Proof of Theorem DC

(⇐) Let $S=\set{\vectorlist{x}{n}}$ be a linearly independent set of eigenvectors of $A$ for the eigenvalues $\scalarlist{\lambda}{n}$. Recall Definition SUV and define \begin{align*} R&=\matrixcolumns{x}{n}\\ D&= \begin{bmatrix} \lambda_1 & 0 & 0 &\cdots & 0\\ 0 &\lambda_2 & 0 &\cdots & 0\\ 0 & 0 &\lambda_3 &\cdots & 0\\ \vdots & \vdots & \vdots & & \vdots\\ 0 & 0 & 0 &\cdots & \lambda_n \end{bmatrix} =[\lambda_1\vect{e}_1|\lambda_2\vect{e}_2|\lambda_3\vect{e}_3|\ldots|\lambda_n\vect{e}_n] \end{align*}

The columns of $R$ are the vectors of the linearly independent set $S$ and so by Theorem NMLIC the matrix $R$ is nonsingular. By Theorem NI we know $\inverse{R}$ exists. \begin{align*} \inverse{R}AR&= \inverse{R}A\matrixcolumns{x}{n}\\ &=\inverse{R}[A\vect{x}_1|A\vect{x}_2|A\vect{x}_3|\ldots|A\vect{x}_n] &&\knowl{./knowls/definition.MM.knowl}{\text{Definition MM}}\\ &=\inverse{R}[\lambda_1\vect{x}_1|\lambda_2\vect{x}_2|\lambda_3\vect{x}_3|\ldots|\lambda_n\vect{x}_n] &&\knowl{./knowls/definition.EEM.knowl}{\text{Definition EEM}}\\ &=\inverse{R}[\lambda_1R\vect{e}_1|\lambda_2R\vect{e}_2|\lambda_3R\vect{e}_3|\ldots|\lambda_nR\vect{e}_n] &&\knowl{./knowls/definition.MVP.knowl}{\text{Definition MVP}}\\ &=\inverse{R}[R(\lambda_1\vect{e}_1)|R(\lambda_2\vect{e}_2)|R(\lambda_3\vect{e}_3)|\ldots|R(\lambda_n\vect{e}_n)] &&\knowl{./knowls/theorem.MMSMM.knowl}{\text{Theorem MMSMM}}\\ &=\inverse{R}R[\lambda_1\vect{e}_1|\lambda_2\vect{e}_2|\lambda_3\vect{e}_3|\ldots|\lambda_n\vect{e}_n] &&\knowl{./knowls/definition.MM.knowl}{\text{Definition MM}}\\ &=I_nD &&\knowl{./knowls/definition.MI.knowl}{\text{Definition MI}}\\ &=D &&\knowl{./knowls/theorem.MMIM.knowl}{\text{Theorem MMIM}} \end{align*}

This says that $A$ is similar to the diagonal matrix $D$ via the nonsingular matrix $R$. Thus $A$ is diagonalizable (Definition DZM).

(⇒) Suppose that $A$ is diagonalizable, so there is a nonsingular matrix of size $n$ \begin{align*} T&=\matrixcolumns{y}{n} \end{align*} and a diagonal matrix (recall Definition SUV) \begin{align*} E&=\begin{bmatrix} d_1 & 0 & 0 &\cdots & 0\\ 0 &d_2 & 0 &\cdots & 0\\ 0 & 0 &d_3 &\cdots & 0\\ \vdots & \vdots & \vdots & & \vdots\\ 0 & 0 & 0 &\cdots & d_n \end{bmatrix} =[d_1\vect{e}_1|d_2\vect{e}_2|d_3\vect{e}_3|\ldots|d_n\vect{e}_n]&&\text{} \end{align*} such that $\inverse{T}AT=E$.

Then consider, \begin{align*} [A\vect{y}_1|A\vect{y}_2|A\vect{y}_3&|\ldots|A\vect{y}_n]\\ &=A\matrixcolumns{y}{n} &&\knowl{./knowls/definition.MM.knowl}{\text{Definition MM}}\\ &=AT\\ &=I_nAT &&\knowl{./knowls/theorem.MMIM.knowl}{\text{Theorem MMIM}}\\ &=T\inverse{T}AT &&\knowl{./knowls/definition.MI.knowl}{\text{Definition MI}}\\ &=TE\\ &=T[d_1\vect{e}_1|d_2\vect{e}_2|d_3\vect{e}_3|\ldots|d_n\vect{e}_n]\\ &=[T(d_1\vect{e}_1)|T(d_2\vect{e}_2)|T(d_3\vect{e}_3)|\ldots|T(d_n\vect{e}_n)] &&\knowl{./knowls/definition.MM.knowl}{\text{Definition MM}}\\ &=[d_1T\vect{e}_1|d_2T\vect{e}_2|d_3T\vect{e}_3|\ldots|d_nT\vect{e}_n] &&\knowl{./knowls/definition.MM.knowl}{\text{Definition MM}}\\ &=[d_1\vect{y}_1|d_2\vect{y}_2|d_3\vect{y}_3|\ldots|d_n\vect{y}_n] &&\knowl{./knowls/definition.MVP.knowl}{\text{Definition MVP}} \end{align*}

This equality of matrices (Definition ME) allows us to conclude that the individual columns are equal vectors (Definition CVE). That is, $A\vect{y}_i=d_i\vect{y}_i$ for $1\leq i\leq n$. In other words, $\vect{y}_i$ is an eigenvector of $A$ for the eigenvalue $d_i$, $1\leq i\leq n$. (Why does $\vect{y}_i\neq\zerovector$?). Because $T$ is nonsingular, the set containing $T$'s columns, $S=\set{\vectorlist{y}{n}}$, is a linearly independent set (Theorem NMLIC). So the set $S$ has all the required properties.