06-240/Classnotes For Tuesday December 5

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In Preparation

The information below is preliminary and cannot be trusted! (v)

Our remaining goal for this semester is to study the following theorem:

Theorem. Let $A$ be an $n\times n$ matrix (with entries in some field $F$) and let $\chi (\lambda ):=\det(A-\lambda I)$ be the characteristic polynomial of $A$. Assume $\chi$ has $n$ distinct roots $\lambda _{1}\ldots \lambda _{n}$, that is, $A$ has $n$ distinct eigenvalues $\lambda _{1}\ldots \lambda _{n}$, and let $v_{1},\ldots ,v_{n}$ be corresponding eigenvectors, so that $Av_{i}=\lambda _{i}v_{i}$ for all $1\leq i\leq n$. Let $D$ be the diagonal matrix that has $\lambda _{1}$ through $\lambda _{n}$ on its main diagonal (in order) and let $P$ be the matrix whose columns are these eigenvectors: $P:=(v_{1}|v_{2}|\cdots |v_{n})$. Then $P$ is invertible and the following equalities hold:

• $D=P^{-1}AP$ and $A=PDP^{-1}$.
• For any positive integer $k$ we have $A^{k}=PD^{k}P^{-1}$ and $D^{k}={\begin{pmatrix}\lambda _{1}^{k}&&0\\&\ddots &\\0&&\lambda _{n}^{k}\end{pmatrix}}$.
• Likewise if $F={\mathbb {R} }$ and $\exp(B):=\sum _{k=0}^{\infty }{\frac {B^{k}}{k!}}$ then $\exp(A)=P\exp(D)P^{-1}$ and $\exp(D)={\begin{pmatrix}e^{\lambda _{1}}&&0\\&\ddots &\\0&&e^{\lambda _{n}}\end{pmatrix}}$.