sum }_{i=1}^n{\lambda }_i$
2. $\det (A)={\Pi }_{i=1}^n{\lambda }_i$

Proof

• By Schur’s Decomposition, $A$ is similar to a diagonal matrix $B$.
• Then, $\det (B-\lambda I)$ is the product of the diagonal entries.
• Thus, $\forall j,\exists i$ s.t $\lambda i=b_{ij}$ and $\forall i,\exists j$ s.t $b_{ii}={\lambda }_j$
  • Reordering the labeling allows us to say that ${\lambda }_i=b_{ii}$
• Now, we can say $trace(A)=trace(B)={\sum }_{i=1}^n{b}_{ii}={\sum }_{i=1}^n{\lambda }_i$
• Now we can say $\det (A)=\det (B)={\Pi }_{i=1}^n{b}_{ii}={\Pi }_{i=1}^n{\lambda }_i$